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CHAPTER 1: Introduction EXERCISES 1.1: Background, page 5 . . . . . . . . . . . . . . . . . . EXERCISES 1.2: Solutions and Initial Value Problems, page 14 . . . EXERCISES 1.3: Direction Fields, page 22 . . . . . . . . . . . . . . . EXERCISES 1.4: The Approximation Method of Euler, page 28 . . . CHAPTER 2: First Order Differential Equations EXERCISES 2.2: Separable Equations, page 46 . . . . . . EXERCISES 2.3: Linear Equations, page 54 . . . . . . . . EXERCISES 2.4: Exact Equations, page 65 . . . . . . . . EXERCISES 2.5: Special Integrating Factors, page 71 . . . EXERCISES 2.6: Substitutions and Transformations, page 78 REVIEW PROBLEMS: page 81 . . . . . . . . . . . . . . . 1 1 3 10 17 27 27 41 59 72 79 90 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . CHAPTER 3: Mathematical Models and Numerical Methods Involving First Order Equations 103 EXERCISES 3.2: Compartmental Analysis, page 98 . . . . . . . . . . . . . . . . . . 103 EXERCISES 3.3: Heating and Cooling of Buildings, page 107 . . . . . . . . . . . . 116 EXERCISES 3.4: Newtonian Mechanics, page 115 . . . . . . . . . . . . . . . . . . . 123 EXERCISES 3.5: Electrical Circuits, page 122 . . . . . . . . . . . . . . . . . . . . . 137 EXERCISES 3.6: Improved Euler's Method, page 132 . . . . . . . . . . . . . . . . . 139 EXERCISES 3.7: Higher Order Numerical Methods: Taylor and RungeKutta, page 142 153 CHAPTER 4: Linear Second Order Equations 167 EXERCISES 4.1: Introduction: The MassSpring Oscillator, page 159 . . . . . . . . 167 EXERCISES 4.2: Homogeneous Linear Equations; The General Solution, page 167 . 169 EXERCISES 4.3: Auxiliary Equations with Complex Roots, page 177 . . . . . . . . 177 iii EXERCISES 4.4: Nonhomogeneous Equations: The Method of Undetermined Coefficients, page 186 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . EXERCISES 4.5: The Superposition Principle and Undetermined Coefficients Revisited, page 192 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . EXERCISES 4.6: Variation of Parameters, page 197 . . . . . . . . . . . . . . . . . EXERCISES 4.7: Qualitative Considerations for VariableCoefficient and Nonlinear Equations, page 208 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . EXERCISES 4.8: A Closer Look at Free Mechanical Vibrations, page 219 . . . . . EXERCISES 4.9: A Closer Look at Forced Mechanical Vibrations, page 227 . . . REVIEW PROBLEMS: page 228 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 . 196 . 211 . . . . 226 232 241 246 259 259 282 293 307 317 325 331 CHAPTER 5: Introduction to Systems and Phase Plane Analysis EXERCISES 5.2: Elimination Method for Systems, page 250 . . . . . . . . . . . . . EXERCISES 5.3: Solving Systems and HigherOrder Equations Numerically, page 261 EXERCISES 5.4: Introduction to the Phase Plane, page 274 . . . . . . . . . . . . . EXERCISES 5.5: Coupled MassSpring Systems, page 284 . . . . . . . . . . . . . . EXERCISES 5.6: Electrical Circuits, page 291 . . . . . . . . . . . . . . . . . . . . . EXERCISES 5.7: Dynamical Systems, Poincar` Maps, and Chaos, page 301 . . . . . e REVIEW PROBLEMS: page 304 . . . . . . . . . . . . . . . . . . . . . . . . . . . . CHAPTER 6: Theory of Higher Order Linear Differential Equations 341 EXERCISES 6.1: Basic Theory of Linear Differential Equations, page 324 . . . . . . 341 EXERCISES 6.2: Homogeneous Linear Equations with Constant Coefficients, page 331 351 EXERCISES 6.3: Undetermined Coefficients and the Annihilator Method, page 337 361 EXERCISES 6.4: Method of Variation of Parameters, page 341 . . . . . . . . . . . . 375 REVIEW PROBLEMS: page 344 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384 CHAPTER 7: Laplace Transforms EXERCISES 7.2: Definition of the Laplace Transform, page 359 . . . . . . . . EXERCISES 7.3: Properties of the Laplace Transform, page 365 . . . . . . . . EXERCISES 7.4: Inverse Laplace Transform, page 374 . . . . . . . . . . . . . EXERCISES 7.5: Solving Initial Value Problems, page 383 . . . . . . . . . . . EXERCISES 7.6: Transforms of Discontinuous and Periodic Functions, page 395 EXERCISES 7.7: Convolution, page 405 . . . . . . . . . . . . . . . . . . . . . EXERCISES 7.8: Impulses and the Dirac Delta Function, page 412 . . . . . . EXERCISES 7.9: Solving Linear Systems with Laplace Transforms, page 416 . REVIEW PROBLEMS: page 418 . . . . . . . . . . . . . . . . . . . . . . . . . iv 389 389 396 402 413 428 450 459 466 481 . . . . . . . . . . . . . . . . . . . . . . . . . . CHAPTER 8: Series Solutions of Differential Equations EXERCISES 8.1: Introduction: The Taylor Polynomial Approximation, page 430 EXERCISES 8.2: Power Series and Analytic Functions, page 438 . . . . . . . . . EXERCISES 8.3: Power Series Solutions to Linear Differential Equations, page 449 EXERCISES 8.4: Equations with Analytic Coefficients, page 456 . . . . . . . . . EXERCISES 8.5: CauchyEuler (Equidimensional) Equations Revisited, page 460 EXERCISES 8.6: Method of Frobenius, page 472 . . . . . . . . . . . . . . . . . EXERCISES 8.7: Finding a Second Linearly Independent Solution, page 482 . . EXERCISES 8.8: Special Functions, page 493 . . . . . . . . . . . . . . . . . . . REVIEW PROBLEMS: page 497 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 491 491 496 505 520 529 534 547 559 563 569 569 570 573 577 584 596 602 617 629 629 635 639 644 653 660 675 675 683 687 690 694 701 v CHAPTER 9: Matrix Methods for Linear Systems EXERCISES 9.1: Introduction, page 507 . . . . . . . . . . . . . . . . . . . . . . . EXERCISES 9.2: Review 1: Linear Algebraic Equations, page 512 . . . . . . . . . EXERCISES 9.3: Review 2: Matrices and Vectors, page 521 . . . . . . . . . . . . EXERCISES 9.4: Linear Systems in Normal Form, page 530 . . . . . . . . . . . . EXERCISES 9.5: Homogeneous Linear Systems with Constant Coefficients, page 541 EXERCISES 9.6: Complex Eigenvalues, page 549 . . . . . . . . . . . . . . . . . . EXERCISES 9.7: Nonhomogeneous Linear Systems, page 555 . . . . . . . . . . . EXERCISES 9.8: The Matrix Exponential Function, page 566 . . . . . . . . . . . CHAPTER 10: Partial Differential Equations EXERCISES 10.2: Method of Separation of Variables, page 587 EXERCISES 10.3: Fourier Series, page 603 . . . . . . . . . . . EXERCISES 10.4: Fourier Cosine and Sine Series, page 611 . EXERCISES 10.5: The Heat Equation, page 624 . . . . . . . . EXERCISES 10.6: The Wave Equation, page 636 . . . . . . . EXERCISES 10.7: Laplace's Equation, page 649 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . CHAPTER 11: Eigenvalue Problems and SturmLiouville Equations EXERCISES 11.2: Eigenvalues and Eigenfunctions, page 671 . . . . . . . . . . . . . EXERCISES 11.3: Regular SturmLiouville Boundary Value Problems, page 682 . . EXERCISES 11.4: Nonhomogeneous Boundary Value Problems and the Fredholm Alternative, page 692 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . EXERCISES 11.5: Solution by Eigenfunction Expansion, page 698 . . . . . . . . . . EXERCISES 11.6: Green's Functions, page 706 . . . . . . . . . . . . . . . . . . . . EXERCISES 11.7: Singular SturmLiouville Boundary Value Problems, page 715 . . EXERCISES 11.8: Oscillation and Comparison Theory, page 725 . . . . . . . . . . . 705 707 707 709 716 718 719 722 725 725 733 741 743 CHAPTER 12: Stability of Autonomous Systems EXERCISES 12.2: Linear Systems in the Plane, page 753 . . . . . . EXERCISES 12.3: Almost Linear Systems, page 764 . . . . . . . . EXERCISES 12.4: Energy Methods, page 774 . . . . . . . . . . . . EXERCISES 12.5: Lyapunov's Direct Method, page 782 . . . . . . EXERCISES 12.6: Limit Cycles and Periodic Solutions, page 791 . EXERCISES 12.7: Stability of HigherDimensional Systems, page 798 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . CHAPTER 13: Existence and Uniqueness Theory EXERCISES 13.1: Introduction: Successive Approximations, page 812 . . . . . EXERCISES 13.2: Picard's Existence and Uniqueness Theorem, page 820 . . . EXERCISES 13.3: Existence of Solutions of Linear Equations, page 826 . . . EXERCISES 13.4: Continuous Dependence of Solutions, page 832 . . . . . . . . . . . . . . . . . . . vi CHAPTER 1: Introduction
EXERCISES 1.1: Background, page 5 1. This equation involves only ordinary derivatives of x with respect to t, and the highest derivative has the second order. Thus it is an ordinary differential equation of the second order with independent variable t and dependent variable x. It is linear because x, dx/dt, and d2 x/dt2 appear in additive combination (even with constant coefficients) of their first powers. 3. This equation is an ODE because it contains no partial derivatives. Since the highest order derivative is dy/dx, the equation is a first order equation. This same term also shows us that the independent variable is x and the dependent variable is y. This equation is nonlinear because of the y in the denominator of the term [y(2  3x)]/[x(1  3y)] . 5. This equation is an ODE because it contains only ordinary derivatives. The term dp/dt is the highest order derivative and thus shows us that this is a first order equation. This term also shows us that the independent variable is t and the dependent variable is p. This equation is nonlinear since in the term kp(P  p) = kP p  kp2 the dependent variable p is squared (compare with equation (7) on page 5 of the text). 7. This equation is an ordinary first order differential equation with independent variable x and dependent variable y. It is nonlinear because it contains the square of dy/dx. 9. This equation contains only ordinary derivative of y with respect to x. Hence, it is an ordinary differential equation of the second order (the highest order derivative is d2 y/dx2) with independent variable x and dependent variable y. This equation is of the form (7) on page 5 of the text and, therefore, is linear. 1 Chapter 1
11. This equation contains partial derivatives, thus it is a PDE. Because the highest order derivative is a second order partial derivative, the equation is a second order equation. The terms N/t and N/r show that the independent variables are t and r and the dependent variable is N. 13. Since the rate of change of a quantity means its derivative, denoting the coefficient proportionality between dp/dt and p(t) by k (k > 0), we get dp = kp. dt 15. In this problem, T M (coffee is hotter than the air), and T is a decreasing function of t, dT = k(M  T ), dt where k > 0 is the proportionality constant. 17. In classical physics, the instantaneous acceleration, a, of an object moving in a straight line is given by the second derivative of distance, x, with respect to time, t; that is d2 x = a. dt2 Integrating both sides with respect to t and using the given fact that a is constant we obtain dx = at + C. dt (1.1) that is dT /dt 0. Thus The instantaneous velocity, v, of an object is given by the first derivative of distance, x, with respect to time, t. At the beginning of the race, t = 0, both racers have zero velocity. Therefore we have C = 0. Integrating equation (1.1) with respect to t we obtain x= 1 2 at + C1 . 2 For this problem we will use the starting position for both competitors to be x = 0 at t = 0. Therefore, we have C1 = 0. This gives us a general equation used for both racers as x= 2 1 2 at 2 or t= 2x , a Exercises 1.2
where the acceleration constant a has different values for Kevin and for Alison. Kevin covers the last
1 4 of the full distance, L, in 3 seconds. This means Kevin's acceleration, aK , is 2L  aK 2(3L/4) , aK determined by: tK  t3/4 = 3 = where tK is the time it takes for Kevin to finish the race. Solving this equation for aK gives, 2 2  3/2 L. aK = 9 Therefore the time required for Kevin to finish the race is given by: tK = 2L 2
2 = L/9 3 2 3/2 2 = 12 + 6 3 22.39 sec. 3/2 Alison covers the last 1/3 of the distance, L, in 4 seconds. This means Alison's acceleration, aA , is found by: tA  t2/3 = 4 = 2L  aA 2(2L/3) , aA where tA is the time required for Alison to finish the race. Solving this equation for aA gives 2 2  4/3 L. aA = 16 Therefore the time required for Alison to finish the race is given by: tA = 2L 2
2 = (L/16) 4 2 4/3 2 = 12 + 4 6 21.80 sec. 4/3 The time required for Alison to finish the race is less than Kevin; therefore Alison wins the race by 6 3  4 6 0.594 seconds. EXERCISES 1.2: Solutions and Initial Value Problems, page 14 1. (a) Differentiating (x) yields (x) = 6x2 . Substitution and for y and y into the given equation, xy = 3y, gives x 6x2 = 3 2x3 , 3 Chapter 1
which is an identity on (, ). Thus (x) is an explicit solution on (, ). (b) We compute d d = (ex  x) = ex  1. dx dx Functions (x) and (x) are defined for all real numbers and d +(x)2 = (ex  1)+(ex  x)2 = (ex  1)+ e2x  2xex + x2 = e2x +(12x)ex +x2 1, dx which is identically equal to the righthand side of the given equation. Thus (x) is an explicit solution on (, ). (c) Note that the function (x) = x2  x1 is not defined at x = 0. Differentiating (x) twice yields d d = x2  x1 = 2x  (1)x2 = 2x + x2 ; dx dx d d d d2 = 2x + x2 = 2 + (2)x3 = 2 1  x3 . = 2 dx dx dx dx Therefore x2 d2 = x2 2 1  x3 = 2 x2  x1 = 2(x), 2 dx and (x) is an explicit solution to the differential equation x2 y = 2y on any interval not containing the point x = 0, in particular, on (0, ). 3. Since y = sin x + x2 , we have y = cos x + 2x and y =  sin x + 2. These functions are defined on (, ). Substituting these expressions into the differential equation y + y = x2 + 2 gives y + y =  sin x + 2 + sin x + x2 = 2 + x2 = x2 + 2 for all x in (, ). Therefore, y = sin x + x2 is a solution to the differential equation on the interval (, ). 5. Differentiating x(t) = cos 2t, we get d dx = (cos 2t) = ( sin 2t)(2) = 2 sin 2t. dt dt 4 Exercises 1.2
So, dx + tx = 2 sin 2t + t cos 2t sin 2t dt on any interval. Therefore, x(t) is not a solution to the given differential equation. 7. We differentiate y = e2x  3ex twice: dy d 2x = e  3ex = e2x (2)  3ex (1) = 2e2x + 3ex ; dx dx d d dy d2 y = 2e2x + 3ex = 2e2x (2) + 3ex (1) = 4e2x  3ex . = 2 dx dx dx dx Substituting y, y , and y into the differential equation and collecting similar terms, we get d2 y dy  2y = 4e2x  3ex  2e2x + 3ex  2 e2x  3ex  2 dx dx = (4  2  2)e2x + (3  3  2(3))ex = 0. Hence y = e2x  3ex is an explicit solution to the given differential equation. 9. Differentiating the equation x2 + y 2 = 6 implicitly, we obtain 2x + 2yy = 0 x y = . y Since there can be no function y = f (x) that satisfies the differential equation y = x/y and the differential equation y = x/y on the same interval, we see that x2 + y 2 = 6 does not define an implicit solution to the differential equation. 11. Differentiating the equation exy + y = x  1 implicitly with respect to x yields d xy d (e + y) = (x  1) dx dx dy d =1 exy (xy) + dx dx dy dy exy y + x =1 + dx dx dy (xexy + 1) = 1 yexy + dx 1  yexy exy  y exy (exy  y) dy = = xy . = xy xy dx 1 + xexy e (e + x) e +x 5 Chapter 1
Therefore, the function y(x) defined by exy + y = x  1 is an implicit solution to the given differential equation. 13. Differentiating the equation sin y + xy  x3 = 2 implicitly with respect to x, we obtain y cos y + xy + y  3x2 = 0 (cos y + x)y = 3x2  y y = 3x2  y . cos y + x Differentiating the second equation above again, we obtain (y sin y + 1)y + (cos y + x)y = 6x  y (cos y + x)y = 6x  y + (y )2 sin y  y = 6x  2y + (y )2 sin y 6x  2y + (y )2 sin y . y = cos y + x 3x2  y , cos y + x Multiplying the righthand side of this last equation by y /y = 1 and using the fact that y = we get y 6x  2y + (y )2 sin y y 2  y)/(cos y + x) cos y + x (3x 2 3 6xy  2(y ) + (y ) sin y . = 3x2  y = Thus y is an implicit solution to the differential equation. 15. We differentiate (x) and substitute and into the differential equation for y and y . This yields (x) = Ce3x + 1 d(x) = Ce3x + 1 = 3Ce3x ; dx d  3 = 3Ce3x  3 Ce3x + 1 = (3C  3C)e3x  3 = 3, dx which holds for any constant C and any x on (, ). Therefore, (x) = Ce3x + 1 is a oneparameter family of solutions to y  3y = 3 on (, ). Graphs of these functions for C = 0, 0.5, 1, and 2 are sketched in Figure 1A. 6 Exercises 1.2
C=2 10 C=1 C=0.5 C=0 0.5 0.5 C=0.5 C=1
10 C=2 Figure 1A: Graphs of the functions y = Ce3x + 1 for C = 0, 0.5, 1, and 2. 17. Differentiating (x), we find that (x) = 2 1  cex = 2 (1  cex )1 (1.2) = 2(1) (1  cex )2 (1  cex ) = 2cex (1  cex )2 . On the other hand, substitution of (x) for y into the righthand side of the given equation yields 2 (x)((x)  2) 1 2 2 = x 2 2 1  ce 1  cex 1 1  (1  cex ) 2 2cex 2 1 = = , = 1  cex 1  cex 1  cex 1  cex (1  cex )2 which is identical to (x) found in (1.2). 19. Squaring and adding the terms dy/dx and y in the equation (dy/dx)2 + y 2 + 3 = 0 gives a nonnegative number. Therefore when these two terms are added to 3, the lefthand side will always be greater than or equal to three and hence can never equal the righthand side which is zero. 7 Chapter 1
21. For (x) = xm , we have (x) = mxm1 and (x) = m(m  1)xm2 . (a) Substituting these expressions into the differential equation, 3x2 y + 11xy  3y = 0, gives 3x2 m(m  1)xm2 + 11x mxm1  3xm = 0 3m(m  1)xm + 11mxm  3xm = 0 [3m(m  1) + 11m  3] xm = 0 3m2 + 8m  3 xm = 0. For the last equation to hold on an interval for x, we must have 3m2 + 8m  3 = (3m  1)(m + 3) = 0. 1 , 3. 3 (b) Substituting the above expressions for (x), (x), and (x) into the differential equaThus either (3m  1) = 0 or (m + 3) = 0, which gives m = tion, x2 y  xy  5y = 0, gives x2 m(m  1)xm2  x mxm1  5xm = 0 m2  2m  5 xm = 0. For the last equation to hold on an interval for x, we must have m2  2m  5 = 0. To solve for m we use the quadratic formula: 2 4 + 20 = 1 6. m= 2 23. In this problem, f (x, y) = x3  y 3 and so (x3  y 3) f = = 3y 2 . y y Clearly, f and f /y (being polynomials) are continuous on the whole xyplane. Thus the hypotheses of Theorem 1 are satisfied, and the initial value problem has a unique solution for any initial data, in particular, for y(0) = 6. 8 Exercises 1.2
25. Writing 4t dx =  = 4tx1 , dt x we see that f (t, x) = 4tx1 and f (t, x)/x = (4tx1 )/x = 4tx2 . The functions f (t, x) and f (t, x)/x are not continuous only when x = 0. Therefore, they are continuous in any rectangle R that contains the point (2, ), but does not intersect the taxis; for instance, R = {(t, x) : 1 < t < 3, 2 < x < 0}. Thus, Theorem 1 applies, and the given initial problem has a unique solution. 26. Here f (x, y) = 3x  3 y  1 and f (x, y)/y =  1 (y  1)2/3 . Unfortunately, f /y is not 3 continuous or defined when y = 1. So there is no rectangle containing (2, 1) in which both f and f /y are continuous. Therefore, we are not guaranteed a unique solution to this initial value problem. 27. Rewriting the differential equation in the form dy/dx = x/y, we conclude that f (x, y) = x/y. Since f is not continuous when y = 0, there is no rectangle containing the point (1, 0) in which f is continuous. Therefore, Theorem 1 cannot be applied. 29. (a) Clearly, both functions 1 (x) 0 and 2 (x) = (x  2)3 satisfy the initial condition, y(2) = 0. Next, we check that they also satisfy the differential equation dy/dx = 3y 2/3 . d d1 = (0) = 0 = 31 (x)2/3 ; dx dx d d2 = (x  2)3 = 3(x  2)2 = 3 (x  2)3 dx dx 2/3 = 32 (x)2/3 . Hence both functions, 1 (x) and 2 (x), are solutions to the initial value problem of Exapmle 9. (b) In this initial value problem, f (x, y) = 3y 2/3 2 f (x, y) 2 = 3 y 2/31 = 1/3 , y 3 y x0 = 0 and y0 = 107 . The function f (x, y) is continuous everywhere; f (x, y)/y is continuous in any region which does not intersect the xaxis (where y = 0). In particular, 9 Chapter 1
both functions, f (x, y) and f (x, y)/y, are continuous in the rectangle R = (x, y) : 1 < x < 1, (1/2)107 < y < (2)107 containing the initial point (0, 107 ). Thus, it follows from Theorem 1 that the given initial value problem has a unique solution in an interval about x0 . 31. (a) To try to apply Theorem 1 we must first write the equation in the form y = f (x, y). Here f (x, y) = 4xy 1 and f (x, y)/y = 4xy 2 . Neither f nor f /y are continuous or defined when y = 0. Therefore there is no rectangle containing (x0 , 0) in which both f and f /y are continuous, so Theorem 1 cannot be applied. (b) Suppose for the moment that there is such a solution y(x) with y(x0 ) = 0 and x0 = 0. Substituting into the differential equation we get y(x0 )y (x0 )  4x0 = 0 or 0 y (x0 )  4x0 = 0 Thus x0 = 0, which is a contradiction. (c) Taking C = 0 in the implicit solution 4x2  y 2 = C given in Example 5 on page 9 gives 4x2  y 2 = 0 or y = 2x. Both solutions y = 2x and y = 2x satisfy y(0) = 0. EXERCISES 1.3: Direction Fields, page 22 4x0 = 0. (1.3) 1. (a) For y = 2x, d dy = (2x) = 2 dx dx and 4x 4x = = 2, y 2x x = 0. Thus y = 2x and y = 2x are solutions to the differential equation dy/dx = 4x/y on any interval not containing the point x = 0. (b) , (c) See Figures B.1 and B.2 in the answers of the text. 10 Exercises 1.3
(d) As x or x , the solution in part (b) increases unboundedly and has the lines y = 2x and y = 2x, respectively, as slant asymptotes. The solution in part (c) also increases without bound as x and approaches the line y = 2x, while it is not even defined for x < 0. 3. From Figure B.3 in the answers section of the text, we conclude that, regardless of the initial velocity, v(0), the corresponding solution curve v = v(t) has the line v = 8 as a horizontal asymptote, that is, limt v(t) = 8. This explains the name "terminal velocity" for the value v = 8. 5. (a) The graph of the directional field is shown in Figure B.4 in the answers section of the text. (b), (c) The direction field indicates that all solution curves (other than p(t) 0) will approach the horizontal line (asymptote) p = 1.5 as t +. Thus limt+ p(t) = 1.5 . (d) No. The direction field shows that populations greater than 1500 will steadily decrease, but can never reach 1500 or any smaller value, i.e., the solution curves cannot cross the line p = 1.5 . Indeed, the constant function p(t) 1.5 is a solution to the given logistic equation, and the uniqueness part of Theorem 1, page 12, prevents intersections of solution curves. 6. (a) The slope of a solution to the differential equation dy/dx = x + sin y is given by dy/dx . Therefore the slope at (1, /2) is equal to dy = 1 + sin = 2. dx 2 (b) The solution curve is increasing if the slope of the curve is greater than zero. From part (a) we know the slope to be x + sin y. The function sin y has values ranging from 1 to 1; therefore if x is greater than 1 then the slope will always have a value greater than zero. This tells us that the solution curve is increasing. (c) The second derivative of every solution can be determined by finding the derivative of 11 Chapter 1
the differential equation dy/dx = x + sin y. Thus d dx dy dx = d (x + sin y); dx dy (chain rule) = 1 + (cos y) dx = 1 + (cos y)(x + sin y) = 1 + x cos y + sin y cos y; 1 = 1 + x cos y + sin 2y. 2 d2 y dx2 d2 y dx2 (d) Relative minima occur when the first derivative, dy/dx, is equal to zero and the second derivative, d2 y/dx2 , is greater than zero. The value of the first derivative at the point (0, 0) is given by dy = 0 + sin 0 = 0. dx This tells us that the solution has a critical point at the point (0, 0). Using the second derivative found in part (c) we have d2 y 1 = 1 + 0 cos 0 + sin 0 = 1. 2 dx 2 This tells us the point (0, 0) is a relative minimum. 7. (a) The graph of the directional field is shown in Figure B.5 in the answers section of the text. (b) The direction field indicates that all solution curves with p(0) > 1 will approach the horizontal line (asymptote) p = 2 as t +. Thus limt+ p(t) = 2 when p(0) = 3. (c) The direction field shows that a population between 1000 and 2000 (that is 1 < p(0) < 2) will approach the horizontal line p = 2 as t +. (d) The direction field shows that an initial population less than 1000 (that is 0 p(0) < 1) will approach zero as t +. (e) As noted in part (d), the line p = 1 is an asymptote. The direction field indicates that a population of 900 (p(0) = 0.9) steadily decreases with time and therefore cannot increase to 1100. 12 Exercises 1.3
9. (a) The function (x), being a solution to the given initial value problem, satisfies d = x  (x), dx Thus d2 d = 2 dx dx d dx (0) = 1. (1.4) = d d (x  (x)) = 1  = 1  x + (x), dx dx where we have used (1.4) substituting (twice) x  (x) for d/dx. (b) First we note that any solution to the given differential equation on an interval I is continuously diferentiable on I. Indeed, if y(x) is a solution on I, then y (x) does exist on I, and so y(x) is continuous on I because it is differentiable. This immediately implies that y (x) is continuous as the difference of two continuous functions, x and y(x). From (1.4) we conclude that d dx = [x  (x)]
x=0 x=0 = 0  (0) = 1 < 0 and so the continuity of (x) implies that, for x small enough, (x) < 0. By the Monotonicity Test, negative derivative of a function results that the function itself is decreasing. When x increases from zero, as far as (x) > x, one has (x) < 0 and so (x) decreases. On the other hand, the function y = x increases unboundedly, as x . Thus, by intermediate value theorem, there is a point, say, x > 0, where the curve y = (x) crosses the line y = x. At this point, (x ) = x and hence (x ) = x  (x ) = 0. (c) From (b) we conclude that x is a critical point for (x) (its derivative vanishes at this point). Also, from part (a), we see that (x ) = 1  (x ) = 1 > 0. Hence, by Second Derivative Test, (x) has a relative minimum at x . (d) Remark that the arguments, used in part (c), can be applied to any point x, where (x) = 0, to conclude that (x) has a relative minimum at x. Since a continuously 13 Chapter 1
differentiable function on an interval cannot have two relative minima on an interval without having a point of relative maximum, we conclude that x is the only point where (x) = 0. Continuity of (x) implies that it has the same sign for all x > x and, therefore, it is positive there since it is positive for x > x and close to x ( (x ) = 0 and (x ) > 0). By Monotonicity Test, (x) increases for x > x . (e) For y = x  1, dy/dx = 1 and x  y = x  (x  1) = 1. Thus the given differential equation is satisfied, and y = x  1 is indeed a solution. To show that the curve y = (x) always stays above the line y = x  1, we note that the initial value problem dy = x  y, dx y(x0 ) = y0 (1.5) has a unique solution for any x0 and y0 . Indeed, functions f (x, y) = xy and f /y 1 are continuous on the whole xyplane, and Theorem 1, Section 1.2, applies. This implies that the curve y = (x) always stays above the line y = x  1: (0) = 1 > 1 = (x  1)
x=0 , and the existence of a point x with (x) (x  1) would imply, by intermediate value theorem, the existence of a point x0 , 0 < x0 x, satisfying y0 := (x0 ) = x0  1 and, therefore, there would be two solutions to the initial value problem (1.5). Since, from part (a), (x) = 1 (x) = 1x+(x) = (x)(x1) > 0, we also conclude that (x) is an increasing function and (x) < 1. Thus there exists limx (x) 1. The strict inequality would imply that the values of the function y = (x), for x large enough, become smaller than those of y = x  1. Therefore, lim (x) = 1 lim [x  (x)] = 1, x x and so the line y = x  1 is a slant asymptote for (x). (f), (g) The direction field for given differential equation and the curve y = (x) are shown in Figure B.6 in the answers of the text. 14 Exercises 1.3
11. For this equation, the isoclines are given by 2x = c. These are vertical lines x = c/2. Each element of the direction field associated with a point on x = c/2 has slope c. (See Figure B.7 in the answers of the text.) 13. For the equation y/x = x/y, the isoclines are the curves x/y = c. These are lines that pass through the origin and have equations of the form y = mx, where m = 1/c , c = 0. If we let c = 0 in x/y = c, we see that the yaxis (x = 0) is also an isocline. Each element of the direction field associated with a point on an isocline has slope c and is, therefore, perpendicular to that isocline. Since circles have the property that at any point on the circle the tangent at that point is perpendicular to a line from that point to the center of the circle, we see that the solution curves will be circles with their centers at the origin. But since we cannot have y = 0 (since x/y would then have a zero in the denominator) the solutions will not be defined on the xaxis. (Note however that a related form of this differential equation is yy + x = 0. This equation has implicit solutions given by the equations y 2 + x2 = C. These solutions will be circles.) The graph of (x), the solution to the equation satisfying the initial condition y(0) = 4, is the upper semicircle with center at the origin and passing through the point (0, 4) (see Figure B.8 in the answers of the text). 15. For the equation dy/dx = 2x2  y, the isoclines are the curves 2x2  y = c, or y = 2x2  c. The curve y = 2x2  c is a parabola which is open upward and has the vertex at (0, c). Three of them, for c = 1, 0, and 2 (dotted curves), as well as the solution curve satisfying the initial condition y(0) = 0, are depicted in Figure B.9. 17. The isoclines for the equation 1 dy =3y+ dx x are given by 3y+ 1 =c x y= 1 + 3  c, x which are hyperbolas having x = 0 as a vertical asymptote and y = 3  c as a horizontal asymptote. Each element of the direction field associated with a point on such a hyperbola has slope c. For x > 0 large enough: if an isocline is located above the line y = 3, then c 0, 15 Chapter 1 c=5 c=4 c=3
5 c=2 c=1 3 c=1 c=2 c=3
0 5 10 c=4 Figure 1B: Isoclines and the direction field for Problem 17. and so the elements of the direction field have negative or zero slope; if an isocline is located below the line y = 3, then c > 0, and so the elements of the direction field have positive slope. In other words, for x > 0 large enough, at any point above the line y = 3 a solution curve decreases passing through this point, and any solution curve increases passing through a point below y = 3. The direction field for this differential equation is depicted in Figure 1B. From this picture we conclude that any solution to the differential equation dy/dx = 3  y + 1/x has the line y = 3 as a horizontal asymptote. 19. Integrating both sides of the equation dy/y = dx/x yields 1 dy =  y y = 1 dx x eC1 x ln y =  ln x + C1 y = C2 , x ln y = ln eC1 x where C1 is an arbitrary constant and so C2 := eC1 is an arbitrary positive constant. The last equality can be written as y= 16 C2 C = , x x Exercises 1.4
where C = C2 is any nonzero constant. The value C = 0 gives y 0 (for x = 0), which is, clearly, also a solution to the given equation. EXERCISES 1.4: The Approximation Method of Euler, page 28 1. In this initial value problem, f (x, y) = x/y, x0 = 0, and y0 = 1. Thus, with h = 0.1, the recursive formulas (2) and (3) on page 25 of the text become xn+1 = xn + h = xn + 0.1 , yn+1 = yn + hf (xn , yn ) = yn + 0.1 We set n = 0 in these formulas and obtain x1 = x0 + 0.1 = 0 + 0.1 = 0.1 , x0 y1 = y0 + 0.1 = 1 + 0.1 y0 Putting n = 1 in the recursive formulas yields x2 = x1 + 0.1 = 0.1 + 0.1 = 0.2 , x1 = 1 + 0.1 y2 = y1 + 0.1 y1 xn yn , n = 0, 1, . . . . 0 1 = 1. 0.1 1 = 1.01 . Continuing in the same manner, we find for n = 2, 3, and 4: x3 = 0.2 + 0.1 = 0.3 , x4 = 0.3 + 0.1 = 0.4 , x5 = 0.4 + 0.1 = 0.5 , y3 = 1.01 + 0.1 0.2 1.01 = 1.02980 ; = 1.05893 ; = 1.09671 , y4 = 1.02980 + 0.1 y5 = 1.05893 + 0.1 0.3 1.02980 0.4 1.05893 where we have rounded off all answers to five decimal places. 2. In this problem, x0 = 0, y0 = 4, h = 0.1, and f (x, y) = x/y. Thus, the recursive formulas given in equations (2) and (3) on page 25 of the text become xn+1 = xn + h = xn + 0.1 , 17 Chapter 1
yn+1 = yn + hf (xn , yn ) = yn + 0.1  xn yn , n = 0, 1, 2, . . . . To find an approximation for the solution at the point x1 = x0 + 0.1 = 0.1, we let n = 0 in the last recursive formula to find y1 = y0 + 0.1  x0 y0 = 4 + 0.1 (0) = 4. To approximate the value of the solution at the point x2 = x1 + 0.1 = 0.2, we let n = 1 in the last recursive formula to obtain y2 = y1 + 0.1  x1 y1 = 4 + 0.1  0.1 4 x2 y2 =4 1 = 3.9975 3.998 . 400 0.2 3.9975 Continuing in this way we find x3 = x2 + 0.1 = 0.3 , x4 = 0.4 , x5 = 0.5 , y3 = y2 + 0.1  y4 3.985 , y5 3.975 , = 3.9975 + 0.1  3.992 , where all of the answers have been rounded off to three decimal places. 3. Here f (x, y) = y(2  y), x0 = 0, and y0 = 3. We again use recursive formulas from Euler's method with h = 0.1. Setting n = 0, 1, 2, 3, and 4 and rounding off results to three decimal places, we get x1 = x0 + 0.1 = 0.1 , x2 = 0.1 + 0.1 = 0.2 , x3 = 0.2 + 0.1 = 0.3 , x4 = 0.3 + 0.1 = 0.4 , x5 = 0.4 + 0.1 = 0.5 , y1 = y0 + 0.1 [y0 (2  y0 )] = 3 + 0.1 [3(2  3)] = 2.700; y2 = 2.700 + 0.1 [2.700(2  2.700)] = 2.511; y3 = 2.511 + 0.1 [2.511(2  2.511)] 2.383; y4 = 2.383 + 0.1 [2.383(2  2.383)] 2.292; y5 = 2.292 + 0.1 [2.292(2  2.292)] 2.225 . 5. In this problem, f (x, y) = (y 2 + y)/x, x0 = y0 = 1, and h = 0.2. The recursive formulas (2) and (3) on page 25 of the text, applied succesively with n = 1, 2, 3, and 4, yield x1 = x0 + 0.2 = 1.2 , 18 y1 = y0 + 0.2
2 y0 + y0 x0 = 1 + 0.2 12 + 1 1 = 1.400; Exercises 1.4
x2 = 1.2 + 0.2 = 1.4 , x3 = 1.4 + 0.2 = 1.6 , x4 = 1.6 + 0.2 = 1.8 , 1.4002 + 1.400 1.2 2 1.960 + 1.960 y3 = 1.960 + 0.2 1.4 2 2.789 + 2.789 y4 = 2.789 + 0.2 1.6 y2 = 1.400 + 0.2 1.960; 2.789; 4.110 . 7. For this problem notice that the independent variable is t and the dependent variable is x. Hence, the recursive formulas given in equations (2) and (3) on page 25 of the text become tn+1 = tn + h and (tn+1 ) xn+1 = xn + hf (tn , xn ), n = 0, 1, 2, . . . . For this problem, f (t, x) = 1+t sin(tx), t0 = 0, and x0 = 0. Thus the second recursive formula above becomes xn+1 = xn + h [1 + tn sin(tn xn )] , n = 0, 1, 2, . . . . For the case N = 1, we have h = (1  0)/1 = 1 which gives us t1 = 0 + 1 = 1 and (1) x1 = 0 + 1 (1 + 0 sin 0) = 1. For the case N = 2, we have h = 1/2 = 0.5 . Thus we have t1 = 0 + 0.5 = 0.5 , and t2 = 0.5 + 0.5 = 1, (1) x2 = 0.5 + 0.5 [1 + 0.5 sin(0.25)] 1.06185 . x1 = 0 + 0.5 (1 + 0 sin 0) = 0.5 , For the case N = 4, we have h = 1/4 = 0.25 , and so the recursive formulas become tn+1 = tn + 0.25 Therefore, we have t1 = 0 + 0.25 = 0.25 , x1 = 0 + 0.25 [1 + 0 sin(0)] = 0.25 . 19 and xn+1 = xn + 0.25 [1 + tn sin(tn xn )] . Chapter 1
Plugging these values into the recursive equations above yields t2 = 0.25 + 0.25 = 0.5 Continuing in this way gives t3 = 0.75 t4 = 1.00 and and x3 = 0.503904 + 0.25 [1 + 0.5 sin(0.251952)] = 0.785066 , (1) x4 = 1.13920 . and x2 = 0.25 + 0.25 [1 + 0.25 sin(0.0625)] = 0.503904 . For N = 8, we have h = 1/8 = 0.125 . Thus, the recursive formulas become tn+1 = tn + 0.125 and xn+1 = xn + 0.125 [1 + tn sin(tn xn )] . Using these formulas and starting with t0 = 0 and x0 = 0, we can fill in Table 1A. From this we see that (1) x8 = 1.19157, which is rounded to five decimal places. Table 1A: Euler's method approximations for the solution of x = 1 + t sin(tx), x(0) = 0, at t = 1 with 8 steps (h = 1/8).
n 1 2 3 4 5 6 7 8 tn 0.125 0.250 0.375 0.500 0.625 0.750 0.875 1.000 xn 0.125 0.250244 0.377198 0.508806 0.649535 0.805387 0.983634 1.191572 9. To approximate the solution on the whole interval [1, 2] by Euler's method with the step h = 0.1, we first approximate the solution at the points xn = 1 + 0.1n, n = 1, . . . , 10. Then, on each subinterval [xn , xn+1 ], we approximate the solution by the linear interval, connecting 20 Exercises 1.4
(xn , yn ) with (xn+1 , yn+1), n = 0, 1, . . . , 9. Since f (x, y) = x2  yx1  y 2 , the recursive formulas have the form xn+1 = xn + 0.1 , yn+1 = yn + 0.1 x0 = 1, y0 = 1. Therefore, x1 = 1 + 0.1 = 1.1 , x2 = 1.1 + 0.1 = 1.2 , x3 = 1.2 + 0.1 = 1.3 , y1 = 1 + 0.1 1 1  (1)2 = 0.9 ;  2 1 1 1 0.9  (0.9)2 0.81653719 ; y2 = 0.9 + 0.1  1.12 1.1 1 0.81653719 y3 = 0.81653719 + 0.1   (0.81653719)2 2 1.2 1.2 0.74572128 ; 1 0.74572128  (0.74572128)2 y4 = 0.74572128 + 0.1  2 1.3 1.3 0.68479653 ; yn 1 2   yn , 2 xn xn n = 0, 1, . . . , 9 , x4 = 1.3 + 0.1 = 1.4 , etc. The results of these computations (rounded to five decimal places) are shown in Table 1B. Table 1B: Euler's method approximations for the solutions of y = x2  yx1  y 2 , y(1) = 1, on [1, 2] with h = 0.1.
n 0 1 2 3 4 5 xn 1.0 1.1 1.2 1.3 1.4 1.5 yn 1.00000 0.90000 0.81654 0.74572 0.68480 0.63176 n 6 7 8 9 10 xn 1.6 1.7 1.8 1.9 2.0 yn 0.58511 0.54371 0.50669 0.47335 0.44314 The function y(x) = 1/x = x1 , obviously, satisfies the initial condition, y(1) = 1. Further 21 Chapter 1 1.2 0 1.4 1.6 1.8 2 Polygonal approximation y=1/x 1 Figure 1C: Polygonal line approximation and the actual solution for Problem 9. we compute both sides of the given differential equation: y (x) = x1 = x2 ,
2 f (x, y(x)) = x2  x1 x1  x1 = x2 + x2  x2 = x2 . Thus, the function y(x) = 1/x is, indeed, the solution to the given initial value problem. The graphs of the obtained polygonal line approximation and the actual solution are sketched in Figure 1C. 11. In this problem, the independent variable is t and the dependent variable is x; f (t, x) = 1+x2 , t0 = 0, and x0 = 0. The function (t) = tan t satisfies the initial condition: (0) = tan 0 = 0. The differential equation is also satisfied: d = sec2 t = 1 + tan2 t = 1 + (t)2 . dt Therefore, (t) is the solution to the given initial value problem. 22 Exercises 1.4
For approximation of (t) at the point t = 1 with N = 20 steps, we take the step size h = (1  t0 )/20 = 0.05. Thus, the recursive formulas for Euler's method are tn+1 = tn + 0.05 , xn+1 = xn + 0.05 1 + x2 . n Applying these formulas with n = 0, 1, . . . , 19, we obtain x1 = x0 + 0.05 1 + x2 = 0.05 , 0 x2 = x1 + 0.05 1 + x2 = 0.05 + 0.05 1 + 0.052 = 0.100125 , 1 x3 = x2 + 0.05 1 + x2 = 0.100125 + 0.05 1 + 0.1001252 0.150626 , 2 . . . x19 = x18 + 0.05 1 + x2 1.328148 , 18 (1) x20 = x19 + 0.05 1 + x2 = 1.328148 + 0.05 1 + 1.3281482 1.466347 , 19 which is a good enough approximation to (1) = tan 1 1.557408. 13. From Problem 12, yn = (1 + 1/n)n and so limn [(e  yn )/(1/n)] is a 0/0 indeterminant. If we let h = 1/n in yn and use L'Hospital's rule, we get e  yn e  (1 + h)1/h g(h) g (h) = lim = lim = lim , n 1/n h0 h0 h h0 h 1 lim where g(h) = e  (1 + h)1/h . Writing (1 + h)1/h as eln(1+h)/h the function g(h) becomes g(h) = e  eln(1+h)/h . The first derivative is given by g (h) = 0  d 1 d ln(1+h)/h e ln(1 + h) . = eln(1+h)/h dh dh h Substituting Maclaurin's series for ln(1 + h) we obtain g (h) = (1 + h)1/h d 1 dh h h 1 2 1 3 1 4 h + h  h + 2 3 4 23 Chapter 1
= (1 + h)1/h = (1 + h)1/h Hence 1 2 3 lim g (h) = lim (1 + h)1/h  + h  h2 + h0 h0 2 3 4 3 1 2 =  lim (1 + h)1/h lim  + h  h2 + h0 h0 2 3 4 From calculus we know that e = lim (1 + h)1/h , which gives
h0 1 1 d 1 1  h + h2  h3 + dh 2 3 4 1 2 3  + h  h2 + . 2 3 4 . h0 lim g (h) = e  1 2 = e . 2 So we have e e  yn = . n 1/n 2 lim 15. The independent variable in this problem is the time t and the dependent variable is the temperature T (t) of a body. Thus, we will use the recursive formulas (2) and (3) on page 25 with x replaced by t and y replaced by T . In the differential equation describing the Newton's Law of Cooling, f (t, T ) = K(M(t)  T ). With the suggested values of K = 1 (min)1 , M(t) 70 , h = 0.1, and the initial condition T (0) = 100 , the initial value problem becomes dT = 70  T, dt and so the recursive formulas are tn+1 = tn + 0.1 , Tn+1 = Tn + 0.1(70  Tn ). For n = 0, t1 = t0 + 0.1 = 0.1 , 24 T1 = T0 + 0.1(70  T0 ) = 100 + 0.1(70  100) = 97 ; T (0) = 100, Exercises 1.4
for n = 1, t2 = t1 + 0.1 = 0.2 , for n = 2, t3 = t2 + 0.1 = 0.3 , T3 = T2 + 0.1(70  T2 ) = 94.3 + 0.1(70  94.3) = 91.87 . T2 = T1 + 0.1(70  T1 ) = 97 + 0.1(70  97) = 94.3 ; Table 1C: Euler's method approximations for the solutions of T = K(M  T ), T (0) = 100, with K = 1, M = 70, and h = 0.1.
n 0 1 2 3 4 5 6 7 8 9 10 tn 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Tn 100.00 97.000 94.300 91.870 89.683 87.715 85.943 84.349 82.914 81.623 80.460 n 11 12 13 14 15 16 17 18 19 20 tn 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 Tn 79.414 78.473 77.626 76.863 76.177 75.559 75.003 74.503 74.053 73.647 By continuing this way and rounding results to three decimal places, we fill in Table 1C. From this table we conclude that (a) the temperature of a body after 1 minute T (1) 80.460 and (b) its temperature after 2 minutes T (2) 73.647 . 16. For this problem notice that the independent variable is t and the dependent variable is T . Hence, in the recursive formulas for Euler's method, the t will take the place of the x and the 25 Chapter 1
T will take the place of the y. Also we see that h = 0.1 and f (t, T ) = K (M 4  T 4 ), where K = 404 and M = 70. The recursive formulas (2) and (3) on page 25 of the text become tn+1 = tn + 0.1 , Tn+1 = Tn + hf (tn , Tn ) = Tn + 0.1 404
4 704  Tn , n = 0, 1, 2, . . . . From the initial condition, T (0) = 100, we see that t0 = 0 and T0 = 100. Therefore, for n = 0, t1 = t0 + 0.1 = 0 + 0.1 = 0.1 , T1 = T0 + 0.1 404
4 704  T0 = 100 + 0.1 404 704  1004 97.0316, where we have rounded off to four decimal places. For n = 1, we have t2 = t1 + 0.1 = 0.1 + 0.1 = 0.2 , T2 = T1 + 0.1 404
4 704  T1 = 97.0316 + 0.1 404 704  97.03164 94.5068 . By continuing this way, we fill in Table 1D. Table 1D: Euler's method approximations for the solution of T = K (M 4  T 4 ), T (0) = 100, with K = 404 , M = 70, and h = 0.1.
n 0 1 2 3 4 5 6 tn 0 0.1 0.2 0.3 0.4 0.5 0.6 Tn 100 97.0316 94.5068 92.3286 90.4279 88.7538 87.2678 n 7 8 9 10 11 12 13 tn 0.7 0.8 0.9 1.0 1.1 1.2 1.3 Tn 85.9402 84.7472 83.6702 82.6936 81.8049 80.9934 80.2504 n 14 15 16 17 18 19 20 tn 1.4 1.5 1.6 1.7 1.8 1.9 2.0 Tn 79.5681 78.9403 78.3613 77.8263 77.3311 76.8721 76.4459 From this table we see that T (1) = T (t10 ) T10 = 82.694 and T (2) = T (t20 ) T20 = 76.446 . 26 CHAPTER 2: First Order Differential Equations
EXERCISES 2.2: Separable Equations, page 46 1. This equation is separable because we can separate variables by multiplying both sides by dx and dividing by 2y 3 + y + 4. 3. This equation is separable because yex+y dy = 2 = dx x +2 5. Writing the equation in the form ds s+1 =  s2 , dt st we see that the righthand side cannot be represented in the form g(t)p(s). Therefore, the equation is not separable. 7. Multiplying both sides of the equation by y 2 dx and integrating yields y 2dy = (1  x2 )dx y 2 dy = (1  x2 )dx y= 3 3x  x3 + C , ex x2 + 2 yey = g(x)p(y). 1 3 1 y = x  x3 + C1 3 3 y 3 = 3x  x3 + C where C := 3C1 is an arbitrary constant. 9. To separate variables, we divide the equation by y and multiply by dx. This results dy dy = y(2 + sin x) = (2 + sin x)dx dx y dy = (2 + sin x)dx ln y = 2x  cos x + C1 y y = e2xcos x+C1 = eC1 e2xcos x = C2 e2xcos x , 27 Chapter 2
where C1 is an arbitrary constant and, therefore, C2 := eC1 is an arbitrary positive constant. We can rewrite the above solution in the form y = C2 e2xcos x = Ce2xcos x , (2.1) with C := C2 or C = C2 . Thus C is an arbitrary nonzero constant. The value C = 0 in (2.1) gives y(x) 0, which is, clearly, is also a solution to the differential equation. Therefore, the answer to the problem is given by (2.1) with an arbitrary constant C. 11. Separating variables, we obtain dy dx = . sec2 y 1 + x2 Using the trigonometric identities sec y = 1/ cos y and cos2 y = (1 + cos 2y)/2 and integrating, we get dx dx dy (1 + cos 2y)dy = = 2y 2 sec 1+x 2 1 + x2 (1 + cos 2y)dy dx = 2 1 + x2 1 1 y + sin 2y = arctan x + C1 2 2 2y + sin 2y = 4 arctan x + 4C1 2y + sin 2y = 4 arctan x + C. The last equation defines implicit solutions to the given differential equation. 13. Writing the given equation in the form dx/dt = x  x2 , we separate the variables to get dx = dt . x  x2 Integrate (the left side is integrated by partial fractions, with 1/(x  x2 ) = 1/x + 1/(1  x)) to obtain: ln x  ln 1  x = t + c 28 ln x =t+c 1x x = et+c = Cet , 1x x = Cet  xCet where C = ec x + xCet = Cet Exercises 2.2 x 1 + Cet = Cet x= Cet . 1 + Cet Note: When C is replaced by K, this answer can also be written as x = Ket /(Ket  1). Further we observe that since we divide by x  x2 = x(1  x), then x 0 and x 1 are also solutions. Allowing K to be zero gives x 0, but no choice for K will give x 1, so we list this as a separate solution. 15. To separate variables, we move the term containin dx to the righthand side of the equation and divide both sides of the result by y. This yields y 1dy = yecos x sin x dx Integrating the last equation, we obtain y 2dy = (ecos x sin x) dx y 1 + C = y= 1 , C  ecos x eu du (u = cos x) y 2dy = ecos x sin x dx. 1  + C = eu = ecos x y where C is an arbitrary constant. 17. First we find a general solution to the equation. Separating variables and integrating, we get dy dy = x3 (1  y) = x3 dx dx 1y dy x4  ln 1  y + C1 = = x3 dx 1y 4 4 x 4 1  y = exp C1  = Cex /4 . 4 To find C, we use the initial condition, y(0) = 3. Thus, substitution 3 for y and 0 for x into the last equation yields 1  3 = Ce0 Therefore, 1  y = 2ex
4 /4 4 /4 2 = C. . Finally, since 1  y(0) = 1  3 < 0, on an interval containing x = 0 one has 1  y(x) < 0 and so 1  y(x) = y(x)  1. The solution to the problem is then y  1 = 2ex
4 /4 or y = 2ex 4 /4 + 1. 29 Chapter 2
19. For a general solution, separate variables and integrate: dy dy = y sin = sin d d y dy = sin d ln y =  cos + C1 y y = e cos +C1 = Ce cos y = Ce cos (because at the initial point, = , y() < 0). We substitute now the initial condition, y() = 3, and obtain 3 = y() = Ce cos = Ce C = 3e1 . Hence, the answer is given by y = 3e1 e cos = 3e1cos . 21. Separate variables to obtain 1 (y + 1)1/2 dy = cos x dx. 2 Integrating, we have (y + 1)1/2 = sin x + C. Using the fact that y() = 0, we find 1 = sin + C Thus (y + 1)1/2 = sin x + 1 23. We have dy = 2x cos2 y dx dy = 2x dx cos2 y 2x dx sec2 y dy = 2x dx y = (sin x + 1)2  1 = sin2 x + 2 sin x . C = 1. sec2 y dy = tan y = x2 + C. Since y = /4 when x = 0, we get tan(/4) = 02 + C and so C = 1. The solution, therefore, is tan y = x2 + 1 30 y = arctan x2 + 1 . Exercises 2.2
25. By separating variables we obtain (1 + y)1 dy = x2 dx. Integrating yields ln 1 + y = x3 +C. 3 (2.2) Substituting y = 3 and x = 0 from the initial condition, we get ln 4 = 0 + C, which implies that C = ln 4. By substituting this value for C into equation (2.2) above, we have ln 1 + y = Hence, eln 1+y = e(x 3 /3)+ln 4 x3 + ln 4 . 3 = ex 3 /3 eln 4 = 4ex 3 /3 3 /3 1 + y = 4ex 3 /3 y = 4ex 1. We can drop the absolute signs above because we are assuming from the initial condition that y is close to 3 and therefore 1 + y is positive. 27. (a) The differential equation dy/dx = ex separates if we multiply by dx. We integrate the separated equation from x = 0 to x = x1 to obtain
x1 x=x1
2 e dx =
0 x=0 x2 dy = y x=x1 x=0 = y(x1 )  y(0). If we let t be the variable of integration and replace x1 by x and y(0) by 0, then we can express the solution to the initial value problem as
x y(x) =
0 et dt. 2 (b) The differential equation dy/dx = ex y 2 separates if we multiply by y 2 and dx. We integrate the separated equation from x = 0 to x = x1 to obtain
x1 x1 2 e dx =
0 0 x2 y 2 dy = 1 3 y 3 x=x1 x=0 = 1 y(x1 )3  y(0)3 . 3 31 Chapter 2
If we let t be the variable of integration and replace x1 by x and y(0) by 1 in the above equation, then we can express the initial value problem as
x et dt =
0 2 1 y(x)3  1 . 3 Solving for y(x) we arrive at y(x) = 1 + 3
0 x 1/3
2 et dt . (2.3) (c) The differential equation dy/dx =
x1 1 + sin x(1 + y 2 ) separates if we divide by (1 + y 2 ) and multiply by dx. We integrate the separated equation from x = 0 to x = x1 and find x=x1 1 + sin x dx =
x=0 (1 + y 2 )1 dy = tan1 y(x1 )  tan1 y(0). 0 If we let t be the variable of integration and replace x1 by x and y(0) by 1 then we can express the solution to the initial value problem by x 1 + sin t dt + . y(x) = tan 4
0 (d) We will use Simpson's rule (Appendix B) to approximate the definite integral found in part (b). (Simpson's rule is implemented on the website for the text.) Simpson's rule requires an even number of intervals, but we don't know how many are required to obtain the desired threeplace accuracy. Rather than make an error analysis, we will compute the approximate value of y(0.5) using 2, 4, 6, . . . intervals for Simpson's rule until the approximate values for y(0.5) change by less than five in the fourth place. For n = 2, we divide [0, 0.5] into 4 equal subintervals. Thus each interval will be of length (0.5  0)/4 = 1/8 = 0.125. Therefore, the integral is approximated by
0.5 ex dx =
0 2 1 0 2 2 2 2 e + 4e(0.125) + 2e(0.25) + 4e(0.325) + e(0.5) 0.544999003 . 24 32 Exercises 2.2
Substituting this value into equation (2.3) from part (b) yields y(0.5) [1 + 3(0.544999003)]1/3 1.38121 . Repeating these calculations for n = 3, 4, and 5 yields Table 2A. Table 2A: Successive approximations for y(0.5) using Simpson's rule.
Number of Intervals 6 8 10 y (0.5) 1.38120606 1.38120520 1.38120497 Since these values do not change by more than 5 in the fourth place, we can conclude that the first three places are accurate and that we have obtained an approximate solution y(0.5) 1.381 . 29. (a) Separating variables and integrating yields dy = dx y 1/3 dy = y 1/3 dx y= 2 2 x + C1 3 3
3/2 1 2/3 y = x + C1 2/3 = 2x +C 3 3/2 . (b) Using the initial condition, y(0) = 0, we find that 2(0) 0 = y(0) = +C 3
3/2 = C 3/2 C = 0, and so y = (2x/3 + 0)3/2 = (2x/3)3/2 , x 0, is a solution to the initial value problem. (c) The function y(x) 0, clearly, satisfies both, the differential equation dy/dx = y 1/3 and the initial condition y(0) = 0. (d) In notation of Theorem 1 on page 12, f (x, y) = y 1/3 and so d 1 1 f = y 1/3 = y 2/3 = 2/3 . y dy 3 3y 33 Chapter 2
Since f /y is not continuous when y = 0, there is no rectangle containing the point (0, 0) in which both, f and f /y, are continuous. Therefore, Theorem 1 does not apply to this initial value problem. 30. (a) Dividing the equation by (y + 1)2/3 and multiplying by dx separate variables. Thus we get dy = (x  3)(y + 1)2/3 dx dy = (y + 1)2/3 y+1= (x  3)dx
3 dy = (x  3)dx (y + 1)2/3 3(y + 1)1/3 = y = 1 + x2  3x + C1 2
3 C1 x2 x+ 6 3 x2 x+C 6 . (2.4) (b) Substitution y(x) 1 into the differential equation gives d(1) = (x  3)[(1) + 1]2/3 dx 0 = (x  3) 0, which is an identity. Therefore, y(x) 1 is, indeed, a solution. (c) With any choice of constant C, x2 /6  x + C is a quadratic polynomial which is not identically zero. So, in (2.4), y = 1 + (x2 /6  x + C)  1 for all C, and the solution y(x) 1 was lost in separation of variables. 31. (a) Separating variables and integrating yields dy dy = x dx = x dx 3 y y3 1 2 1 2 y 2 = x2  2C1 y = x + C1 2 2 x2 + y 2 = C, where C := 2C1 is an arbitrary constant. (b) To find the solution satisfying the initial condition y(0) = 1, we substitute in (2.5) 0 for x and 1 for y and obtain 02 + 12 = C 34 C=1 x2 + y 2 = 1.
3 (2.5) Exercises 2.2
Solving for y yields 1 . (2.6) 1  x2 Since, at the initial point, x = 0, y(0) = 1 > 1, we choose the positive sign in the above y = expression for y. Thus, the solution is y= 1 . 1  x2 y= y= 1 ; 4  x2 1 (1/4)  x2 Similarly we find solutions for the other two initial conditions: y(0) = 1 2 C=4 C= 1 4 y(0) = 2 . (c) For the solution to the first initial problem in (b), y(0) = 1, the domain is the set of all values of x satisfying two conditions 1  x2 0 (for existence of the square root) 1  x = 0 (for existence of the quotient) Solving for x, we get x2 < 1 x < 1 or  1 < x < 1.
2 1  x2 > 0. In the same manner, we find domains for solutions to the other two initial value problems: y(0) = 1 2 2 < x < 2 ; 1 1  <x< . 2 2 y(0) = 2 (d) First, we find the solution to the initial value problem y(0) = a, a > 0, and its domain. Following the lines used in (b) and (c) for particular values of a, we conclude that y(0) = a 02 + a2 = C y= a2 and so its domain is  x2 1 1  <x< . a a 1 a2  x2 > 0 x2 < a2 As a +0, 1/a +, and the domain expands to the whole real line; as a +, 1/a 0, and the domain shrinks to x = 0. 35 Chapter 2
1 a= 2 a=1 a=2 4 2 2 0 2 2 4 1 a= 2 a=1 a=2 Figure 2A: Solutions to the initial value problem y = xy 3 , y(0) = a, a 0.5, 1, and 2. (e) For the values a = 1/2, 1, and 2 the solutions are found in (b); for a = 1, we just have to choose the negative sign in (2.6); similarly, we reverse signs in the other two solutions in (b) to obtain the answers for a = 1/2 and 2. The graphs of these functions are shown in Figure 2A. 33. Let A(t) be the number of kilograms of salt in the tank at t minutes after the process begins. Then we have dA(t) = rate of salt in  rate of salt out. dt rate of salt in = 10 L/min 0.3 kg/L = 3 kg/min. Since the tank is kept uniformly mixed, A(t)/400 is the mass of salt per liter that is flowing out of the tank at time t. Thus we have rate of salt out = 10 L/min 36 A(t) A(t) kg/L = kg/min. 400 40 Exercises 2.2
dA A 120  A =3 = . dt 40 40 Separating this differential equation and integrating yield 40 dA = dt 120  A 40 ln 120  A = t + C Therefore, ln 120  A =  C t + C, where  is replaced by C 40 40 120  A = Cet/40 , where C can now be positive or negative A = 120  Cet/40 . There are 2 kg of salt in the tank initially, thus A(0) = 2. Using this initial condition, we find 2 = 120  C C = 118 . Substituting this value of C into the solution, we have A = 120  118et/40 . Thus A(10) = 120  118e10/40 28.1 kg. Note: There is a detailed discussion of mixture problems in Section 3.2. 35. In Problem 34 we saw that the differential equation dT /dt = k(M  T ) can be solved by separation of variables to yield T = Cekt + M. When the oven temperature is 120 we have M = 120. Also T (0) = 40. Thus 40 = C + 120 Because T (45) = 90, we have 90 = 80e45k + 120 3 = e45k 8 45k = ln 3 . 8 C = 80. Thus k = ln(3/8)/45 0.02180. This k is independent of M. Therefore, we have the general equation T (t) = Ce0.02180t + M. 37 Chapter 2
(a) We are given that M = 100. To find C we must solve the equation T (0) = 40 = C + 100. This gives C = 60. Thus the equation becomes T (t) = 60e0.02180t + 100. We want to solve for t when T (t) = 90. This gives us 90 = 60e0.02180t + 100 0.0218t = ln 1 6 1 = e0.02180t 6 0.0218t = ln 6 . Therefore t = ln 6/0.0218 82.2 min. (b) Here M = 140, so we solve T (0) = 40 = C + 140 As above, solving for t in the equation T (t) = 100e0.02180t + 140 = 90 (c) With M = 80, we solve 40 = C + 80, yielding C = 40. Setting T (t) = 40e0.02180t + 80 = 90 1  = e0.02180t . 4 t 31.8 . C = 100. This last equation is impossible because an exponential function is never negative. Hence it never attains desired temperature. The physical nature of this problem would lead us to expect this result. A further discussion of Newton's law of cooling is given in Section 3.3. 37. The differential equation dP r = P dt 100 38 Exercises 2.2
separates if we divide by P and multiply by dt. 1 r dP = P 100 dt ln P = r t+C 100 P (t) = Kert/100 , where K is the initial amount of money in the savings account, K = $1000, and r% is the interest rate, r = 5. This results in P (t) = 1000e5t/100 . (2.7) (a) To determine the amount of money in the account after 2 years we substitute t = 2 into equation (2.7), which gives P (2) = 1000e10/100 = $1105.17 . (b) To determine when the account will reach $4000 we solve equation (2.7) for t with P = $4000: 4000 = 1000e5t/100 e5t/100 = 4 t = 20 ln 4 27.73 years. (c) To determine the amount of money in the account after 3 1 years we need to determine 2 the value of each $1000 deposit after 3 1 years has passed. This means that the initial 2 $1000 is in the account for the entire 3 1 years and grows to the amount which is given 2 by P0 = 1000e5(3.5)/100 . For the growth of the $1000 deposited after 12 months, we take t = 2.5 in equation (2.7) because that is how long this $1000 will be in the account. This gives P1 = 1000e5(2.5)/100 . Using the above reasoning for the remaining deposits we arrive at P2 = 1000e5(1.5)/100 and P3 = 1000e5(0.5)/100 . The total amount is determined by the sum of the Pi 's. P = 1000 e5(3.5)/100 + e5(2.5)/100 + e5(1.5)/100 + e5(0.5)/100 $4, 427.59 . 39. Let s(t), t > 0, denote the distance traveled by driver A from the time t = 0 when he ran out of gas to time t. Then driver A's velocity vA (t) = ds/dt is a solution to the initial value problem dvA 2 = kvA , dt vA (0) = vB , 39 Chapter 2
where vB is driver B's constant velocity, and k > 0 is a positive constant. Separating variables we get dvA = k dt 2 vA dvA = 2 vA k dt 1 = kt + C . vA (t) From the initial condition we find 1 1 =k0+C =C = vB vA (0) Thus vA (t) = The function s(t) therefore satisfies vB ds = , dt vB kt + 1 Integrating we obtain s(t) = vB 1 dt = ln (vB kt + 1) + C1 . vB kt + 1 k s(0) = 0. 1 vB . = kt + 1/vB vB kt + 1 C= 1 . vB To find C1 we use the initial condition: 0 = s(0) = So, s(t) = 1 ln (vB kt + 1) . k 1 ln (vB k 0 + 1) + C1 = C1 k C1 = 0. At the moment t = t1 when driver A's speed was halved, i.e., vA (t1 ) = vA (0)/2 = vB /2, we have 1 vB 1 vB = vA (t1 ) = and 1 = s(t1 ) = ln (vB kt1 + 1) 2 vB kt1 + 1 k and so k = ln (vB kt1 + 1) = ln 2 vB kt1 + 1 = 2 1 ln (vB t ln 2 + 1) . s(t) = ln 2 40 Exercises 2.3
Since driver B was 3 miles behind driver A at time t = 0, and his speed remained constant, he finished the race at time tB = (3 + 2)/vB = 5/vB . At this moment, driver A had already gone s(tB ) = 1 1 5 vB ln 2 + 1 ln (vB tB ln 2 + 1) = ln ln 2 ln 2 vB 1 ln (5 ln 2 + 1) 2.1589 > 2 miles, = ln 2 i.e., A won the race. EXERCISES 2.3: 1. Writing dy  x2 y = x2 cos x , dx we see that this equation has the form (4) on page 50 of the text with P (x) = x2 and Q(x) = x2 cos x. Therefore, it is linear. y  cos x dy = . dx x2 Since the righthand side cannot be represented as a product g(x)p(y), the equation is not separable. 3. In this equation, the independent variable is t and the dependent variable is x. Dividing by x, we obtain sin t dx =  t2 . dt x Therefore, it is neither linear, because of the sin t/x term, nor separable, because the righthand side is not a product of functions of single variables x and t. 5. This is a linear equation with independent variable t and dependent variable y. This is also a separable equation because dy y(t  1) = 2 = dt t +1 t1 y = g(t)p(y). t2 + 1 41 Isolating dy/dx yields Linear Equations, page 54 Chapter 2
7. In this equation, P (x) 1 and Q(x) = e3x . Hence the integrating factor (x) = exp P (x)dx = exp (1)dx = ex . Multiplying both sides of the equation by (x) and integrating, we obtain dy d (ex y) x x 3x 2x e e y =e e =e = e2x dx dx 1 2x x 2x e y = e dx = e + C 2 1 2x e3x e + C ex = + Cex . y= 2 2
x 9. This is a linear equation with dependent variable r and independent variable . The method we will use to solve this equation is exactly the same as the method we use to solve an equation in the variables x and y since these variables are just dummy variables. Thus we have P () = tan and Q() = sec which are continuous on any interval not containing odd multiples of /2. We proceed as usual to find the integrating factor (). We have () = exp Thus we have () = sec , where we can drop the absolute value sign by making K = 1 if is in an interval where sec is positive or by making K = 1 if sec is negative. Multiplying the equation by the integrating factor yields sec dr + (sec tan )r = sec2 d D (r sec ) = sec2 . tan d = e ln  cos +C = K 1 = K sec ,  cos  where K = eC . Integrating with respect to yields r sec = sec2 d = tan + C r = cos tan + C cos r = sin + C cos . Because of the continuity of P () and Q() this solution is valid on any open interval that has end points that are consecutive odd multiples of /2. 42 Exercises 2.3
11. Choosing t as the independent variable and y as the dependent variable, we put the equation put into standard form: t+y+1 dy =0 dt dy  y = t + 1. dt (2.8) Thus P (t) 1 and so (t) = exp (1)dt = et . We multiply both sides of the second equation in (2.8) by (t) and integrate. This yields et dy  et y = (t + 1)et dt et y = d t e y = (t + 1)et dt et dt (t + 1)et dt = (t + 1)et + = (t + 1)et  et + C = (t + 2)et + C y = et (t + 2)et + C = t  2 + Cet , where we have used integration by parts to find (t + 1)et dt. 13. In this problem, the independent variable is y and the dependent variable is x. So, we divide the equation by y to rewrite it in standard form. y dx + 2x = 5y 2 dy dx 2 + x = 5y 2. dy y Therefore, P (y) = 2/y and the integrating factor, (y), is (y) = exp 2 dy y = exp (2 ln y) = y2 = y 2 . Multiplying the equation (in standard form) by y 2 and integrating yield y2 dx + 2y x = 5y 4 dy y 2x = d y 2x = 5y 4 dy x = y 2 y 5 + C = y 3 + Cy 2. 5y 4 dy = y 5 + C 15. To put this linear equation in standard form, we divide by (x2 + 1) to obtain x x dy + 2 y= 2 . dx x + 1 x +1 (2.9) 43 Chapter 2
Here P (x) = x/(x2 + 1), so P (x) dx = Thus the integrating factor is (x) = e(1/2) ln(x
2 +1) 1 x dx = ln(x2 + 1). x2 + 1 2 = eln[(x 2 +1)1/2 = (x2 + 1)1/2 . Multiplying equation (2.9) by (x) yields (x2 + 1)1/2 which becomes d x (x2 + 1)1/2 y = 2 . dx (x + 1)1/2 Now we integrate both sides and solve for y to find (x2 + 1)1/2 y = (x2 + 1)1/2 + C y = 1 + C(x2 + 1)1/2 . x dy x + 2 y= 2 , 1/2 dx (x + 1) (x + 1)1/2 This solution is valid for all x since P (x) and Q(x) are continuous for all x. 17. This is a linear equation with P (x) = 1/x and Q(x) = xex which is continuous on any interval not containing 0. Therefore, the integrating factor is given by (x) = exp  1 x dx = e ln x = 1 , x for x > 0. Multiplying the equation by this integrating factor yields y 1 dy  2 = ex x dx x Integrating gives y = ex + C x y = xex + Cx. Dx y = ex . x Now applying the initial condition, y(1) = e  1, we have e1=e+C 44 C = 1. Exercises 2.3
Thus, the solution is y = xex  x, on the interval (0, ). Note: This interval is the largest interval containing the initial value x = 1 in which P (x) and Q(x) are continuous. 19. In this problem, t is the independent variable and x is the dependent variable. One can notice that the lefthand side is the derivative of xt3 with respect to t. Indeed, using product rule for differentiation, we get dx 3 dx d (t3 ) d xt3 = t +x = t3 + 3t2 x. dt dt dt dt Thus the equation becomes d xt3 = t dt x = t3 t2 +C 2 xt3 = = t dt = t2 +C 2 1 C + 3. 2t t (Of course, one could divide the given equation by t3 to get standard form, conclude that P (t) = 3/t, find that (t) = t3 , multiply by t3 back, and come up with the original equation.) We now use the initial condition, x(2) = 0, to find C. 0 = x(2) = 1 C + 3 2(2) 2 1 C + =0 4 8 C = 2. Hence, the solution is x = 1/(2t)  2/(t3 ). 21. Putting the equation in standard form yields sin x dy + y = 2x cos x dx cos x Therefore, P (x) = tan x and so (x) = exp tan x dx = exp ( ln  cos x) =  cos x1 . 45 dy + (tan x)y = 2x cos x. dx Chapter 2
At the initial point, x = /4, cos(/4) > 0 and, therefore, we can take (x) = (cos x)1 . Multiplying the standard form of the given equation by (x) gives sin x 1 dy + y = 2x cos x dx cos2 x 1 y = 2x dx = x2 + C cos x From the initial condition, we find C: 15 2 2 =y = cos 32 4 4 d dx 1 y cos x = 2x y = cos x x2 + C . 4 2 +C C =  2 . Hence, the solution is given by y = cos x (x2  2 ). 23. We proceed similarly to Example 2 on page 52 and obtain an analog of the initial value problem (13), that is, dy + 5y = 40e20t , dt Thus P (t) 5 and (t) = exp by (t) and integrating, we obtain e5t dy + 5e5t y = 40e20t e5t = 40e15t dt d (e5t y) e5t y = = 40e15t dt y(0) = 10. (2.10) 5dt = e5t . Multiplying the differential equation in (2.10) 40e15t dt = 40 15t + C. e 15 Therefore, a general solution to the differential equation in (2.10) is y = e5t 40 15t e +C 15 = Ce5t  8 20t e . 3 Finally, we find C using the initial condition. 10 = y(0) = Ce50  8 200 8 e =C 3 3 C = 10 + 8 38 = . 3 3 Hence, the mass of RA2 for t 0 is given by y(t) = 46 38 5t 8 20t e  e . 3 3 Exercises 2.3
25. (a) This is a linear problem and so an integrating factor is (x) = exp 2x dx = exp x2 . Multiplying the equation by this integrating factor yields ex
2 dy 2 2 + 2xex y = ex dx
x x Dx yex
2 2 = ex 2 2 Dt ye t2 dt =
2 et dt, where we have changed the dummy variable x to t and integrated with respect to t from 2 (since the initial value for x in the initial condition is 2) to x. Thus, since y(2) = 1, x x x yex  e4 =
2
2 et dt 2 2 y = ex e4 + 2 2 2 et dt = e4x + ex et dt .
2 2 2 (b) We will use Simpson's rule (page A.3 of the Appendix B) to approximate the definite integral found in part (a) with upper limit x = 3. Simpson's rule requires an even number of intervals, but we don't know how many are required to obtain the desired 3 place accuracy. Rather than make an error analysis, we will compute the approximate value of y(3) using 4, 6, 8, 10, 12, . . . intervals for Simpson's rule until the approximate values for y(3) change by less than 5 in the fourth place. For n = 2 we divide [2, 3] into 4 equal subintervals. Thus, each subinterval will be of length (3  2)/4 = 1/4. Therefore, the integral is approximated by
3 et dt 2 2 1 (2)2 2 2 2 2 e + e(2.25) + e(2.5) + e(2.75) + e(3) 1460.354350 . 12
2 2 Dividing this by e(3) and adding e43 = e5 , gives y(3) 0.186960 . Doing calculations for 6, 8, 10, and 12 intervals yields Table 2B. 47 Chapter 2 Table 2B: Successive approximations for y(3) using Simpson's rule.
Number of Intervals 6 8 10 12 y (3) 0.183905 0.183291 0.183110 0.183043 Since the last 3 approximate values do not change by more than 5 in the fourth place, it appears that their first three places are accurate and the approximate solution is y(3) 0.183 . 27. (a) The given differential equation is in standard form. Thus P (x) = we cannot express calculus to conclude that, with any fixed constant a, a x 1 + sin2 x. Since P (x) dx as an elementary function, we use fundamental theorem of P (t)dt = P (x), that is, the above definite integral with variable upper bound is an antiderivative of P (x). Since, in the formula for (x), one can choose any antiderivative of P (x), we take the above definite integral with a = 0. (Such a choice of a comes from the initial point x = 0 and makes it easy to satisfy the initial condition.) Therefore, the integrating factor (x) can be chosen as (x) = exp 0 x 1 + sin2 t dt . Multiplying the differential equaion by (x) and integrating from x = 0 to x = s, we obtain d[(x)y] = (x)x dx 48 d[(x)y] = (x)x dx Exercises 2.3
s s s 0 d[(x)y] =
0 (x)x dx
s x=s (x)y(x)
x=0 =
0 (x)x dx (s)y(s)  (0)y(0) =
0 (x)x dx . From the initial condition, y(0) = 2. Also, note that 0 (0) = exp 0 1 + sin2 t dt = e0 = 1. This yields (0)y(0) = 2 and so
s (s)y(s) =
0 (x)x dx + 2 . Dividing by (s) and interchanging x and s give the required. (b) The values of (x), x = 0.1, 0.2, . . ., 1.0, approximated by using Simpson's rule, are given in Table 2C.
x 0 Table 2C: Approximations of (x) = rule.
x 0.0 0.1 0.2 0.3 0.4 0.5 (x) 0.0 0.100166 0.201315 0.304363 0.410104 0.519172 (x) 1.0000 1.105354 1.223010 1.355761 1.506975 1.680635 1 + sin2 t dt and (x) = e(x) using Simpson's
x 0.6 0.7 0.8 0.9 1.0 (x) 0.632016 0.748903 0.869917 0.994980 1.123865 (x) 1.881401 2.114679 2.386713 2.704670 3.076723 We now use these values of (x) to approximate again. With n = 5 and h= 1 0 (s)s ds by applying Simpson's rule 10 = 0.1 2n 49 Chapter 2
the Simpson's rule becomes
1 (s)s ds 0 0.1 [(0)(0) + 4(0.1)(0.1) + 2(0.2)(0.2) + 4(0.3)(0.3) 3 +2(0.4)(0.4) + 4(0.5)(0.5) + 2(0.6)(0.6) + 4(0.7)(0.7) +2(0.8)(0.8) + 4(0.9)(0.9) + (1.0)(1.0)] 1.064539 . Therefore,
1 1 y(1) (1) (s)s ds +
0 1 2 2 = 1.064539 + = 0.9960 . (1) 3.076723 3.076723 (c) We rewrite the differential equation in the form used in Euler's method, dy = x  1 + sin2 x y , y(0) = 2, dx and conclude that f (x, y) = x  1 + sin2 xy. Thus the recursive formulas (2) and (3) on page 25 of the text become xn+1 = xn + h, yn+1 = yn + h xn  1 + sin2 xn yn , n = 0, 1, . . . , x0 = 0, y0 = 2. With h = 0.1 we need (1  0)/0.1 steps to get an approximation at x = 1. n = 0 : x1 = 0.1 , y1 = (2) + 0.1[(0)  1 + sin2 (0) (2)] = 1.8000; 1 + sin2 (0.1) (1.8)] 1.6291; 1 + sin2 (0.2) (1.6291)] 1.4830; n = 1 : x2 = 0.2 , y2 = (1.8) + 0.1[(0.1)  n = 2 : x3 = 0.3 , y3 = (1.6291) + 0.1[(0.2)  . . . Results of these computations, rounded off to four decimal places, are given in Table 2D. Thus Euler's method with step h = 0.1 gives y(1) 0.9486 . Next we take h = 0.05 and fill in the Table 2E. So, with step h = 0.05, we have y(1) 0.9729 . 50 Exercises 2.3 Table 2D: Euler's method approximations for the solution of y + y 1 + sin2 x = x, y(0) = 2, at x = 1 with h = 0.1.
k 0 1 2 3 xk 0.0 0.1 0.2 0.3 yk 2.0000 1.8000 1.6291 1.4830 k 4 5 6 7 xk 0.4 0.5 0.6 0.7 yk 1.3584 1.2526 1.1637 1.0900 k 8 9 10 xk 0.8 0.9 1.0 yk 1.0304 0.9836 0.9486 Table 2E: Euler's method approximations for the solution of y + y 1 + sin2 x = x, y(0) = 2, at x = 1 with h = 0.05.
n 0 1 2 3 4 5 6 xn 0.00 0.05 0.10 0.15 0.20 0.25 0.30 yn 2.0000 1.9000 1.8074 1.7216 1.6420 1.5683 1.5000 n 7 8 9 10 11 12 13 xn 0.35 0.40 0.45 0.50 0.55 0.60 0.65 yn 1.4368 1.3784 1.3244 1.2747 1.2290 1.1872 1.1490 n 14 15 16 17 18 19 20 xn 0.70 0.75 0.80 0.85 0.90 0.95 1.00 yn 1.1144 1.0831 1.0551 1.0301 1.0082 0.9892 0.9729 29. In the presented form, the equation dy 1 = 4y dx e + 2x is, clearly, not linear. But, if we switch the roles of variables and consider y as the independent variable and x as the dependent variable (using the connection between derivatives of inverse functions, that is, the formula y (x) = 1/x (y)), then the equation transforms to dx = e4y + 2x dy dx  2x = e4y . dy This is a linear equation with P (y) = 2. Thus the integrating factor is (y) = exp (2)dy = e2y 51 Chapter 2
d 2y e2y e x = e2y e4y = e2y + C. e2y x = e2y dy = dy 2 Solving for x yields e2y e4y x = e2y +C = + Ce2y . 2 2 31. (a) On the interval 0 x 2, we have P (x) = 1. Thus we are solving the equation dy + y = x, dx The integrating factor is given by (x) = exp dx = ex . y(0) = 1. and so Multiplying the equation by the integrating factor, we obtain ex dy + ex y = xex dx Dx [ex y] = xex ex y = xex dx . Calculating this integral by parts and dividing by ex yields y = ex (xex  ex + C) = x  1 + Cex . (b) Using the initial condition, y(0) = 1, we see that 1 = y(0) = 0  1 + C = 1 + C Thus the solution becomes y = x  1 + 2ex . (c) In the interval x > 2, we have P (x) = 3. Therefore, the integrating factor is given by (x) = exp 3 dx = e3x . C = 2. Multiplying the equation by this factor and solving yields e3x dy + 3e3x y = xe3x dx Dx e3x y = xe3x e3x y = xe3x dx . Integrating by parts and dividing by e3x gives y = e3x 1 3x 1 3x x 1 xe  e + C =  + Ce3x . 3 9 3 9 52 Exercises 2.3
(d) We want the value of the initial point for the solution in part (c) to be the value of the solution found in part (b) at the point x = 2. This value is given by y(2) = 2  1 + 2e2 = 1 + 2e2 . Thus the initial point we seek is y(2) = 1 + 2e2 . Using this initial point to find the constant C given in part (c) yields 1 + 2e2 = y(2) = 2 1  + Ce6 3 9 C= 4 6 e + 2e4 . 9 Thus, the solution of the equation on the interval x > 2 is given by y= 4 6 x 1  + e + 2e4 e3x . 3 9 9 Patching these two solutions together gives us a continuous solution to the original equation on the interval x 0: x  1 + 2ex , 0 x 2; y= x 1  + 4 e6 + 2e4 e3x , 2 < x. 3 9 9 (e) The graph of the solution is given in Figure B.18 of the answers in the text. 33. (a) Writing the equation in standard form yields 2 dy + y = 3. dx x Therefore, P (x) = 2/x and (x) = exp Hence d x2 y = 3x2 dx x2 y = 3x2 dx = x3 + C y =x+ C x2 53 2 dx x = exp (2 ln x) = x2 = x2 . Chapter 2
is a general solution to the given differential equation. Unless C = 0 and so y = x, the function y = x + C/x2 is not defined when x = 0. Therefore, among all solutions, the only function defined at x = 0 is (x) = x, and the initial value problem with y(0) = y0 has a solution (and unique) if and only if y0 = (x) = 0.
x=0 (b) Standard form of the equation xy  2y = 3x is 2 dy  y = 3. dx x This gives P (x) = 2/x, (x) = exp d x2 y = 3x2 dx x2 y = (2/x)dx = x2 , and 3x2 dx = 3x1 + C y = 3x + Cx2 . Therefore, any solution is a polynomial and so is defined for all real numbers. Moreover, any solution satisfies the initial condition y(0) = 0 because 3x + Cx2
x=0 = 3(0) + C(0)2 = 0 and, therefore, is a solution to the initial value problem. (This also implies that the initial value problem with y(0) = y0 = 0 has no solution.) 35. (a) This part of the problem is similar to Problem 33 in Section 2.2. So, we proceed in the same way. Let A(t) denote the mass of salt in the tank at t minutes after the process begins. Then we have rate of input = 5 L/min 0.2 kg/L = 1 kg/min , A(t) A(t) rate of exit = 5 L/min kg/L = kg/min , 500 100 dA A 100  A =1 = . dt 100 100 Separating this differential equation yields dA/(100  A) = dt/100. Integrating, we obtain  ln 100  A = 54 t + C1 100 100  A = et/100C1 = eC1 et/100 Exercises 2.3 100  A = Cet/100 C = eC1 A = 100  Cet/100 . The initial condition, A(0) = 5 (initially, there were 5 kg of salt in the tank) implies that 5 = A(0) = 100  C C = 95. Substituting this value of C into the solution, we have A(t) = 100  95et/100 . Thus the mass of salt in the tank after 10 min is A(10) = 100  95e10/100 14.04 kg , which gives the concentration 14.04 kg/500 L 0.0281 kg/L. (b) After the leak develops, the system satisfies a new differential equation. While the rate of input remains the same, 1 kg/min, the rate of exit is now different. Since, every minute, 5 liters of the solution is coming in and 5 + 1 = 6 liters are going out, the volume of the solution in the tank decreases by 6  5 = 1 liter per minute. Thus, for t 10, the volume of the solution in the tank is 500  1 (t  10) = 510  t liters. This gives the concentration of salt in the tank A(t) kg/L 510  t and rate of exit = 6 L/min A(t) 6A(t) kg/L = kg/min . 510  t 510  t dA 6A + =1 dt 510  t (2.11) Hence, the differential equation, for t > 10, becomes dA 6A =1 dt 510  t with the initial condition A(10) = 14.04 (the value found in (a) ). This equation is a linear equation. We have (t) = exp 6 dt 510  t = exp (6 ln 510  t) = (510  t)6 55 Chapter 2 d (510  t)6 A = 1 (510  t)6 = (510  t)6 dt 1 (510  t)6 A = (510  t)6 dt = (510  t)5 + C 5 1 A = (510  t) + C(510  t)6 . 5 Using the initial condition, A(10) = 14.04, we compute C. 14.04 = A(10) = Therefore, A(t) = 1 85.96 1 (510  t)  (510  t)6 = (510  t)  85.96 6 5 (500) 5 510  t 500
6 1 (510  10) + C(510  10)6 5 C= 85.96 . (500)6 and, according to (2.11), the concentration of salt is given by A(t) 1 85.96 =  510  t 5 510  t
6 510  t 500 6 . 20 minutes after the leak develops, that is, when t = 30, the concentration will be 1 85.96  5 510  30 37. We are solving the equation t dx + 2x = 1  cos dt 12 , x(0) = 10. 510  30 500 0.0598 kg/L . This is a linear problem with dependent variable x and independent variable t so that P (t) = 2. Therefore, to solve this equation we first must find the integrating factor (t). (t) = exp Multiplying the equation by this factor yields e2t 56 dx + 2xe2t = e2t 1  cos dt xe2t = e2t dt  t 12 e2t cos = e2t  e2t cos t 12 t 12 e2t cos t dt. 12 2 dt = e2t . 1 dt = e2t  2 Exercises 2.3
The last integral can be found by integrating by parts twice which leads back to an integral similar to the original. Combining these two similar integrals and simplifying, we obtain t e cos 12
2t dt = e2t 2 cos + 12 sin 4 + ( 12 )2 t 12 t 12 t 12 + C. Thus we see that t 1 2 cos 12 + 12 sin x(t) =  2 4 + ( 12 )2 + Ce2t . Using the initial condition, t = 0 and x = 10, to solve for C, we obtain C= Therefore, the desired solution is x(t) =
t 1 2 cos 12 + 12 sin  2 4 + ( 12 )2 t 12 2 19 + . 2 4 + ( 12 )2 + 19 2 e2t . + 2 4 + ( 12 )2 39. Let Tj (t), j = 0, 1, 2, . . ., denote the temperature in the classroom for 9 + j t < 10 + j, where t = 13 denotes 1 : 00 p.m., t = 14 denotes 2 : 00 p.m., etc. Then T (9) = 0, and the continuity of the temperature implies that lim = Tj+1 (10 + j), j = 0, 1, 2, . . . . (2.13) (2.12) t10+j According to the work of the heating unit, the temperature satisfies the equation dTj = dt 1  Tj , if j = 2k Tj , if j = 2k + 1 , 9 + j < t < 10 + j k = 0, 1, . . . . The general solutions of these equations are: for j even dTj dTj = 1  Tj = dt dt 1  Tj ln 1  Tj  = t + cj Tj (t) = 1  Cj et ; 57 Chapter 2
for j odd dTj dTj = Tj = dt dt Tj ln Tj  = t + cj where Cj = 0 are constants. From (2.12) we have: 0 = T0 (9) = 1  C0 et = 1  C0 e9 C0 = e9 . Tj (t) = Cj et ; t=9 Also from (2.13), for even values of j (say, j = 2k) we get 1  C2k et = C2k+1 et = C2k+1 e t=9+(2k+1) t=9+(2k+1) 1  C2k e (10+2k) (10+2k) C2k+1 = e10+2k  C2k . Similarly from (2.13) for odd values of j (say, j = 2k + 1) we get C2k+1 et = 1  C2k+2 et = 1  C2k+2 e t=9+(2k+2) t=9+(2k+2) C2k+1 e (11+2k) (11+2k) C2k+2 = e11+2k  C2k+1 . In general we see that for any integer j (even or odd) the following formula holds: Cj = e9+j  Cj1. Using this recurrence formula we successively compute C1 = e10  C0 = e10  e9 = e9 (e  1) C2 = e11  C1 = e11  e10 + e9 = e9 (e2  e + 1) . . .
j Cj = e 58 9 k=0 (1)jk ek . Exercises 2.4
Therefore, the temperature at noon (when t = 12 and j = 3) is
3 T3 (12) = C3 e12 = e12 e9
k=0 (1)3k ek = 1  e1 + e2  e3 0.718 = 71.8 F. At 5 p.m.(when t = 17 and j = 8), we find
8 8 T8 (17) = 1  C8 e17 = 1  e17 e9
k=0 (1)8k ek =
k=1 (1)k+1 ek 1  (e1 )8 = e1 0.269 = 26.9 F. 1 1+e EXERCISES 2.4: Exact Equations, page 65 1. In this equation, M(x, y) = x2 y + x4 cos x and N(x, y) = x3 . Taking partial derivatives, we obtain N M x2 y + x4 = x2 = 3x2 = = . y y x Therefore, according to Theorem 2 on page 61 of the text, the equation is not exact. Rewriting the equation in the form x2 y + x4 cos x 1 dy = = y + x cos x, 3 dx x x (2.14) we conclude that it is not separable because the righthand side in (2.14) cannot be factored as p(x)q(y). We also see that the equation is linear with y as the dependent variable. 3. Here M(x, y) = yexy + 2x, N(x, y) = xexy  2y. Thus M xy = (yexy + 2x) = exy + y (e ) = exy + yexy x = exy (1 + yx), y y y N xy = (xexy  2y) = exy + x (e ) = exy + xexy y = exy (1 + xy), x x x M/y = N/x, and the equation is exact. We write the equation in the form yexy + 2x dy =  xy dx xe  2y 59 Chapter 2
and conclude that it is not separable because the righthand side cannot be represented as a product of two functions of single variables x and y. Also, the righthand side is not linear with respect to y which implies that the equation is not linear with y as the dependent variable. Similarly, choosing x as the dependent variable (taking the reciprocals of both sides) we conclude that the equation is not linear either. 5. The differential equation is not separable because (2xy + cos y) cannot be factored. This equation can be put in standard form by defining x as the dependent variable and y as the independent variable. This gives dx 2  cos y , + x= dy y y2 so we see that the differential equation is linear. If we set M(x, y) = y 2 and N(x, y) = 2xy + cos y we are able to see that the differential equation is also exact because My (x, y) = 2y = Nx (x, y). 7. In this problem, the variables are r and , M(r, ) = , and N(r, ) = 3r   1. Because N M =1=3= , r the equation is not exact. With r as the dependent variable, the equation takes the form dr 3r   1 3 +1 = = r+ , d and it is linear. Since the righthand side in the above equation cannot be factored as p()q(r), the equation is not separable. 9. We have that M(x, y) = 2xy + 3 and N(x, y) = x2  1. Therefore, My (x, y) = 2x = Nx (x, y) and so the equation is exact. We will solve this equation by first integrating M(x, y) with respect to x, although integration of N(x, y) with respect to y is equally easy. Thus F (x, y) = 60 (2xy + 3) dx = x2 y + 3x + g(y). Exercises 2.4
Differentiating F (x, y) with respect to y gives Fy (x, y) = x2 + g (y) = N(x, y) = x2  1. From this we see that g = 1. (As a partial check we note that g (y) does not involve x.) Integrating gives g(y) = (1) dy = y. Since the constant of integration will be incorporated into the parameter of the solution, it is not written here. Substituting this expression for g(y) into the expression that we found for F (x, y) yields F (x, y) = x2 y + 3x  y. Therefore, the solution of the differential equation is x2 y + 3x  y = C y= C  3x . x2  1 The given equation could be solved by the method of grouping. To see this, express the differential equation in the form (2xy dx + x2 dy) + (3 dx  dy) = 0. The first term of the lefthand side we recognize as the total differential of x2 y. The second term is the total differential of (3x  y). Thus we again find that F (x, y) = x2 y + 3x  y and, again, the solution is x2 y + 3x  y = C. 11. Computing partial derivatives of M(x, y) = cos x cos y + 2x and N(x, y) = (sin x sin y + 2y), we obtain M = (cos x cos y + 2x) =  cos x sin y , y y N = [ (sin x sin y + 2y)] =  cos x sin y , x x M N = , y x and the equation is exact. 61 Chapter 2
Integrating M(x, y) with respect to x yields F (x, y) = M(x, y)dx = (cos x cos y + 2x) dx cos x dx + 2x dx = sin x cos y + x2 + g(y). = cos y To find g(y), we compute the partial derivative of F (x, y) with respect to y and compare the result with N(x, y). F = sin x cos y + x2 + g(y) =  sin x sin y + g (y) =  (sin x sin y + 2y) y y g (y) = 2y g(y) = (2y)dy = y 2 . (We take the integration constant C = 0.) Therefore, F (x, y) = sin x cos y + x2  y 2 = c is a general solution to the given equation. 13. In this equation, the variables are y and t, M(y, t) = t/y, N(y, t) = 1 + ln y. Since M = t t the equation is exact. Integrating M(y, t) with respect to y, we get F (y, t) = t dy = t ln y + g(t) = t ln y + g(t). y t y = 1 y and N 1 = (1 + ln y) = , y y y (From N(y, t) = 1 + ln y we conclude that y > 0.) Therefore, F = [t ln y + g(t)] = ln y + g (t) = 1 + ln y t t g(t) = t g (t) = 1 F (y, t) = t ln y + t, and a general solution is given by t ln y + t = c (or, explicitly, t = c/(ln y + 1)). 62 Exercises 2.4
15. This differential equation is expressed in the variables r and . Since the variables x and y are dummy variables, this equation is solved in exactly the same way as an equation in x and y. We will look for a solution with independent variable and dependent variable r. We see that the differential equation is expressed in the differential form M(r, ) dr + N(r, ) d = 0, This implies that M (r, ) =  sin = Nr (r, ), and so the equation is exact. Therefore, to solve the equation we need to find a function F (r, ) that has cos dr + (r sin + e ) d as its total differential. Integrating M(r, ) with respect to r we see that F (r, ) = Thus we have that g () = e g() = e , cos dr = r cos + g() where M(r, ) = cos and N(r, ) = r sin + e . F (r, ) = r sin + g () = N(r, ) = r sin + e . where the constant of integration will be incorporated into the parameter of the solution. Substituting this expression for g() into the expression we found for F (r, ) yields F (r, ) = r cos + e . From this we see that the solution is given by the one parameter family r cos + e = C, or, solving for r, r= C  e = (C  e ) sec . cos 17. Partial derivatives of M(x, y) = 1/y and N(x, y) =  (3y  x/y 2 ) are M = y y 1 y = 1 y2 and N = x x 3y + x y2 = 1 . y2 Since M/y = N/x, the equation is not exact. 63 Chapter 2
19. Taking partial derivatives of M(x, y) = 2x + y/(1 + x2 y 2 ) and N(x, y) = 2y + x/(1 + x2 y 2 ) with respect to y and x, respectively, we get M 1  x2 y 2 (1)(1 + x2 y 2 )  yx2 (2y) y = = , = 2x + y y 1 + x2 y 2 (1 + x2 y 2)2 (1 + x2 y 2 )2 1  x2 y 2 N (1)(1 + x2 y 2 )  xy 2 (2x) x = = . = 2y + x x 1 + x2 y 2 (1 + x2 y 2 )2 (1 + x2 y 2)2 Therefore, the equation is exact. F (x, y) = d(xy) = x2 + arctan(xy) + g(y) 1 + (xy)2 F x x = x2 + arctan(xy) + g(y) = + g (y) = 2y + 2 y y 1 + (xy) 1 + x2 y 2 g (y) = 2y g(y) = y 2 2x + y 1 + x2 y 2 dx = x2 + F (x, y) = x2  y 2 + arctan(xy) and a general solution then is given implicitly by x2  y 2 + arctan(xy) = c. 21. We check the equation for exactness. We have M(x, y) = 1/x + 2y 2x, N(x, y) = 2yx2  cos y, 1 M = + 2y 2x = 4yx, y y x N = 2yx2  cos y = 4yx. x x Thus M/y = N/x. Integrating M(x, y) with respect to x yields F (x, y) = Therefore, F = ln x + x2 y 2 + g(y) = 2x2 y + g (y) = N(x, y) = 2yx2  cos y y y g (y) =  cos y g(y) = ( cos y)dy =  sin y 1 + 2y 2x dx = ln x + x2 y 2 + g(y). x F (x, y) = ln x + x2 y 2  sin y, and a general solution to the given differential equation is ln x + x2 y 2  sin y = c. 64 Exercises 2.4
Substituting the initial condition, y = when x = 1, we find c. ln 1 + 12 2  sin = c c = 2. Therefore, the answer is given implicitly by ln x + x2 y 2  sin y = 2 . (We also used the fact that at the initial point, (1, ), x > 0 to skip the absolute value sign in the logarithmic term.) 23. Here M(t, y) = et y + tet y and N(t, y) = tet + 2. Thus My (t, y) = et + tet = Nt (t, y) and so the equation is exact. To find F (t, y) we first integrate N(t, y) with respect to y to obtain F (t, y) = (tet + 2) dy = (tet + 2)y + h(t), where we have chosen to integrate N(t, y) because this integration is more easily accomplished. Thus Ft (t, y) = et y + tet y + h (t) = M(t, y) = et y + tet y h (t) = 0 h(t) = C. We will incorporate this constant into the parameter of the solution. Combining these results gives F (t, y) = tet y + 2y. Therefore, the solution is given by tet y + 2y = C. Solving for y yields y = C/(tet + 2). Now we use the initial condition y(0) = 1 to find the solution that passes through the point (0, 1). Thus y(0) = C = 1 0+2 C = 1 2 2 . tet + 2 C = 2. This gives us the solution y= 25. One can check that the equation is not exact (M/y = N/x), but it is separable because it can be written in the form y 2 sin x dx + 1y dy = 0 x y1 dy. x sin x dx = y2 y 2 sin x dx = y1 dy x 65 Chapter 2
Integrating both sides yields x sin x dx = y1 dy y2 x( cos x)  1 + C, y x sin x dx. Substitution of the initial condition, 1 +C 1 ( cos x)dx = 1 1  2 y y dy x cos x + sin x = ln y + where we applied integration by parts to find y() = 1, results  cos + sin = ln 1 + C =  1. So, the solution to the initial value problem is x cos x + sin x = ln y + 1/y +  1 . (Since y() = 1 > 0, we have removed the absolute value sign in the logarithmic term.) 27. (a) We want to find M(x, y) so that for N(x, y) = sec2 y  x/y we have 1 My (x, y) = Nx (x, y) =  . y Therefore, we must integrate this last expression with respect to y. That is, M(x, y) =  1 y dy =  ln y + f (x), where f (x), the "constant" of integration, is a function only of x. (b) We want to find M(x, y) so that for N(x, y) = sin x cos y  xy  ey we have My (x, y) = Nx (x, y) = cos x cos y  y. Therefore, we must integrate this last expression with respect to y. That is M(x, y) = (cos x cos y  y) dy = cos x y2 + f (x), 2 cos y dy  y dy = cos x sin y  where f (x), a function only of x, is the "constant" of integration. 66 Exercises 2.4
29. (a) We have M(x, y) = y 2 + 2xy and N(x, y) = x2 . Therefore My (x, y) = 2y + 2x and Nx (x, y) = 2x. Thus My (x, y) = Nx (x, y), so the differential equation is not exact. (b) If we multiply (y 2 + 2xy)dx  x2 dy = 0 by y 2 , we obtain 1+ 2x x2 dx  2 dy = 0. y y In this equation we have M(x, y) = 1 + 2xy 1 and N(x, y) = x2 y 2. Therefore, M(x, y) N(x, y) 2x = 2 = . y y x So the new differential equation is exact. (c) Following the method for solving exact equations we integrate M(x, y) in part (b) with respect to x to obtain F (x, y) = 1+2 x2 x + g(y) . dx = x + y y To determine g(y), take the partial derivative of both sides of the above equation with respect to y to obtain F x2 =  2 + g (y) . y y x2 x2 =  2 + g (y) y2 y Substituting N(x, y) (given in part (b)) for F/y, we can now solve for g (y) to obtain N(x, y) =  g (y) = 0 . The integral of g (y) will yield a constant and the choice of the constant of integration is not important so we can take g(y) = 0. Hence we have F (x, y) = x + x2 /y and the solution to the equation is given implicitly by x+ x2 =C. y Solving the above equation for y, we obtain y= x2 . C x 67 Chapter 2
(d) By dividing both sides by y 2 we lost the solution y 0. 31. Following the proof of Theorem 2, we come to the expression (10) on page 63 of the text for g (y), that is
x g (y) = N(x, y)  y
x0 M(s, y) ds (2.15) (where we have replaced the integration variable t by s). In other words, g(y) is an antiderivative of the righthand side in (2.15). Since an antiderivative is defined up to an additive constant and, in Theorem 2, such a constant can be chosen arbitrarily (that is, g(y) can be any antiderivative), we choose g(y) that vanishes at y0 . According to fundamental theorem of calculus, this function can be written in the form y y N(x, t)  g(y) = g (t) dt = t
y0 y0 y y x M(s, t) ds dt x0 x M(s, t) ds dt =
y0 y N(x, t) dt 
y0 t x0 x t=y =
y0 y N(x, t) dt  x0 x M(s, t) ds
t=y0 x =
y0 N(x, t) dt 
x0 M(s, y) ds +
x0 M(s, y0 ) ds . Substituting this function into the formula (9) on page 63 of the text, we conclude that y x x x F (x, y) =
x0 y M(t, y) dt + y0 x N(x, t) dt 
x0 M(s, y) ds +
x0 M(s, y0 ) ds =
y0 N(x, t) dt +
x0 M(s, y0 ) ds . (a) In the differential form used in Example 1, M(x, y) = 2xy 2 + 1 and N(x, y) = 2x2 y. 68 Exercises 2.4
Thus, N(x, t) = 2x2 t and M(s, y0 ) = 2s 02 + 1 = 1, and (18) yields
y x y x F (x, y) =
0 2x t dt +
t=y 0 s=x 2 1 ds = x 2 0 2t dt +
0 ds = x2 t2 +s
t=0 s=0 = x2 y 2 + x. (b) Since M(x, y) = 2xy  sec2 x and N(x, y) = x2 + 2y, we have N(x, t) = x2 + 2t and M(s, y0 ) = 2s 0  sec2 s =  sec2 s,
y x F (x, y) =
0 2 x + 2t dt +
0 2 t=y t=0 2  sec2 s ds
s=x s=0 = x t+t  tan s = x2 y + y 2  tan x. (c) Here, M(x, y) = 1 + ex y + xex y and N(x, y) = xex + 2. Therefore, N(x, t) = xex + 2 and M(s, y0 ) = 1 + es 0 + ses 0 = 1,
y x F (x, y) =
0 (xe + 2) dt +
0 t=y x 1 ds
s=x = (xex + 2) t +s
t=0 s=0 = (xex + 2) y + x, which is identical to F (x, y) obtained in Example 3. 32. (a) The slope of the orthogonal curves, say m , must be 1/m, where m is the slope of the original curves. Therefore, we have m = Fy (x, y) Fx (x, y) dy Fy (x, y) = dx Fx (x, y) Fy (x, y) dx  Fx (x, y) dy = 0. (b) Let F (x, y) = x2 + y 2 . Then we have Fx (x, y) = 2x and Fy (x, y) = 2y. Plugging these expressions into the final result of part (a) gives 2y dx  2x dy = 0 y dx  x dy = 0. 69 Chapter 2
To find the orthogonal trajectories, we must solve this differential equation. To this end, note that this equation is separable and thus 1 1 dx = dy x y eln xC = eln y ln x = ln y + C y = kx, where k = eC . Therefore, the orthogonal trajectories are lines through the origin. (c) Let F (x, y) = xy. Then we have Fx (x, y) = y and Fy (x, y) = x. Plugging these expressions into the final result of part (a) gives x dx  y dy = 0. To find the orthogonal trajectories, we must solve this differential equation. To this end, note that this equation is separable and thus x dx = y dy x2 y2 = +C 2 2 x2  y 2 = k , where k := 2C. Therefore, the orthogonal trajectories are hyperbolas. 33. We use notations and results of Problem 32, that is, for a family of curves given by F (x, y) = k, the orthogonal trajectories satisfy the differential equation F (x, y) F (x, y) dx  dy = 0. y x (a) In this problem, F (x, y) = 2x2 + y 2 and the equation (2.16) becomes (2x2 + y 2) (2x2 + y 2 ) dx  dy = 0 y x Separating variables and integrating yield 2y dx = 4x dy 70 2dy dx = x y ln x = 2 ln y + c1 x = ec1 y2 = c2 y 2 dx = x 2dy y 2y dx  4x dy = 0. (2.17) (2.16) eln x = e2 ln y+c1 x = c2 y 2 = cy 2 , Exercises 2.4
where c as any nonzero constant. Separating variables, we divided the equation (2.17) by xy. As a result, we lost two constant solutions x 0 and y 0 (see the discussion on pages 4445 of Section 2.2 of the text). Thus the orthogonal trajectories for the family 2x2 + y 2 = k are x = cy 2 , c = 0, x 0, and y 0. (Note that x 0 can be obtained from x = cy 2 by taking c = 0 while y 0 cannot.) (b) First we rewrite the equation defining the family of curves in the form F (x, y) = k by dividing it by x4 . This yields yx4 = k. We use (2.17) to set up an equation for the orthogonal trajectories: F = 4yx5 , x F = x4 y x4 dx  4yx5 dy = 0 . Solving this separable equation yields x4 dx = 4yx5 dy = 0 x dx = (4y)dy x dx = 4y dy x2 = 2y 2 + c1 2 x2 + 4y 2 = c. Thus, the family of orthogonal trajectories is x2 + 4y 2 = c. (c) Taking logarithm of both sides of the equation, we obtain ln y = kx ln y = k, x and so F (x, y) = (ln y)/x, F/x = (ln y)/x2 , F/y = 1/(xy). The equation (2.17) becomes ln y 1 dx   2 dy = 0 xy x Separating variables and integrating, we obtain x dx = y ln y dy ln y 1 dx =  2 dy. xy x x dx =  y ln y dy y2 1 x2 y2 y2 y2 =  ln y + dy =  ln y + + c1 2 2 2 y 2 4 x2 y 2 y2 2x2 + 2y 2 ln y  y 2 = c, + ln y  = c1 2 2 4 y ln y dy. 71 where c := 4c1 , and we have used integration by parts to find Chapter 2
(d) We divide the equation, y 2 = kx, by x and get y 2 /x = k. Thus, F (x, y) = y 2 /x and F F y2 2y = 2, = x x y x 2 y 2y dx   2 dy = 0 x x 2x dx = y dy 2 2y dx = x  y2 x2 dy 2x2 + y 2 = c. y2 x =  + c1 2 35. Applying Leibniz's theorem, we switch the order of differentiation (with respect to y) and integration. This yields g = N(x, y) 
x0 x M(t, y) y dt. Therefore, g is differentiable (even continuously) with respect to x as a difference of two (continuously) differentiable functions, N(x, y) and an integral with variable upper bound of a continuous function My (t, y). Taking partial derivatives of both sides with respect to x and using fundamental theorem of calculus, we obtain x (g ) = N(x, y)  M(t, y) dt x x y x0 x N(x, y)  M(t, y) dt = N(x, y)  M(x, y) = 0 = x x y x y
x0 due to (5). Thus (g ) /x 0 which implies that g does not depend on x (a consequence of mean value theorem). EXERCISES 2.5: Special Integrating Factors, page 71 1. Here M(x, y) = 2y 3 + 2y 2 and N(x, y) = 3y 2 x + 2xy. Computing M = 6y 2 + 4y y and N = 3y 2 + 2y , x we conclude that this equation is not exact. Note that these functions, as well as M itself, depend on y only. Then, clearly, so does the expression (N/x  M/y)/M, and the 72 Exercises 2.5
equation has an integrating factor depending on y alone. Also, since M/y  N/x (6y 2 + 4y)  (3y 2 + 2y) 3y 2 + 2y 1 = = = , 2 x + 2xy 2 + 2y) N 3y x(3y x the equation has an integrating factor depending on x. Writing the equation in the form 3y 2 x + 2xy y(3y + 2) xy(3y + 2) dx = = 2 x = 2 dy 2y 3 + y 2 2y (y + 1) 2y (y + 1) we conclude that it is separable and linear with x as the dependent variable. 3. This equation is not separable because of the factor (y 2 + 2xy). It is not linear because of the factor y 2 . To see if it is exact, we compute My (x, y) and Nx (x, y), and see that My (x, y)2y + 2x = 2x = Nx (x, y). Therefore, the equation is not exact. To see if we can find an integrating factor of the form (x), we compute N M  2y + 4x y x = , N x2 which is not a function of x alone. To see if we can find an integrating factor of the form (y), we compute M N  4x  2y 2(2x + y) 2 x y = 2 = = . M y + 2xy y(y + 2x) y Thus the equation has an integrating factor that is a function of y alone. 5. In this problem, M(x, y) = 2y 2x  y and N(x, y) = x. Therefore, M = 4yx  1 y and N =1 x N M  = 2  4yx . x y The equation is not exact, because M/y = N/x, but it has an integrating factor depending just on y since N/x  M/y 2  4yx 2(2yx  1) 2 = 2 = = . M 2y x  y y(2yx  1) y 73 Chapter 2
dy y  2y 2x y = =  2y 2 . dx x x The righthand side cannot be factorized as p(x)q(y), and so the equation is not separable. Also, it is not linear with y as the dependent variable (because of 2y 2 term). By taking the reciprocals we also conclude that it is not linear with the dependent variable x. 7. The equation (3x2 + y) dx + (x2 y  x) dy = 0 is not separable or linear. To see if it is exact, we compute M N = 1 = 2xy  1 = . y x Thus, the equation is not exact. To see if we can find an integrating factor, we compute M/y  N/x 2  2xy 2(xy  1) 2 = 2 = = . N x yx x(xy  1) x From this we see that the integrating factor will be (x) = exp 2 dx x = exp (2 ln x) = x2 . Isolating dy/dx, we obtain To solve the equation, we multiply it by the integrating factor x2 to obtain (3 + yx2 ) dx + (y  x1 ) dy = 0. This is now exact. Thus, we want to find F (x, y). To do this, we integrate M(x, y) = 3 + yx2 with respect to x to get F (x, y) = Therefore, F (x, y) = 3x  yx1 + And so we see that an implicit solution is y2 y  + 3x = C. 2 x 74 3 + yx2 dx = 3x  yx1 + g(y) Fy (x, y) = x1 + g (y) = N(x, y) = y  x1 y2 . g (y) = y g(y) = 2 y2 . 2 Exercises 2.5
Since (x) = x2 we must check to see if the solution x 0 was either gained or lost. The function x 0 is a solution to the original equation, but is not given by the above implicit solution for any choice of C. Hence, y2 y  + 3x = C 2 x are solutions. 9. We compute partial derivatives of M(x, y) = 2y 2 + 2y + 4x2 and N(x, y) = 2xy + x. M 2y 2 + 2y + 4x2 = 4y + 2, = y y N = (2xy + x) = 2y + 1. x x and x0 Although the equation is not exact (M/y = N/x), the quotient (4y + 2)  (2y + 1) 2y + 1 1 M/y  N/x = = = N 2xy + x x(2y + 1) x depends on x only, and so the equation has an integrating factor, which can be found by applying formula (8) on page 70 of the text. Namely, (x) = exp 1 dx x = exp (ln x) = x. Note that if is an integrating factor, then  is an integrating factor as well. This observation allows us to take (x) = x. Multiplying given differential equation by x yields an exact equation 2y 2 + 2y + 4x2 x dx + x2 (2y + 1) dy = 0. Therefore, F (x, y) = x2 (2y + 1) dy = x2 y 2 + y + h(x) F = 2x y 2 + y + h (x) = 2y 2 + 2y + 4x2 x x h (x) = 4x3 h(x) = 4x3 dx = x4 F (x, y) = x2 y 2 + y + x4 = x2 y 2 + x2 y + x4 , and x2 y 2 + x2 y + x4 = c is a general solution. 75 Chapter 2
11. In this differential equation, M(x, y) = y 2 + 2xy, N(x, y) = x2 . Therefore, M = 2y + 2x, y N = 2x, x and so (N/x  M/y)/M = (4x  2y)/(y 2 + 2xy) = 2/y is a function of y. Then (y) = exp  2 y dy = exp (2 ln y) = y 2 . Multiplying the differential equation by (y) and solving the obtained exact equation, we get y 2 y 2 + 2xy dx  y 2x2 dy = 0 F (x, y) = y 2 x2 dy = y 1 x2 + h(x) F = y 1 x2 + h(x) = 2y 1x + h (x) = y 2 y 2 + 2xy = 1 + 2xy 1 x x h (x) = 1 h(x) = x F (x, y) = y 1x2 + x. Since we multiplied given equation by (y) = y 2 (in fact, divided by y 2 ) to get an exact equation, we could lose the solution y 0, and this, indeed, happened: y 0 is, clearly, a solution to the original equation. Thus a general solution is y 1x2 + x = c and y 0. 13. We will multiply the equation by the factor xn y m and try to make it exact. Thus, we have 2xn y m+2  6xn+1 y m+1 dx + 3xn+1 y m+1  4xn+2 y m dy = 0. We want My (x, y) = Nx (x, y). Since My (x, y) = 2(m + 2)xn y m+1  6(m + 1)xn+1 y m , Nx (x, y) = 3(n + 1)xn y m+1  4(n + 2)xn+1 y m , we need 2(m + 2) = 3(n + 1) 76 and 6(m + 1) = 4(n + 2). Exercises 2.5
Solving these equations simultaneously, we obtain n = 1 and m = 1. So, (x, y) = xy. With these choices for n and m we obtain the exact equation (2xy 3  6x2 y 2 ) dx + (3x2 y 2  4x3 y) dy = 0. Solving this equation, we have F (x, y) = (2xy 3  6x2 y 2 ) dx = x2 y 3  2x3 y 2 + g(y) Fy (x, y) = 3x2 y 2  4x3 y + g (y) = N(x, y) = 3x2 y 2  4x3 y. Therefore, g (y) = 0. Since the constant of integration can be incorporated into the constant C of the solution, we can pick g(y) 0. Thus, we have F (x, y) = x2 y 3  2x3 y 2 and the solution becomes x2 y 3  2x3 y 2 = C. Since we have multiplied the original equation by xy we could have added the extraneous solutions y 0 or x 0. But, since y 0 implies that dy/dx 0 or x 0 implies that dx/dy 0, y 0 and x 0 are solutions of the original equation as well as the transformed equation. 15. Assume that, for a differential equation M(x, y)dx + N(x, y)dy = 0, N/x  M/y = H(xy) xM  yN is a function of xy only. Denoting (z) = exp H(z)dz 77 the expression (2.18) (2.19) Chapter 2
and multiplying (2.18) by (xy), we get a differential equation (xy)M(x, y)dx + (xy)N(x, y)dy = 0. Let us check it for exactness. First we note that (z) = exp H(z)dz = exp H(z)dz H(z)dz = (z)H(z). (2.20) Next, using this fact, we compute partial derivatives of the coefficients in (2.20). M(x, y) (xy) {(xy)M(x, y)} = (xy) M(x, y) + (xy) y y y M(x, y) = (xy)H(xy) xM(x, y) + (xy) y M(x, y) = (xy) H(xy) xM(x, y) + , y N(x, y) (xy) {(xy)N(x, y)} = (xy) N(x, y) + (xy) x x x N(x, y) = (xy)H(xy) yN(x, y) + (xy) x N(x, y) = (xy) H(xy) yN(x, y) + . x But (2.19) implies that M N  = (xM  yN)H(xy) x y and, therefore, [(xy)M(x, y)] [(xy)N(x, y)] = . y x This means that the equation (2.20) is exact. 17. (a) Expressing the family y = x  1 + kex in the form (y  x + 1)ex = k, we have (with notation of Problem 32) F (x, y) = (y  x + 1)ex . We compute F (y  x + 1) x d(ex ) = [(y  x + 1)ex ] = e + (y  x + 1) x x x dx = ex + (y  x + 1)ex = (y  x)ex , 78 yNH(xy) + N M = xMH(xy) + , x y Exercises 2.6 (y  x + 1) x F = [(y  x + 1)ex ] = e = ex . y y y Now we can use the result of Problem 32 to derive an equation for the orthogonal trajectories (i.e., velocity potentials) of the given family of curves: F F dx  dy = 0 y x ex dx  (y  x)ex dy = 0 dx + (x  y)dy = 0. (b) In the differential equation dx + (x  y)dy = 0, M = 1 and N = x  y. Therefore, (x  y)/x  (1)/y N/x  M/y = = 1, M (1) and an integrating factor (y) is given by (y) = exp the form G(x, y) = c. ey dx + (x  y)ey dy = 0 G(x, y) = ey dx = xey + g(y) ey dy g (y) = yey = yey + ey . (1)dy = ey . Multiplying the equation from part (a) by (y) yields an exact equation, and we look for its solutions of G = xey + g (y) = (x  y)ey y g(y) = (yey )dy =  yey  Thus, the velocity potentials are given by G(x, y) = xey  yey + ey = c EXERCISES 2.6: or x = y  1 + cey . Substitutions and Transformations, page 78 1. We can write the equation in the form dy = (y  4x  1)2 = [(y  4x)  1]2 = G(y  4x), dx where G(t) = (t  1)2 . Thus, it is of the form dy/dx = G(ax + by). 3. In this equation, the variables are x and t. Its coefficients, t + x + 2 and 3t  x  6, are linear functions of x and t. Therefore, given equation is an equation with linear coefficients. 79 Chapter 2
5. The given differential equation is not homogeneous due to the e2x terms. The equation (ye2x + y 3) dx  e2x dy = 0 is a Bernoulli equation because it can be written in the form dy/dx + P (x)y = Q(x)y n as follows: dy  y = e2x y 3 . dx The differential equation does not have linear coefficients nor is it of the form y = G(ax + by). 7. Here, the variables are y and . Writing dy y 3  y 2 (y/)3  (y/)2 = = , d 22 y 2(y/) we see that the righthand side is a function of y/ alone. Hence, the equation is homogeneous. 9. First, we write the equation in the form 3x2 + y 2 y 3  3x2 y (y/x)3  3(y/x) dy = . = = dx xy  x3 y 1 xy 2  x3 (y/x)2  1 Therefore, it is homogeneous, and we we make a substitution y/x = u or y = xu. Then y = u + xu , and the equation becomes u+x Separating variables and integrating yield u3  3u 2u u2  1 2 du = 2 u=  2 du =  dx dx u 1 u 1 u x 2 1 u 1 2 dx u du =  dx du = 2 u x u x 1 2 u  ln u = 2 ln x + C1 u2  ln u2 + ln(x4 ) = C. 2 Substituting back y/x for u and simplifying, we finally get y x
2 u3  3u du = 2 . dx u 1  ln y2 x2 + ln(x4 ) = C y2 x6 y2  2 = K. x y2 + ln x2 x6 y2 = C, which can also be written as ln 80 Exercises 2.6
11. From dx xy  y 2 y 2 y = =  dy x2 x x we conclude that given equation is homogeneous. Let u = y/x. Then y = xu and y = u + xu . Substitution yields u+x du du = u  u2 = u2 x dx dx dx du 1 =  = ln x + C 2 u x u x x = ln x + C y= . y ln x + C  du dx = 2 u x Note that, solving this equation, we have performed two divisions: by x2 and u2 . In doing this, we lost two solutions, x 0 and u 0. (The latter gives y 0.) Therefore, a general solution to the given equation is y= x , ln x + C x 0, and y 0. 13. Since we can express f (t, x) in the form G(x/t), that is, (dividing numerator and denominator by t2 ) (x/t)2 + (x/t)2 x2 + t t2 + x2 = , tx (x/t) the equation is homogeneous. Substituting v = x/t and dx/dt = v + tdv/dt into the equation yields dv dv 1 + v2 1 + v2 v+t =v+ t = . dt v dt v This transformed equation is separable. Thus we have 1 v dv = dt 2 t 1+v 1 + v 2 = ln t + C, where we have integrated with the integration on the left hand side being accomplished by the substitution u = 1 + v 2 . Substituting x/t for v in this equation gives the solution to the original equation which is 1+ x2 = ln t + C. t2 81 Chapter 2
15. This equation is homogeneous because x2  y 2 1  (y/x)2 dy = = . dx 3xy 3(y/x) Thus, we substitute u = y/x (y = xu and so y = u + xu ) to get u+x 1  u2 du = dx 3u 3u du = 1  4u2 3 ln 1  4 dx x x du 1  4u2 3u du dx = = dx 3u 1  4u2 x 3  ln 1  4u2 = ln x + C1 8 y 2 = 8 ln x + C2 x 3 ln(x2 )  3 ln x2  4y 2 = 8 ln x + C2 ,
3 which, after some algebra, gives (x2  4y 2) x2 = C. 17. With the substitutions z = x + y and dz/dx = 1 + dy/dx or dy/dx = dz/dx  1 this equation becomes the separable equation dz dz 1 = z1 = z dx dx 2z 1/2 = x + C . z 1/2 dz = dx Substituting x + y for z in this solution gives the solution of the original equation 2 x+y =x+C which, on solving for y, yields y= Thus, we have y= x C + 2 2 2  x. (x + C)2  x. 4 19. The righthand side of this equation has the form G(x  y) with G(t) = (t + 5)2 . Thus we substitute t=xy 82 y =xt y = 1t, Exercises 2.6
separate variables, and integrate. 1 dt = (t + 5)2 dx dt = 1  (t + 5)2 = (1  t  5)(1 + t + 5) = (t + 4)(t + 6) dx dt dt = dx =  dx (t + 4)(t + 6) (t + 4)(t + 6) 1 t+4 1 1 = 2x + C1  dt =  dx ln 2 t+4 t+6 t+6 xy+4 xy+6 = 2x + C1 = C2 e2x ln xy+6 xy+4 2 2 = C2 e2x . 1+ y =x+4+ 2x + 1 xy+4 Ce t+40 has been lost in separation variables. 21. This is a Bernoulli equation with n = 2. So, we make a substitution u = y 1n = y 1 . We have y = u1 , y = u2 u , and the equation becomes  1 du 1 x2 + = 2 u2 dx ux u du 1  u = x2 . dx x Also, the solution y =x+4 The last equation is a linear equation with P (x) = 1/x. Following the procedure of solving linear equations, we find an integrating factor (x) = 1/x and multiply the equation by (x) to get 1 du 1 1 d  2 u = x u x dx x dx x 1 1 u = (x)dx =  x2 + C1 x 2 1 2 y= . = 3 /2 + C x x Cx  x3 1 y, divided by y 2 ) to obtain a linear equation. 83 = x 1 u =  x3 + C1 x 2 Also, y 0 is a solution which was lost when we multiplied the equation by u2 (in terms of Chapter 2
23. This is a Bernoulli equation with n = 2. Dividing it by y 2 and rewriting gives y 2 dy  2x1 y 1 = x2 . dx Making the substitution v = y 1 and hence dv/dx = y 2 dy/dx, the above equation becomes v dv + 2 = x2 . dx x This is a linear equation in v and x. The integrating factor (x) is given by (x) = exp 2 dx x = exp (2 ln x) = x2 . Multiplying the linear equation by this integrating factor and solving, we have x2 dv + 2vx = x4 dx x2 v = Dx x2 v = x4 x5 + C1 5 v= x3 C1 + 2. 5 x x4 dx = Substituting y 1 for v in this solution gives a solution to the original equation. Therefore, we find x3 C1 y = + 2 5 x Letting C = 5C1 and simplifying yields
1 y= x5 + 5C1 5x2 1 . y= 5x2 . x5 + C Note: y 0 is also a solution to the original equation. It was lost in the first step when we divided by y 2 . 25. In this Bernoulli equation, n = 3. Dividing the equation by x3 , we obtain x3 dx 1 2 + x = t. dt t Now we make a substitution u = x2 to obtain a linear equation. Since u = 2x3 x , the equation becomes  84 1 du 1 + u = t 2 dt t du 2  u = 2t dt t Exercises 2.6 (t) = exp  2 dt t = t2 t2 u = x2 2 dt = 2 ln t + C t = 2t2 ln t + Ct2 . 2 d (t2 u) = dt t u = 2t2 ln t + Ct2 x 0 is also a solution, which we lost dividing the equation by x3 . 27. This equation is a Bernoulli equation with n = 2, because it can be written in the form dr 2  r = r 2 2 . d Dividing by r 2 and making the substitution u = r 1 , we obtain a linear equation. r 2 dr 2 1 du 2   u = 2  r = 2 d d du 2 2 + u = 2 d () = exp d d (2 u) = 1 2 u =  + C d 2 r= C 29. Solving for h and k in the linear system 3h + k  1 = 0 h+k+3=0 gives h = 1 and k = 2. Thus, we make the substitutions x = u  1 and y = v  2, so that dx = du and dy = dv, to obtain (3u + v) du + (u + v) dv = 0. This is the same transformed equation that we encountered in Example 4 on page 77 of the text. There we found that its solution is v 2 + 2uv  3u2 = C. 85 = 2 u=  + C . 2 Making back substitution (and adding the lost solution r 0), we obtain a general solution and r 0. Chapter 2
Substituting x + 1 for u and y + 2 for v gives the solution to the original equation (y + 2)2 + 2(x + 1)(y + 2)  3(x + 1)2 = C. 31. In this equation with linear coefficients, we make a substitution x = u + h, y = v + k, where h and k satisfy 2h  k = 0 4h + k = 3 k = 2h 4h + 2h = 3 k = 1, h = 1/2. Thus x = u + 1/2, y = v + 1. As dx = du and dy = dv, substitution yields (2u  v)du + (4u + v)dv = 0 z= u v u = vz du 4u + v 4(u/v) + 1 = = dv 2u  v 2(u/v)  1 du dz =z+v dv dv dz 4z + 1 (2z + 1)(z + 1) v = z = dv 2z  1 2z  1 2z  1 1 dz =  dv . (2z + 1)(z + 1) v z+v 4z + 1 dz = dv 2z  1 2z  1 1 dz =  dv (2z + 1)(z + 1) v To find the integral in the lefthand side of the above equation, we use the partial fraction decomposition 4 3 2z  1 = + . (2z + 1)(z + 1) 2z + 1 z + 1 Therefore, the integration yields 2 ln 2z + 1 + 3 ln z + 1 =  ln v + C1 z + 13 v = eC1 2z + 12 3 2 u u + 1 v = C2 2 + 1 (u + v)3 = C2 (2u + v)2 v v (x  1/2 + y  1)3 = C2 (2x  1 + y  1)2 (2x + 2y  3)3 = C(2x + y  2)2 . 33. In Problem 1, we found that the given equation is of the form dy/dx = G(y  4x) with G(u) = (u  1)2 . Thus we make a substitution u = y  4x to get dy = (y  4x  1)2 dx 86 4+ du = (u  1)2 dx Exercises 2.6 du = (u  1)2  4 = (u  3)(u + 1) dx du = (u  3)(u + 1) dx . To integrate the lefthand side, we use partial fractions: 1 1 = (u  3)(u + 1) 4 Thus 1 u3 ln (ln u  3  ln u + 1) = x + C1 = 4x + C2 4 u+1 u3 Ce4x + 3 u= = Ce4x u+1 1  Ce4x Ce4x + 3 y = 4x + , 1  Ce4x 1 1  . u3 u+1 (2.21) where C = 0 is an arbitrary constant. Separating variables, we lost the constant solutions u 3 and u 1, that is, y = 4x + 3 and y = 4x  1. While y = 4x + 3 can be obtained from (2.21) by setting C = 0, the solution y = 4x  1 is not included in (2.21). Therefore, a general solution to the given equation is y = 4x + Ce4x + 3 1  Ce4x and y = 4x  1. 35. This equation has linear coefficients. Thus we make a substitution t = u + h and x = v + k with h and k satisfying h+k+2 = 0 3h  k  6 = 0 As dt = du and dx = dv, the substitution yields (u + v)dv + (3u  v)du = 0 du u+v (u/v) + 1 = = . dv 3u  v 3(u/v)  1 h = 1, k = 3. With z = u/v, we have u = vz, u = z + vz , and the equation becomes z+v z+1 dz = dv 3z  1 1 3z  1 dz =  dv 3z 2 + 1 v dz 3z 2 + 1 = dv 3z  1 3z  1 dz =  3z 2 + 1 v 1 dv v 87 Chapter 2 dz =  ln v + C1 +1 1 1 ln 3z 2 + 1  arctan z 3 =  ln v + C1 2 3 2 ln (3z 2 + 1)v 2  arctan z 3 = C2 . 3 3zdz  3z 2 + 1 3z 2 x+3 2 ln 3(t  1)2 + (x + 3)2 + arctan = C. 3 3(t  1) 37. In Problem 5, we have written the equation in the form dy  y = e2x y 3 dx y 3 dy  y 2 = e2x . dx Making back substitution, after some algebra we get Making a substitution u = y 2 (and so u = 2y 3y ) in this Bernoulli equation, we get du + 2u = 2e2x (x) = exp 2dx = e2x dx 1 d (e2x u) 2e4x dx =  e4x + C = 2e2x e2x = 2e4x e2x u = dx 2 1 2x 1 2x u =  e + Ce2x y 2 =  e + Ce2x . 2 2 The constant function y 0 is also a solution, which we lost dividing the equation by y 3 . 39. Since the equation is homogeneous, we make a substitution u = y/. Thus we get (y/)3  (y/)2 dy = d 2(y/) du u2 + u = d 2 2du = u(u + 1) u2 C = , 2 (u + 1) Back substitution u = y/ yields y2 C = 2 (y + ) 88 y 2 = C(y + )2 , C = 0. d C = 0. du u3  u2 u2  u = = d 2u 2 2du d = u(u + 1) 2 u ln =  ln  + C1 (u + 1)2 u+ Exercises 2.6
When C = 0, the above formula gives 0 or y 0, which were lost in separating variables. Also, we lost another solution, u + 1 0 or y = . Thus, the answer is y 2 = C(y + )2 where C is an arbitrary constant. 41. The righthand side of (8) from Example 2 of the text can be written as y  x  1 + (x  y + 2)1 = (x  y + 2) + 1 + (x  y + 2)1 = G(x  y + 2) with G(v) = v + v 1 + 1. With v = x  y + 2, we have y = 1  v , and the equation becomes dv = v + v 1 + 1 dx ln v 2  1 = 2x + C1 1 v2  1 v dv = dv = dx 21 dx v v v 2  1 = Ce2x , C = 0. and y = , Dividing by v 2  1, we lost constant solutions v = 1, which can be obtained by taking C = 0 in the above formula. Therefore, a general solution to the given equation is (x  y + 2)2 = Ce2x + 1, where C is an arbitrary constant. 43. (a) If f (tx, ty) = f (x, y) for any t, then, substituting t = 1/x, we obtain f (tx, ty) = f 1 1 x, y x x = f 1, y , x which shows that f (x, y) depends, in fact, on y/x alone. (b) Since M(x, y) dy = =: f (x, y) dx N(x, y) and the function f (x, y) satisfies M(tx, ty) tn M(x, y) M(x, y) f (tx, ty) =  = n = = f (x, y), N(tx, ty) t N(x, y) N(x, y) we apply (a) to conclude that the equation M(x, y)dx + N(x, y)dy = 0 is homogeneous. 89 Chapter 2
45. To obtain (17), we divide given equations: 4x + y 4 + (y/x) dy = = . dx 2x  y (y/x)  2 Therefore, the equation is homogeneous, and the substitution u = y/x yields u+x du 4+u du 4+u u2 + 3u + 4 = x = u = dx u2 dx u2 u2 u2 1 u2 du =  dx du =  u2  3u  4 x u2  3u  4 1 dx . x Using partial fractions, we get u2 and so 3 2 ln u  4 + ln u + 1 =  ln x + C1 5 5 (u  4)2 (u + 1)3 x5 = C 2 y 3 y 4 + 1 x5 = C x x REVIEW PROBLEMS: page 81 2 1 3 1 u2 = + ,  3u  4 5 u4 5 u+1 (y  4x)2 (y + x)3 = C. 1. Separation variables yields y1 dy = ex dx ey (y  1)ey dy = ex dx ex dx (y  1)ey  ey = ex + C (y  1)ey dy = (y  1)ey + ex + yey = C, ey dy = ex + C and we can replace C by K. 3. The differential equation is an exact equation with M = 2xy 3x2 and N = x2 2y 3 because My = 2x = Nx . To solve this problem we will follow the procedure for solving exact equations 90 Review Problems
given in Section 2.4. First we integrate M(x, y) with respect to x to get F (x, y) = 2xy  3x2 dx + g(y) (2.22) F (x, y) = x2 y  x3 + g(y). To determine g(y) take the partial derivative with respect to y of both sides and substitute N(x, y) for F (x, y)/y to obtain N = x2  2y 3 = x2 + g (y). Solving for g (y) yields g (y) = 2y 3 . Since the choice of the constant of integration is arbitrary we will take g(y) = y 2. Hence, from equation (2.22) we have F (x, y) = x2 y  x3 + y 2 and the solution to the differential equation is given implicitly by x2 y  x3 + y 2 = C. 5. In this problem, M(x, y) = sin(xy) + xy cos(xy), We check the equation for exactness: M = [x cos(xy)] + [x cos(xy)  xy sin(xy)x] = 2x cos(xy)  x2 y sin(xy), y N = 0 + [2x cos(xy)  x2 sin(xy)y] = 2x cos(xy)  x2 y sin(xy). x Therefore, the equation is exact. So, we use the method discussed in Section 2.4 and obtain F (x, y) = N(x, y)dy = 1 + x2 cos(xy) dy = y + x sin(xy) + h(x) N(x, y) = 1 + x2 cos(xy). F = sin(xy) + x cos(xy)y + h (x) = M(x, y) = sin(xy) + xy cos(xy) x h(x) 0, h (x) = 0 and a general solution is given implicitly by y + x sin(xy) = c. 91 Chapter 2
7. This equation is separable. Separating variables and integrating, we get t3 y 2 dt = t4 y 6 dy ln t + C1 = 1 7 y 7 dy dt = 8 t y y = (7 ln t + C)1/7 . The function t 0 is also a solution. (We lost it when divided the equation by t4 .) 9. The given differential equation can be written in the form dy 1 x + y =  y 1 . dx 3x 3 This is a Bernoulli equation with n = 1, P (x) = 1/(3x), and Q(x) = x/3. To transform this equation into a linear equation, we first multiply by y to obtain y 1 2 1 dy + y =  x. dx 3x 3 Next we make the substitution v = y 2. Since v = 2yy , the transformed equation is 1 1 1 v + v =  x, 2 3x 3 2 2 v + v =  x. 3x 3 (2.23) The above equation is linear, so we can solve it for v using the method for solving linear equations discussed in Section 2.3. Following this procedure, the integrating factor (x) is found to be (x) = exp 2 dx 3x = exp 2 ln x 3 = x2/3 . Multiplying equation (2.23) by x2/3 gives x2/3 v + 2 2 v =  x5/3 3x1/3 3 x2/3 v 2 =  x5/3 . 3 1 2 x + Cx2/3 . 4 We now integrate both sides and solve for v to find x2/3 v = 2 5/3 1 8/3 x dx = x +C 3 4 1 y 2 =  x2 + Cx2/3 4 v= Substituting v = y 2 gives the solution (x2 + 4y 2)x2/3 = 4C or, cubing both sides, (x2 + 4y 2)3 x2 = C1 , where C1 := (4C)3 is an arbitrary constant. 92 Review Problems
11. The righthand side of this equation is of the form G(t  x) with G(u) = 1 + cos2 u. Thus we make a substitution tx =u which yields 1 du = 1 + cos2 u dt sec2 u du = dt tan u = t + C du =  cos2 u dt sec2 u du =  tan(t  x) + t = C. dt x=tu x =1u, 13. This is a linear equation with P (x) = 1/x. Following the method for solving linear equations given on page 51 of the text, we find that an integrating factor (x) = 1/x, and so d[(1/x)y] 1 = x2 sin 2x = x sin 2x dx x y 1 1 = x sin 2x dx =  x cos 2x + x 2 2 2 x x y =  cos 2x + sin 2x + Cx. 2 4 1 1 cos 2x dx =  x cos 2x + sin 2x + C 2 4 15. The righthand side of the differential equation y = 2  2x  y + 3 is a function of 2x  y and so can be solved using the method for equations of the form y = G(ax + by) on page 74 of the text. By letting z = 2x  y we can transform the equation into a separable one. To solve, we differentiate z = 2x  y with respect to x to obtain dz dy =2 dx dx dy dz =2 . dx dx Substituting z = 2x  y and y = 2  z into the differential equation yields dz dz =2 z+3 = z +3. or dx dx To solve this equation we divide by z + 3, multiply by dx, and integrate to obtain 2 (z + 3)1/2 dz = dx 2(z + 3)1/2 = x + C . 93 Chapter 2
Thus we get z+3= Finally, replacing z by 2x  y yields 2x  y + 3 = Solving for y, we obtain y = 2x + 3  (x + C)2 . 4 (x + C)2 . 4 (x + C)2 . 4 17. This equation is a Bernoulli equation with n = 2. So, we divide it by y 2 and substitute u = y 1 to get  du du + 2u = 1  2u = 1 () = exp (2)d = e2 d d d e2 u e2 e2 u = e2 d = = e2 + C1 d 2 2 1 1 + Ce2 1 + C1 e2 = y= y = . 2 2 1 + Ce2 This formula, together with y 0, gives a general solution to the given equation. 19. In the differential equation M(x, y) = x2  3y 2 and N(x, y) = 2xy. The differential equation is not exact because M N = 6y = 2x = . y x However, because (M/y  N/x) /N = (8y)/(2xy) = 4/x depends only on x, we can determine (x) from equation (8) on page 70 of the text. This gives (x) = exp 4 dx x = x4 . When we multiply the differential equation by (x) = x4 we get the exact equation (x2  3x4 y 2) dx + 2x3 y dy = 0. To find F (x, y) we integrate (x2  3x4 y 2) with respect to x: F (x, y) = 94 x2  3x4 y 2 dx = x1 + x3 y 2 + g(y). Review Problems
Next we take the partial derivative of F with respect to y and substitute 2x3 y for F/y: 2x3 y = 2x3 y + g (y). Thus g (y) = 0 and since the choice of the constant of integration is not important, we will take g(y) 0. Hence, we have F (x, y) = x1 + x3 y 2 and the implicit solution to the differential equation is x1 + x3 y 2 = C. Solving for y 2 yields y 2 = x2 + Cx3 . Finally we check to see if any solutions were lost in the process. We multiplied by the integrating factor (x) = x4 so we check x 0. This is also a solution to the original equation. 21. This equation has linear coefficients. Therefore, we are looking for a substitution x = u + h and y = v + k with h and k satisfying 2h + k  1 = 0 h+k4 = 0 h = 1, k = 3. So, x = u + 1 (dx = du) and y = v + 3 (dy = dv), and the equation becomes (2u + v)du + (u + v)dv = 0 With z = v/u, we have v = z + uz , and so z+u 2z dz 2z z 2  2z + 2 dz = u = z = du 1+z du 1+z 1+z z+1 du 1+z du dz =  dz =  2 + 2z  2 2 + 2z  2 z u z u 1 z 2 + 2z  2 u2 = C2 . ln z 2 + 2z  2 =  ln u + C1 2 dv 2u  v 2  (v/u) = = . du u+v 1 + (v/u) Back substitution, z = v/u = (y  3)/(x  1), yields v 2 + 2uv  2u2 = C2 (y  3)2 + 2(x  1)(y  3)  2(x  1)2 = C2 y 2  8y  2x2  2x + 2xy = C. 95 Chapter 2
23. Given equation is homogeneous because dy xy 1  (y/x) = = . dx x+y 1 + (y/x) Therefore, substituting u = y/x, we obtain a separable equation. u+x du 1u du u2  2u + 1 = x = dx 1+u dx 1+u u+1 dx 1+u dx du =  du =  2 + 2u  1 2 + 2u  1 u x u x 1 ln u2 + 2u  1 =  ln x + C1 u2 + 2u  1 x2 = C, 2 and, substituting back u = y/x, after some algebra we get a general solution y 2 +2xyx2 = C. 25. In this differential form, M(x, y) = y(x  y  2) and N(x, y) = x(y  x + 4). Therefore, N M = x  2y  2, = y  2x + 4 y x (y  2x + 4)  (x  2y  2) 3(x  y  2) 3 N/x  M/y = = = , M y(x  y  2) y(x  y  2) y which is a function of y alone. Therefore, the equation has a special integrating factor (y). We use formula (9) on page 70 of the text to find that (y) = y 3. Multiplying the equation by (y) yields y 2(x  y  2) dx + xy 3 (y  x + 4) dy = 0 y 2x2  y 1 + 2y 2 x + g(y) F (x, y) = y 2(x  y  2) dx = 2 F = y 3 x2  y 2  4y 3 x + g (y) = N(x, y) = xy 3 (y  x + 4) y g(y) 0, g (x) = 0 and so F (x, y) = y 2x2  x y 1 + 2y 2 = C1 2 x2 y 2  2xy 1  4xy 2 = C is a general solution. In addition, y 0 is a solution that we lost when multiplied the equation by (y) = y 3 (i.e., divided by y 3 ). 96 Review Problems
27. This equation has linear coefficients. Thus we make a substitution x = u + h, y = v + k with h and k satisfying 3h  k  5 = 0 hk+1 = 0 With this substitution, (3u  v)du + (u  v)dv = 0 z= 3u  v 3  (v/u) dv = = du uv 1  (v/u) h = 3, k = 4. v , v = uz, v = z + uz u 3z dz 3z z2  3 dz = u = z = z+u du 1z du 1z z1 du z1 du z1 dz =  dz =  . z2  3 u z2  3 u We use partial fractions to find the integral in the lefthand side. Namely, z1 B A + , = 2 3 z z 3 z+ 3 Therefore, integration yields A ln z  3 + B ln z + 11/ 3 A= 1 1  , 2 2 3 B= 1 1 + . 2 2 3 3 =  ln u + C1 z+ 1+1/ 3 z 3 11/ 3 3 u2 = C =C vu 3 v  3u
2 2 v+u 3 1/ 3 1+1/ 3 v+u 3 vu 3 =C (y  4) + (x  3) 3 (y  4)  (x  3) 3 1/ 3 (y  4)2  3(x  3)2 = C. 29. Here M(x, y) = 4xy 3  9y 2 + 4xy 2 and N(x, y) = 3x2 y 2  6xy + 2x2 y. We compute N M = 12xy 2  18y + 8xy, = 6xy 2  6y + 4xy, y x (12xy 2  18y + 8xy)  (6xy 2  6y + 4xy) 2y(3xy  6 + 2x) 2 M/y  N/x = = = , 2 y 2  6xy + 2x2 y N 3x xy(3xy  6 + 2x) x 97 Chapter 2
which is a function of x alone. Thus, the equation has a special integrating factor (x) = exp Multiplying the equation by (x), we find that F (x, y) = x2 4xy 3  9y 2 + 4xy 2 dx = x4 y 3  3x3 y 2 + x4 y 2 + g(y) 2 dx x = x2 . F = 3x4 y 2  6x3 y + 2x4 y + g (y) = x2 N(x, y) = x2 3x2 y 2  6xy + 2x2 y y g(y) 0 g (y) = 0 F (x, y) = x4 y 3  3x3 y 2 + x4 y 2 = C is a general solution. 31. In this problem, M = 1, y N = 1, x and so M/y  N/x 2 = . N x Therefore, the equation has a special integrating factor (x) = exp 2 x dx = x2 . We multiply the given equation by (x) to get an exact equation. x 1 y dx + dy = 0 x2 x 1 y F (x, y) = dy = + h(x) x x y F y =  2 + h (x) = x  2 x x x y x2 + =C x 2 h (x) = x h(x) = x2 , 2 and a general solution is given by F (x, y) = and x 0. (The latter has been lost in multiplication by (x).) Substitution the initial values, y = 3 when x = 1, yields 3 12 + =C 1 2 C= 7 . 2 98 Review Problems
Hence, the answer is y x2 7 + = x 2 2 x3 7x + . 2 2 y= 33. Choosing x as the dependent variable, we transform the equation to dx + x = (t + 3). dt This equation is linear, P (t) 1. So, (t) = exp d (et x) = (t + 3)et dt et x =  (t + 3)et dt = (t + 3)et + et dt = (t + 2)et + C x = (t + 2) + Cet . dt = et and Using the initial condition, x(0) = 1, we find that 1 = x(0) = (0 + 2) + Ce0 and so x = t  2 + 3et . 35. For M(x, y) = 2y 2 + 4x2 and N(x, y) = xy, we compute M = 4y, y N = y x M/y  N/x 4y  (y) 5 = = , N xy x C = 3, which is a function of x only. Using (8) on page 70 of the text, we find an integrating factor (x) = x5 and multiply the equation by (x) to get an exact equation, x5 2y 2 + 4x2 dx  x4 y dy = 0. Hence, F (x, y) = x4 y dy =  x4 y 2 + h(x) 2 4x5 y 2 F = + h (x) = x5 M(x, y) = 2x5 y 2 + 4x3 x 2 h (x) = 4x3 h(x) = 2x2 99 Chapter 2 F (x, y) =  x4 y 2  2x2 = C. 2 We find C by substituting the initial condition, y(1) = 2:  So, the solution is  x4 y 2  2x2 = 4 2 y 2 + 4x2 = 8x4 y 2 = 8x4  4x2 = 4x2 2x2  1 y = 2x 2x2  1 , (1)4 (2)2  2(1)2 = C 2 C = 4 . where, taking the square root, we have chosen the negative sign because of the initial negative value for y. 37. In this equation with linear coefficients we make a substitution x = u + h, y = v + k with h and k such that 2h  k = 0 h+k = 3 Therefore, (2u  v)du + (u + v)dv = 0 v  2u (v/u)  2 dv = = du v+u (v/u) + 1 z = v/u, v = uz, v = z + uz z2 dz z2 + 2 dz = u = z+u du z+1 du z+1 z+1 du dz =  . z2 + 2 u Integration yields z+1 dz =  z2 + 2 100 du u z dz + z2 + 2 dz = z2 + 2 du u k = 2h h + (2h) = 3 k = 2, h = 1. Review Problems z 1 1 ln z 2 + 2 + arctan =  ln u + C1 2 2 2 z =C ln z 2 + 2 u2 + 2 arctan 2 v ln v 2 + 2u2 + 2 arctan =C u 2 y2 = C. ln (y  2)2 + 2(x  1)2 + 2 arctan (x  1) 2 The initial condition, y(0) = 2, gives C = ln 2, and so the answer is ln (y  2)2 + 2(x  1)2 + 39. Multiplying the equation by y, we get y We substitute u = y 2 and obtain 1 1 du 2  u= 2 dx x x which is linear and has an integrating factor (x) = exp Hence, d (x4 u) = 2x5 dx Substitution y(1) = 3 yields 1 32 =  + C(1)4 2 or C= 19 . 2 101 x4 u = x4 y 2 =  2x5 dx =  x4 +C 2  4 dx = x4 . x du 4 2  u= , dx x x 2 1 dy  y2 = . dx x x 2 arctan y2 = ln 2 . (x  1) 2 x4 +C 2 1 y 2 =  + Cx4 . 2 Chapter 2
Therefore, the solution to the given initial value problem is 1 19x4 y2 =  + 2 2 or y= 19x4  1 . 2 102 CHAPTER 3: Mathematical Models and Numerical Methods Involving First Order Equations
EXERCISES 3.2: Compartmental Analysis, page 98 1. Let x(t) denote the mass of salt in the tank at time t with t = 0 denoting the moment when the process started. Thus we have x(0) = 0.5 kg. We use the mathematical model described by equation (1) on page 90 of the text to find x(t). Since the solution is entering the tank with rate 8 L/min and contains 0.05 kg/L of salt, input rate = 8 (L/min) 0.05 (kg/L) = 0.4 (kg/min). We can determine the concentration of salt in the tank by dividing x(t) by the volume of the solution, which remains constant, 100 L, because the flow rate in is the same as the flow rate out. Therefore, the concentration of salt at time t is x(t)/100 kg/L and output rate = Then the equation (1) yields 2x dx = 0.4  dt 25 dx 2x + = 0.4 , dt 25 x(0) = 0.5 . x(t) 2x(t) (kg/L) 8 (L/min) = (kg/min). 100 25 This equation is linear, has integrating factor (t) = exp d e2t/25 x = 0.4e2t/25 dt e2t/25 x = 0.4 25 2 e2t/25 + C = 5e2t/25 + C (2/25)dt = e2t/25 , and so x = 5 + Ce2t/25 . Using the initial condition, we find C. 0.5 = x(0) = 5 + C C = 4.5 , 103 Chapter 3
and so the mass of salt in the tank after t minutes is x(t) = 5  4.5e2t/25 . If the concentration of salt in the tank is 0.02 kg/L, then the mass of salt is 0.02 100 = 2 kg, and, to find this moment, we solve 5  4.5e2t/25 = 2 e2t/25 = 2 3 t= 25 ln(3/2) 5.07 (min). 2 3. Let x(t) be the volume of nitric acid in the tank at time t. The tank initially held 200 L of a 0.5% nitric acid solution; therefore, x(0) = 200 0.005 = 1. Since 6 L of 20% nitric acid solution are flowing into the tank per minute, the rate at which nitric acid is entering is 6 0.2 = 1.2 L/min. Because the rate of flow out of the tank is 8 L/min and the rate of flow in is only 6 L/min, there is a net loss in the tank of 2 L of solution every minute. Thus, at any time t, the tank will be holding 200  2t liters of solution. Combining this with the fact that the volume of nitric acid in the tank at time t is x(t), we see that the concentration of nitric acid in the tank at time t is x(t)/(200  2t). Here we are assuming that the tank is kept well stirred. The rate at which nitric acid flows out of the tank is, therefore, 8 [x(t)/(200  2t)] L/min. From all of these facts, we see that input rate = 1.2 L/min, 8x(t) output rate = L/min. 200  2t We know that dx = input rate  output rate . dt Thus we must solve the differential equation 4x(t) dx = 1.2  , dt 100  t This is the linear equation dx 4 + x = 1.2 , dt 100  t 104 x(0) = 1. x(0) = 1. Exercises 3.2
An integrating factor for this equation has the form (t) = exp 4 dt 100  t = e4 ln(100t) = (100  t)4 . Multiplying the linear equation by the integrating factor yields (100  t)4 dx + 4x(100  t)5 = (1.2)(100  t)4 dt Dt (100  t)4 x = (1.2)(100  t)4 1.2 (100  t)4 x = 1.2 (100  t)4 dt = (100  t)3 + C 3 x(t) = (0.4)(100  t) + C(100  t)4 . To find the value of C, we use the initial condition x(0) = 1. Therefore, x(0) = (0.4)(100) + C(100)4 = 1 This means that at time t there is x(t) = (0.4)(100  t)  (3.9 107)(100  t)4 liters of nitric acid in the tank. When the percentage of nitric acid in the tank is 10%, the concentration of nitric acid is 0.1. Thus we want to solve the equation x(t) = 0.1 . 200  2t Therefore, we divide the solution x(t) that we found above by 2(100  t) and solve for t. That is, we solve (0.2)  (1.95 107)(100  t)3 = 0.1 107 t =  0.1 1.95
1/3 C= 39 = 3.9 107 . 4 100 + 100 19.96 (min). 5. Let x(t) denote the volume of chlorine in the pool at time t. Then in the formula rate of change = input rate  output rate 105 Chapter 3
we have 0.001% = 5 105 (gal/min), 100% x(t) (gal) output rate = 5 (gal/min) = 5 104 x(t) (gal/min), 10, 000 (gal) input rate = 5 (gal/min) and the equation for x(t) becomes dx = 5 105  5 104 x dt This is a linear equation. Solving yields x(t) = 0.1 + Ce510 Using the initial condition, x(0) = 10, 000 (gal) we find the value of C: 1 = 0.1 + Ce0.00050 C = 0.9 . 0.01% = 1 (gal), 100%
4 t dx + 5 104 x = 5 105 . dt = 0.1 + Ce0.0005t . Therefore, x(t) = 0.1 + 0.9e0.0005t and the concentration of chlorine, say, c(t), in the pool at time t is c(t) = x(t) (gal) x(t) 100% = % = 0.001 + 0.009e0.0005t %. 10, 000 (gal) 100 After 1 hour (i.e., t = 60 min), c(60) = 0.001 + 0.009e0.000560 = 0.001 + 0.009e0.03 0.0097 %. To answer the second question, we solve the equation c(t) = 0.001 + 0.009e0.0005t = 0.002 t= ln(1/9) 4394.45 (min) 73.24 (h). 0.0005 7. Let x(t) denote the mass of salt in the first tank at time t. Assuming that the initial mass is x(0) = x0 , we use the mathematical model described by equation (1) on page 90 of the text to 106 Exercises 3.2
find x(t). We can determine the concentration of salt in the first tank by dividing x(t) by the its volume, i.e., x(t)/60 kg/gal. Note that the volume of brine in this tank remains constant because the flow rate in is the same as the flow rate out. Then output rate1 = (3 gal/min) x(t) kg/gal 60 = x(t) kg/min. 20 Since the incoming liquid is pure water, we conclude that input rate1 = 0. Therefore, x(t) satisfies the initial value problem x dx = input rate1  output rate1 =  , dt 20 x(0) = x0 . This equation is linear and separable. Solving and using the initial condition to evaluate the arbitrary constant, we find x(t) = x0 et/20 . Now, let y(t) denote the mass of salt in the second tank at time t. Since initially this tank contained only pure water, we have y(0) = 0. The function y(t) can be described by the same mathematical model. We get input rate2 = output rate1 = x0 t/20 x(t) = e kg/min. 20 20 Further since the volume of the second tank also remains constant, we have output rate2 = (3 gal/min) y(t) kg/gal 60 = y(t) kg/min. 20 Therefore, y(t) satisfies the initial value problem dy x0 t/20 y(t) = input rate2  output rate2 = e ,  dt 20 20 or dy y(t) x0 t/20 + = e , dt 20 20 y(0) = 0 . 107 y(0) = 0 . Chapter 3
This is a linear equation in standard form. Using the method given on page 51 of the text we find the general solution to be y(t) = x0 t/20 te + Cet/20 . 20 The constant C can be found from the initial condition: 0 = y(0) = x0 0 e0/20 + Ce0/20 20 C = 0. Therefore, y(t) = (x0 /20) tet/20 . To investigate y(t) for maximum value we calculate t dy x0 t/20 y(t) x0 t/20  1 = e = e . dt 20 20 20 20 Thus dy t =0 1 =0 t = 20, dt 20 which is the point of global maximum (notice that dy/dt > 0 for t < 20 and dy/dt < 0 for t > 20). In other words, at this moment the water in the second tank will taste saltiest, and comparing concentrations, it will be y(20)/60 y(20) 1 = 20 e20/20 = e1 = x0 /60 x0 20 times as salty as the original brine. 9. Let p(t) be the population of splake in the lake at time t. We start counting the population in 1980. Thus, we let t = 0 correspond to the year 1980. By the Malthusian law stated on page 93 of the text, we have p(t) = p0 ekt . Since p0 = p(0) = 1000, we see that p(t) = 1000ekt . To find k we use the fact that the population of splake was 3000 in 1987. Therefore, p(7) = 3000 = 1000ek7 108 3 = ek7 k= ln 3 . 7 Exercises 3.2 < < < 0 > > > > > < < < < < a/b Figure 3A: The phase line for p = (a  bp)p. Putting this value for k into the equation for p(t) gives p(t) = 1000e(t ln 3)/7 = 1000 3t/7 . To estimate the population in 2010 we plug t = 2010  1980 = 30 into this formula to get p(30) = 1000 330/7 110, 868 splakes. 11. In this problem, the dependent variable is p, the independent variable is t, and the function f (t, p) = (abp)p. Since f (t, p) = f (p), i.e., does not depend on t, the equation is autonomous. To find equilibrium solutions, we solve f (p) = 0 (a  bp)p = 0 p1 = 0, p2 = a . b Thus, p1 (t) 0 and p2 (t) a/b are equilibrium solutions. For p1 < p < p2 , f (p) > 0, and f (p) < 0 when p > p2 . (Also, f (p) < 0 for p < p1 .) Thus the phase line for the given equation is as it is shown in Figure 3A. From this picture, we conclude that the equilibrium p = p1 is a source while p = p2 is a sink. Thus, regardless of an initial point p0 > 0, the solution to the corresponding initial value problem will approach p2 = a/b as t . 13. With year 1980 corresponding to t = 0, the data given can be written as t0 = 0, ta = 1987  1980 = 7, p0 = p(t0 ) = 1000; pa = p(ta ) = 3000; tb = 1994  1980 = 14, pb = p(tb ) = 5000. Since tb = 2ta , we can use formulas in Problem 12 to compute parameters p1 and A in the logistic model (14) on page 94 of the text. We have: p1 = (3000)(5000)  2(1000)(5000) + (1000)(3000) (3000) = 6000; (3000)2  (1000)(5000) 109 Chapter 3
A= 5000(3000  1000) ln 5 1 ln = . (6000)7 1000(5000  3000) 42000 Thus the formula (15) on page 95 of the text becomes p(t) = p0 p1 (1000)(6000) 6000 = = . Ap1 t (ln 5/42000)6000t p0 + (p1  p0 )e (1000) + (6000  1000)e 1 + 51t/7 6000 5970. 1 + 5130/7 (3.1) In the year 2010, t = 2010  1980 = 30, and the estimated population of splake is p(30) = Taking the limit in (3.1), as t , yields
t lim p(t) = lim t 6000 6000 = = 6000. 1t/7 1+5 1 + lim 51t/7
t Therefore, the predicted limiting population is 6000. 15. Counting time from the year 1970, we have the following data: t0 = 0, ta = 1975  1970 = 5, p0 = p(t0 ) = 300; pa = p(ta ) = 1200; tb = 1980  1970 = 10, pb = p(tb ) = 1500. Since tb = 2ta , we use the formulas in Problem 12 to find parameters in the logistic model. p1 = (1200)(1500)  2(300)(1500) + (300)(1200) 16800 ; (1200) = 2  (300)(1500) (1200) 11 (1500)(1200  300) 1 11 ln(15) ln . A= = (16800/11)5 (300)(1500  1200) 84000 p(t) = Therefore, 300(16800/11) 16800 = .  ln(15)t/5 300 + [(16800/11)  300]e 11 + 3 151t/5 In the year 2010, t = 2010  1970 = 40, and so the estimated population of alligators is p(40) = 16800 16800 = 1527. 140/5 11 + 3 15 11 + 3 157 Taking the limit of p(t), as t , we get the predicted limiting population of 16800 16800 1527. = 1t/5 t 11 + 3 15 11 lim 110 Exercises 3.2
16. By definition, p(t + h)  p(t) . h0 h p (t) = lim Replacing h by h in the above equation, we obtain p (t) = lim p(t  h)  p(t) p(t)  p(t  h) = lim . h0 h0 h h Adding the previous two equations together yields 2p (t) = lim p(t + h)  p(t) p(t)  p(t  h) + h0 h h p(t + h)  p(t  h) = lim . h0 h p(t + h)  p(t  h) . 2h Thus p (t) = lim
h0 19. This problem can be regarded as a compartmental analysis problem for the population of fish. If we let m(t) denote the mass in million tons of a certain species of fish, then the mathematical model for this process is given by dm = increase rate  decrease rate. dt The increase rate of fish is given by 2m million tons/yr. The decrease rate of fish is given as 15 million tons/yr. Substituting these rates into the above equation we obtain dm = 2m  15, dt m(0) = 7 (million tons). This equation is linear and separable. Using the initial condition, m(0) = 7 to evaluate the arbitrary constant we obtain 1 15 m(t) =  e2t + . 2 2 Knowing this equation we can now find when all the fish will be gone. To determine when all the fish will be gone we set m(t) = 0 and solve for t. This gives 1 15 0 =  e2t + 2 2 111 Chapter 3
and, hence, t= 1 ln(15) 1.354 (years). 2 To determine the fishing rate required to keep the fish mass constant we solve the general problem dm = 2m  r, dt with r as the fishing rate. Thus we obtain m(t) = Ke2t + r . 2 m(0) = 7, The initial mass was given to be 7 million tons. Substituting this into the above equation we can find the arbitrary constant K: m(0) = 7 = K + Thus m(t) is given by m(t) = 7  r 2t r e + . 2 2 r 2 K =7 r . 2 A fishing rate of r = 14 million tons/year will give a constant mass of fish by canceling out the coefficient of the e2t term. 21. Let D = D(t), S(t), and V (t) denote the diameter, surface area, and volume of the snowball at time t, respectively. From geometry, we know that V = D 3 /6 and S = D 2 . Since we are given that V (t) is proportional to S(t), the equation describing the melting process is d 3 dV = kS D = k D 2 dt dt 6 dD 2 dD D = kD 2 = 2k = const. 2 dt dt Solving, we get D = 2kt + C. Initially, D(0) = 4, and we also know that D(30) = 3. These data allow us to find k and C. 4 = D(0) = 2k 0 + C C = 4; 2k =  1 . 30 3 = D(30) = 2k 30 + C = 2k 30 + 4 112 Exercises 3.2
Thus D(t) =  t +4. 30 The diameter D(t) of the snowball will be 2 inches when  t +4=2 30 t = 60 (min) = 1 (h), and the snowball will disappear when  t +4=0 30 t = 120 (min) = 2 (h). 23. If m(t) (with t measured in "days") denotes the mass of a radioactive substance, the law of decay says that dm = km(t) , dt with the decay constant k depending on the substance. Solving this equation yields m(t) = Cekt . If the initial mass of the substance is m(0) = m0 , then, similarly to the equation (11) on page 93 of the text, we find that m(t) = m0 ekt . In this problem, m0 = 50 g, and we know that m(3) = 10 g. These data yield 10 = m(3) = 50 ek(3) and so the decay is governed by the equation m(t) = 50e(ln 5)t/3 = (50)5t/3 . After 4 days, the remaining amount will be m(4) = (50)54/3 g, which is (50)54/3 100% = 54/3 100% 11.7% 50 of the original amount. 113 k= ln 5 , 3 (3.2) Chapter 3
25. Let M(t) denote the mass of carbon14 present in the burnt wood of the campfire. Then since carbon14 decays at a rate proportional to its mass, we have dM = M, dt where is the proportionality constant. This equation is linear and separable. Using the initial condition, M(0) = M0 we obtain M(t) = M0 et . Given the halflife of carbon14 to be 5600 years, we solve for since we have 1 M0 = M0 e(5600) 2 which yields = Thus, M(t) = M0 e0.000123776t . Now we are told that after t years 2% of the original amount of carbon14 remains in the campfire and we are asked to determine t. Thus 0.02 = e0.000123776t 0.02M0 = M0 e0.000123776t ln 0.02 31, 606 (years). t= 0.000123776 27. The element Hh decays according to the general law of a radioactive decay, which is described by (3.2) (this time, with t measured in "years"). Since the initial mass of Hh is m0 = 1 kg and the decay constant k = kHh = 2/yr, we get Hh(t) = ekHh t = e2t . (3.3) ln(0.5) 0.000123776 . 5600 1 = e(5600) , 2 For It, the process is more complicated: it has an incoming mass from the decay of Hh and, at the same, looses its mass decaying to Bu. (This process is very similar to "brine solution" 114 Exercises 3.2
problems.) Thus we use the general idea in getting a differential equation describing this process: rate of change = input rate  output rate. (3.4) The "input rate" is the rate of mass coming from Hh's decay, which is opposite to the rate of decay of Hh (Hh looses the mass but It gains it), i.e., input rate =  dHh = 2e2t , dt (3.5) where we have used (3.3). The "output rate" is the rate with which It decays, which (again, according to the general law of a radioactive decay) is proportional to its current mass. Since the decay constant for It is k = kIt = 1/yr, output rate = kIt It(t) = It(t). Therefore, combining (3.4)(3.6) we get the equation for It, that is, dIt(t) = 2e2t  It(t) dt dIt(t) + It(t) = 2e2t . dt (1)dt = et . (3.6) This is a linear equation with P (t) 1 and an integrating factor (t) = exp Multiplying the equation by (t) yields d [et It(t)] = 2et dt et It(t) = 2et + C It(t) = 2e2t + Cet . Initially, there were no It, which means that It(0) = 0. With this initial condition we find that 0 = It(0) = 2e2(0) + Ce(0) = 2 + C and the mass of It remaining after t years is It(t) = 2e2t + 2et = 2 et  e2t . (3.7) C = 2, The element Bu only gains its mass from It, and the rate with which it does this is opposite to the rate with which It looses its mass. Hence (3.6) yields dBu(t) = It(t) = 2 et  e2t . dt 115 Chapter 3
Integrating, we obtain Bu(t) = 2 et  e2t dt = 2et + e2t + C, and the initial condition Bu(0) = 0 gives C = 1. Therefore, Bu(t) = 2et + e2t + 1. EXERCISES 3.3: Heating and Cooling of Buildings, page 107 1. Let T (t) denote the temperature of coffee at time t (in minutes). According to the Newton's Law (1) on page 102 of the text, dT = K[21  T (t)], dt where we have taken H(t) U(t) 0, M(t) 21 C, with the initial condition T (0) = 95 C. Solving this initial value problem yields dT = K dt  ln T  21 = Kt + C1 21  T 95 = T (0) = 21 + CeK(0) C = 74 T (t) = 21 + CeKt ; T (t) = 21 + 74eKt . To find K, we use the fact that after 5 min the temperature of coffee was 80 C. Thus 80 = T (5) = 21 + 74eK(5) and so T (t) = 21 + 74e ln(74/59)t/5 = 21 + 74 K= ln(74/59) , 5
t/5 74 . 59 Finally, we solve the equation T (t) = 50 to find the time appropriate for drinking coffee: 74 50 = 21 + 74 59
t/5 74 59 t/5 = 29 74 t= 5 ln(74/29) 20.7 (min). ln(74/59) 3. This problem is similar to one of cooling a building. In this problem we have no additional heating or cooling so we can say that the rate of change of the wine's temperature, T (t) is given by Newton's law of cooling dT = K[M(t)  T (t)], dt 116 Exercises 3.3
where M(t) = 32 is the temperature of ice. This equation is linear and is rewritten in the standard form as dT + KT (t) = 32K. dt We find that the integrating factor is eKt . Multiplying both sides by eKt and integrating gives eKt dT + eKt KT (t) = 32KeKt dt eKt T (t) = 32eKt + C eKt T (t) = 32KeKt dt T (t) = 32 + CeKt . By setting t = 0 and using the initial temperature 70 F, we find the constant C. 70 = 32 + C C = 38 . Knowing that it takes 15 minutes for the wine to chill to 60 F, we can find the constant, K: 60 = 32 + 38eK(15) . Solving for K yields K= Therefore, T (t) = 32 + 38e0.02035t . We can now determine how long it will take for the wine to reach 56 F. Using our equation for temperature T (t), we set 56 = 32 + 38e0.02035t and, solving for t, obtain t= 1 ln 0.02035 56  32 38 22.6 min. 1 ln 15 60  32 38 0.02035 . 5. This problem can be treated as one similar to that of a cooling building. If we assume the air surrounding the body has not changed since the death, we can say that the rate of change of the body's temperature, T (t) is given by Newton's law of cooling: dT = K[M(t)  T (t)], dt 117 Chapter 3
where M(t) represents the surrounding temperature which we've assumed to be a constant 16 C. This differential equation is linear and is solved using an integrating factor of eKt . Rewriting the above equation in standard form, multiplying both sides by eKt and integrating gives dT + KT (t) = K(16) dt eKt T (t) = 16eKt + C dT + eKt KT (t) = 16KeKt dt T (t) = 16 + CeKt . eKt Let us take t = 0 as the time at which the person died. Then T (0) = 37 C (normal body temperature) and we get 37 = 16 + C C = 21 . Now we know that at sometime, say X hours after death, the body temperature was measured to be 34.5 C and that at X + 1 hours after death the body temperature was measured to be 33.7 C. Therefore, we have 34.5 = 16 + 21eKX and 33.7 = 16 + 21eK(X+1) . Solving the first equation for KX we arrive at KX =  ln 34.5  16 21 = 0.12675 . (3.8) Substituting this value into the second equation we, can solve for K as follows: 33.7 = 16 + 21e0.12675K K =  0.12675 + ln 33.7  16 21 = 0.04421 . This results in an equation for the body temperature of T (t) = 16 + 21e0.04421t . From equation (3.8) we now find the number of hours X before 12 Noon when the person died. 0.12675 0.12675 = 2.867 (hours). K 0.04421 Therefore, the time of death is 2.867 hours (2 hours and 52 min) before Noon or 9 : 08 a.m. X= 118 Exercises 3.3
7. The temperature function T (t) changes according to Newton's law of cooling (1) on page 102 of the text. Similarly to Example 1 we conclude that, with H(t) U(t) 0 and the outside temperature M(t) 35 C, a general solution formula (4) on page 102 becomes T (t) = 35 + CeKt . To find C, we use the initial condition, T (0) = T (at noon) = 24C , and get 24 = T (0) = 35 + CeK(0) C = 24  35 = 11 T (t) = 35  11eKt . The time constant for the building 1/K = 4 hr; so K = 1/4 and T (t) = 35  11et/4 . At 2 : 00 p.m. t = 2, and t = 6 at 6 : 00 p.m. Substituting this values into the solution, we obtain that the temperature at 2 : 00 p.m. will be at 6 : 00 p.m. will be Finally, we solve the equation T (t) = 35  11et/4 = 27 to find the time when the temperature inside the building reaches 27 C. 35  11et/4 = 27 11et/4 = 8 t = 4 ln 11 8 1.27 . T (2) = 35  11e2/4 28.3 C; T (6) = 35  11e6/4 32.5 C. Thus, the temperature inside the building will be 27 C at 1.27 hr after noon, that is, at 1 : 16 : 12 p.m. 9. Since we are evaluating the temperature in a warehouse, we can assume that any heat generated by people or equipment in the warehouse will be negligible. Therefore, we have H(t) = 0. 119 Chapter 3
Also, we are assuming that there is no heating or air conditioning in the warehouse. Therefore, we have that U(t). We are also given that the outside temperature has a sinusoidal fluctuation. Thus, as in Example 2, page 103, we see that M(t) = M0  B cos t , where M0 is the average outside temperature, B is a positive constant for the magnitude of the temperature shift from this average, and = /2 radians per hour. To find M0 and B, we are given that at 2 : 00 a.m., M(t) reaches a low of 16 C and at 2 : 00 p.m. it reaches a high of 32 C. This gives M0 = 16 + 32 = 24 C. 2 By letting t = 0 at 2 : 00 a.m. (so that low for the outside temperature corresponds to the low for the negative cosine function), we can calculate the constant B. That is 16 = 24  B cos 0 = 24  B Therefore, we see that M(t) = 24  8 cos t, where = /12. As in Example 2, using the fact that B0 = M0 + H0 /K = M0 + 0/K = M0 , we see that T (t) = 24  8F (t) + CeKt , where F (t) = cos t + (/K) sin t = 1+ 1 + (/K)2 K
2 1/2 B = 8. cos(t  ). In the last expression, is chosen such that tan = /K. By assuming that the exponential term dies off, we obtain T (t) = 24  8 1 + K
2 1/2 cos(t  ). This function will reach a minimum when cos(t  ) = 1 and it will reach a maximum when cos(t  ) = 1. 120 Exercises 3.3
For the case when the time constant for the building is 1, we see that 1/K = 1 which implies that K = 1. Therefore, the temperature will reach a maximum of K T = 24 + 8 1 + 12 It will reach a minimum of T = 24  8 1 + 12
2 1/2 2 1/2 31.7 C. 16.3 C. For the case when the time constant of the building is 5, we have 1 =5 K K= 1 . 5 Then, the temperature will reach a maximum of T = 24 + 8 1 + and a minimum of T = 24  8 1 + 5 12 5 12
2 1/2 28.9C, 2 1/2 19.1 C. 11. As in Example 3, page 105 of the text, this problem involves a thermostat to regulate the temperature in the van. Hence, we have U(t) = KU [TD  T (t)] , where TD is the desired temperature 16 C and KU is a proportionality constant. We will assume that H(t) = 0 and that the outside temperature M(t) is a constant 35 C. The time constant for the van is 1/K = 2 hr, hence K = 0.5. Since the time constant for the van with its air conditioning system is 1/K1 = 1/3 hr, then K1 = K + KU = 3. Therefore, KU = 3  0.5 = 2.5. The temperature in the van is governed by the equation dT = (0.5)(35  T ) + (2.5)(16  T ) = 57.5  3T. dt 121 Chapter 3
Solving this separable equation yields T (t) = 19.17 + Ce3t . When t = 0 we are given T (0) = 55. Using this information to solve for C gives C = 35.83. Hence, the van temperature is given by T (t) = 19.17 + 35.83e3t . To find out when the temperature in the van will reach 27 C, we let T (t) = 27 and solve for t. Thus, we see that 27 = 19.17 + 35.83e3t t e3t = 7.83 0.2185 35.83 30.4 min. ln(0.2185) 0.5070 (hr) 3 or 13. Since the time constant is 64, we have K = 1/64. The temperature in the tank increases at the rate of 2 F for every 1000 Btu. Furthermore, every hour of sunlight provides an input of 2000 Btu to the tank. Thus, H(t) = 2 2 = 4 F per hr. We are given that T (0) = 110, and that the temperature M(t) outside the tank is a constant 80 F. Hence the temperature in the tank is governed by dT 1 1 = [80  T (t)] + 4 =  T (t) + 5.25 , dt 64 64 Solving this separable equation gives T (t) = 336 + Cet/64 . To find C, we use the initial condition to see that T (0) = 110 = 336 + C This yields the equation T (t) = 336  226et/64 . 122 C = 226. T (0) = 110. Exercises 3.4
After 12 hours of sunlight, the temperature will be T (12) = 336  226e12/64 148.6 F. 15. The equation dT /dt = k (M 4  T 4 ) is separable. Separation variables yields T4 dT = k dt  M4 T4 dT =  M4 k dt = kt + C1 . (3.9) Since T 4  M 4 = (T 2  M 2 ) (T 2 + M 2 ), we have 1 1 1 1 (M 2 + T 2 ) + (M 2  T 2 ) 1 = =  2 , 4  M4 2 (T 2  M 2 ) (T 2 + M 2 ) 2 T 2  M2 T 2M 2M T + M2 and the integral in the lefthand side of (3.9) becomes T4 1 dT = 4 M 2M 2 T2 dT   M2 T2 dT 1 T M  2 arctan = ln 2 3 +M 4M T +M T M . Thus a general solution to Stefan's equation is given implicitly by T M 1 ln  2 arctan 3 4M T +M or T  M = C(T + M) exp 2 arctan When T is close to M, M 4  T 4 = (M  T )(M + T ) M 2 + T 2 (M  T )(2M) 2M 2 4M 3 (M  T ), and so dT k 4M 3 (M  T )4M 3 = k1 (M  T ) dt with k1 = 4M 3 k, which constitutes Newton's law. EXERCISES 3.4: Newtonian Mechanics, page 115 T M  4M 3 kt . T M = kt + C1 1. This problem is a particular case of Example 1 on page 110 of the text. Therefore, we can use the general formula (6) on page 111 with m = 5, b = 50, and v0 = v(0) = 0. But let us follow the general idea of Section 3.4, find an equation of the motion, and solve it. 123 Chapter 3
With given data, the force due to gravity is F1 = mg = 5g and the air resistance force is F2 = 50v. Therefore, the velocity v(t) satisfies m dv = F1 + F2 = 5g  50v dt dv = g  10v, dt v(0) = 0. Separating variables yields dv = dt 10v  g v(t) = 1 ln 10v  g = t + C1 10 g + Ce10t . 10 Substituting the initial condition, v(0) = 0, we get C = g/10, and so v(t) = Integrating this equation yields x(t) = v(t) dt = g g 1  e10t dt = 10 10 t+ 1 10t e 10 + C, g 1  e10t . 10 and we find C using the initial condition x(0) = 0: g g 1 C= 0 + e10(0) + C 10 10 100 g g x(t) = t+ e10t  1 = (0.981)t + (0.0981)e10t  0.0981 (m). 10 100 0= When the object hits the ground, x(t) = 1000 m. Thus we solve (0.981)t + (0.0981)e10t  0.0981 = 1000, which gives (t is nonnegative!) t 1019.468 1019 sec. 3. For this problem, m = 500 kg, v0 = 0, g = 9.81 m/sec2 , and b = 50 kg/sec. We also see that the object has 1000 m to fall before it hits the ground. Plugging these variables into equation (6) on page 111 of the text gives the equation x(t) = 124 500 (500)(9.81) t+ 50 50 0 (500)(9.81) 50 1  e50t/500 Exercises 3.4 x(t) = 98.1t + 981et/10  981. To find out when the object will hit the ground, we solve x(t) = 1000 for t. Therefore, we have 1000 = 98.1t + 981et/10  981 98.1t + 981et/10 = 1981. In this equation, if we ignore the term 981et/10 we will find that t 20.2. But this means that we have ignored the term similar to 981e2 132.8 which we see is to large to ignore. Therefore, we must try to approximate t. We will use Newton's method on the equation f (t) = 98.1t + 981et/10  1981 = 0. (If we can find a root to this equation, we will have found the t we want.) Newton's method generates a sequence of approximations given by the formula tn+1 = tn  f (tn ) . f (tn ) Since f (t) = 98.1  98.1et/10 = 98.1 1  et/10 , the recursive equation above becomes tn+1 = tn  tn + 10etn /10  (1981/98.1) . 1  etn /10 (3.10) To start the process, let t0 = 1981/98.1 20.19368, which was the approximation we obtained when we neglected the exponential term. Then, by equation (3.10) above we have t1 = 20.19368  20.19368 + 10e2.019368  20.19368 1  e2.019368 t1 18.663121 . To find t2 we plug this value for t1 into equation (3.10). This gives t2 18.643753. Continuing this process, we find that t3 18.643749. Since t2 and t3 agree to four decimal places, an approximation for the time it takes the object to strike the ground is t 18.6437 sec. 5. We proceed similarly to the solution of Problem 1 to get F1 = 5g, F2 = 10g 125 Chapter 3 5 dv = F1 + F2 = 5g  10v dt dv = g  2v, v(0) = 50. dt Solving this iniial value problem yields g + Ce2t ; 2 g 50 = v(0) = + Ce2(0) 2 g 100  g 2t v(t) = + e . 2 2 v(t) = C= 100  g 2 We now integrate v(t) to obtain the equation of the motion of the object: x(t) = v(t) dt = 100  g 2t g 100  g 2t g + e e + C, dt = t  2 2 2 4 where C is such that x(0) = 0. Computing 0 = x(0) = g 100  g 2(0) +C (0)  e 2 4 C= 100  g , 4 we answer the first question in this problem, that is, x(t) = 100  g 2t 100  g g t e + 4.905t + 22.5475  22.5475 e2t. 2 4 4 Answering the second question, we solve the equation x(t) = 500 to find time t when the object passes 500 m, and so strikes the ground. 4.905t + 22.5475  22.5475 e2t = 500 t 97.34 (sec). 7. Since the air resistance force has different coefficients of proportionality for closed and for opened chute, we need two differential equations describing the motion. Let x1 (t), x1 (0) = 0, denote the distance the parachutist has fallen in t seconds, and let v1 (t) = dx/dt denote her velocity. With m = 75, b = b1 = 30 Nsec/m, and v0 = 0 the initial value problem (4) on page 111 of the text becomes 75 126 dv1 = 75g  30v1 dt dv1 2 + v1 = g, dt 5 v1 (0) = 0. Exercises 3.4
This is a linear equation. Solving yields d e2t/5 v1 = e2t/5 g dt 0 = v1 (0) = v1 (t) = 5g + C1 e2t/5 ; 2 5g C1 =  2 5g 5g + C1 e0 = + C1 2 2 5g 1  e2t/5 v1 (t) = 2
t x1 (t) =
0 v1 (s)ds = 5g 2 s+ 5 2s/5 e 2 s=t =
s=0 5g 2 t+ 5 2t/5 5  e 2 2 . To find the time t when the chute opens, we solve 20 = v1 (t ) 20 = 5g 1  e2t /5 2 t =  5 8 ln 1  2 g 4.225 (sec). By this time the parachutist has fallen x1 (t ) = 5g 2 t + 5 2t /5 5 e  2 2 5g 2 4.225 + 5 24.225/5 5 e  2 2 53.62 (m), and so she is 2000  53.62 = 1946.38 m above the ground. Setting the second equation, we for convenience reset the time t. Denoting by x2 (t) the distance passed by the parachutist from the moment when the chute opens, and by v2 (t) := x2 (t) her velocity, we have 75 Solving, we get 5g + C2 e6t/5 ; 6 5g 20 = v2 (0) = + C2 6 5g 5g v2 (t) = + 20  6 6 v2 (t) =
t dv2 = 75g  90v2 , dt v2 (0) = v1 (t ) = 20, x2 (0) = 0. C2 = 20  e6t/5 20  5g 6 x2 (t) =
0 v2 (s)ds = 5 5g t+ 6 6 5g 5 s 6 6 5g 6 5g 6 s=t e6s/5
s=0 = 20  1  e6t/5 . 127 Chapter 3
With the chute open, the parachutist falls 1946.38 m. It takes t seconds, where t satisfies x2 (t ) = 1946.38. Solving yields 5g 5 t + 6 6 20  5g 6 1  e6t /5 = 1946.38 t 236.884 (sec). Therefore, the parachutist will hit the ground after t + t 241.1 seconds. 9. This problem is similar to Example 1 on page 110 of the text with the addition of a buoyancy force of magnitude (1/40)mg. If we let x(t) be the distance below the water at time t and v(t) the velocity, then the total force acting on the object is F = mg  bv  1 mg. 40 We are given m = 100 kg, g = 9.81 m/sec2 , and b = 10 kg/sec. Applying Newton's Second Law gives 100 dv 10 = (100)(9.81)  10v  (9.81) dt 4 dv = 9.56  (0.1)v . dt Solving this equation by separation of variables, we have v(t) = 95.65 + Cet/10 . Since v(0) = 0, we find C = 95.65 and, hence, v(t) = 95.65  95.65et/10 . Integrating yields x(t) = 95.65t  956.5et/10 + C1 . Using the fact that x(0) = 0, we find C1 = 956.5. Therefore, the equation of motion of the object is x(t) = 95.65t  956.5et/10  956.5 . To determine when the object is traveling at the velocity of 70 m/sec, we solve v(t) = 70. That is, 70 = 95.65  95.65et/10 = 95.65 1  et/10 70 t = 10 ln 1  13.2 sec. 95.65 128 Exercises 3.4
11. Let v(t) = V [x(t)]. Then, using the chain rule, we get dv dV dx dV = = V dt dx dt dx and so, for V (x), the initial value problem (4) on page 111 of the text becomes m dV V = mg  bV, dx V (0) = V [x(0)] = v(0) = v0 . This differential equation is separable. Solving yields m g V dV = dx  1 dV = dx g  (b/m)V b g  (b/m)V m g  1 dV = dx b g  (b/m)V m mg  ln g  (b/m)V   V = x + C b b b2 x + C1 . mg ln mg  bV  + bV =  m Substituting the initial condition, V (0) = v0 , we find that C1 = mg ln mg  bv0  + bv0 and hence mg ln mg  bV  + bV =  b2 x + mg ln mg  bv0  + bv0 m
2 x/m ebV mg  bV mg = ebv0 mg  bv0 mg eb . 13. There are two forces acting on the shell: a constant force due to the downward pull of gravity and a force due to air resistance that acts in opposition to the motion of the shell. All of the motion occurs along a vertical axis. On this axis, we choose the origin to be the point where the shell was shot from and let x(t) denote the position upward of the shell at time t. The forces acting on the object can be expressed in terms of this axis. The force due to gravity is F1 = mg, where g is the acceleration due to gravity near Earth. Note we have a minus force because our coordinate system was chosen with up as positive and gravity acts in a downward direction. The force due to air resistance is F2 = (0.1)v 2 . 129 Chapter 3
The negative sign is present because air resistance acts in opposition to the motion of the object. Therefore the net force acting on the shell is F = F1 + F2 = mg  (0.1)v 2. We now apply Newton's second law to obtain m dv =  mg + (0.1)v 2 . dt Because the initial velocity of the shell is 500 m/sec, a model for the velocity of the rising shell is expressed as the initialvalue problem m dv =  mg + (0.1)v 2 , dt v(t = 0) = 500, (3.11) where g = 9.81. Separating variables, we get dv dt = 10mg + v 2 10m and so dv = 10mg + v 2 dt 10m 1 tan1 10mg v 10mg = t + C. 10m Setting m = 3, g = 9.81 and v = 500 when t = 0, we find C= 1 10(3)(9.81) tan1 500 10(3)(9.81) 0.08956 . Thus the equation of velocity v as a function of time t is 1 tan1 10mg v 10mg = t + 0.08956 . 10m From physics we know that when the shell reaches its maximum height the shell's velocity will be zero; therefore tmax will be tmax = 10(3) 1 tan1 10(3)(9.81) 0 10(3)(9.81)  0.08956 = (30)(0.08956) 2.69 (seconds). 130 Exercises 3.4
Using equation (3.11) and noting that dv/dt = (dv/dx)(dx/dt) = (dv/dx)v, we can determine the maximum height attained by the shell. With the above substitution, equation (3.11) becomes dv =  mg + 0.1v 2 , dx Using separation of variables and integration, we get mv v dv dx = 10mg + v 2 10m v(0) = 500. 1 x ln 10mg + v 2 =  +C 2 10m 10mg + v 2 = Kex/(5m) . Setting v = 500 when x = 0, we find K = e0 10(3)(9.81) + (500)2 = 250294.3 . Thus the equation of velocity as a function of distance is v 2 + 10mg = (250294.3)ex/(5m) . The maximum height will occur when the shell's velocity is zero, therefore xmax is xmax = 5(3) ln 0 + 10(3)(9.81) 250294.3 101.19 (meters). 15. The total torque exerted on the flywheel is the sum of the torque exerted by the motor and the retarding torque due to friction. Thus, by Newton's second law for rotation, we have I d = T  k dt with (0) = 0 , where I is the moment of inertia of the flywheel, (t) is the angular velocity, d/dt is the angular acceleration, T is the constant torque exerted by the motor, and k is a positive constant of proportionality for the torque due to friction. Solving this separable equation gives T + Cekt/I . k Using the initial condition (0) = 0 we find C = (0  T /k). Hence, (t) = (t) = T T + 0  k k ekt/I . 131 Chapter 3
17. Since the motor is turned off, its torque is T = 0, and the only torque acting on the flywheel is the retarding one, 5 . Then Newton's second law for rotational motion becomes I d = 5 dt with (0) = 0 = 225 (rad/sec) and I = 50 kg/m2 . The general solution to this separable equation is (t) =  Using the initial condition, we find (0) = 0.05 0 + C Thus 1 15  (t) = 20 15  (t) . 0.05 when the flywheel stops rotating we have (tstop ) = 0 and so t= tstop = 20(15  0) = 300 (sec). C= (0) = 225 = 15. 5 t + C = 0.05t + C. 2I At the moment t = tstop 19. There are three forces acting on the object: F1 , the force due to gravity, F2 , the air resistance force, and F3 , the friction force. Using Figure 3.11 (with 30 replaced by 45 ), we obtain F1 = mg sin 45 = mg 2/2 , F3 = N = mg cos 45 = mg 2/2 , and so the equation describing the motion is mg 2 mg 2 dv =   3v m dt 2 2 with the initial condition v(0) = 0. Solving yields v(t) = 9.5g 2 + Cet/20 ; 0 = v(0) = 9.5g 2 + C 132 C = 9.5g 2 F2 = 3v, dv v = 0.475g 2  dt 20 Exercises 3.4 v(t) = 9.5g 2 1  et/20 . Since x(0) = 0, integrating the above equation, we obtain
t t x(t) =
0 v(s)ds =
0 9.5g 2 1  es/20 ds = 9.5g 2 s + 20es/20 s=t s=0 = 9.5g 2 t + 20et/20  20 131.8t + 2636et/20  2636. The object reaches the end of the inclined plane when x(t) = 131.8t + 2636et/20  2636 = 10 t 1.768 (sec). 21. In this problem there are two forces acting on a sailboat: A constant horizontal force due to the wind and a force due to the water resistance that acts in opposition to the motion of the sailboat. All of the motion occurs along a horizontal axis. On this axis, we choose the origin to be the point where the hard blowing wind begins and x(t) denotes the distance the sailboat travels in time t. The forces on the sailboat can be expressed in terms of this axis. The force due to the wind is F1 = 600 N. The force due to water resistance is F2 = 100v N. Applying Newton's second law we obtain m dv = 600  100v. dt Since the initial velocity of the sailboat is 1 m/sec, a model for the velocity of the moving sailboat is expressed as the initialvalue problem m dv = 600  100v, dt v(0) = 1 . Using separation of variables, we get, with m = 50 kg, dv = 2dt 6v 6 ln(6  v) = 2t + C. 133 Chapter 3
Therefore, the velocity is given by v(t) = 6  Ke2t . Setting v = 1 when t = 0, we find that 1=6K K = 5. Thus the equation for velocity v(t) is v(t) = 6  5e2t . The limiting velocity of the sailboat under these conditions is found by letting time approach infinity: lim v(t) = lim 6  5e2t = 6 (m/sec).
t t To determine the equation of motion we will use the equation of velocity obtained previously and substitute dx/dt for v(t) to obtain dx = 6  5e2t , dt Integrating this equation we obtain x(t) = 6t + Setting x = 0 when t = 0, we find 0=0+ 5 + C1 2 5 C1 =  . 2 5 2t e + C1 . 2 x(0) = 0. Thus the equation of motion for the sailboat is given by x(t) = 6t + 5 2t 5 e  . 2 2 23. In this problem, there are two forces acting on a boat: the wind force F1 and the water resistance force F2 . Since the proportionality constant in the water resistance force is different for the velocities below and above of a certain limit (5 m/sec for the boat A and 6 m/sec for the boat B), for each boat we have two differential equations. (Compare with Problem 7.) Let x1 (t) denote the distance passed by the boat A for the time t, v1 (t) := dx1 (t)/dt. Then the equation describing the motion of the boat A before it reaches the velocity 5 m/sec is dv m 1 dt 134
(A) (A) (A) (A) = F1 + F2 = 650  (A) b1 v1 dv1 dt (A) = 65 4 (A)  v1 . 6 3 (3.12) Exercises 3.4
Solving this linear equation and using the initial condition, v1 (0) = 2, we get v1 (t) = and so x1 (t) =
0 (A) t (A) (A) 65 49 4t/3  e , 8 8 65 49 4s/3 65 147 4t/3 e 1 . ds =  e t 8 8 8 32 The boat A will have the velocity 5 m/sec at t = t satisfying 65 49 4t /3  e =5 8 8 and it will be 65 147 4t /3 t  e  1 1.85 (m) 8 32 away from the starting point or, equivalently, 5001.85 = 498.15 meters away from the finish. x1 (t ) =
(A) t =  3 ln(25/49) 0.5 (sec), 4 Similarly to (3.12), resetting the time, we obtain an equation of the motion of the boat A starting from the moment when its velocity reaches 5 m/sec. Denoting by x2 (t) the distance passed by the boat A and by v2 (t) its velocity, we get x2 (0) = 0, v2 (0) = 5, and m (A) (A) (A) (A) (A) dv2 dt (A) = 650  b2 v2 dv2 dt
(A) (A) = 65 (A)  v2 6
t v2 (t) = (A) 65 35 t  e 6 6 (A) x2 (t) =
0 65 35 s 65 35 t ds =  e t+ e 1 . 6 6 6 6 Solving the equation x2 (t) = 498.15, we find the time (counting from the moment when the boat A's velocity has reached 5 m/sec) t 46.5 sec, which is necessary to come to the end of the first leg. Therefore, the total time for the boat A is t + t 0.5 + 46.5 = 47 sec. Similarly, for the boat B, we find that v1 (t) = v2 (t) =
(B) (B) 147 5t/3 65 49 5t/3 65 3 ln(17/49) (B)  e t+ e 0.635 ; , x1 (t) =  1 , t =  8 8 8 40 5 65 35 5t/6 65 42 5t/6 (B) e , x2 (t) =  1 , t 38.895 .  e t+ 5 5 5 5 Thus, t + t < 40 sec, and so the boat B will be leading at the end of the first leg. 135 Chapter 3
25. (a) From Newton's second law we have m GMm dv = . dt r2 dv gR2 , = dt r2 where g is the gravitational force of Earth, R is the radius of Earth and r is the distance between Earth and the projectile. (b) Using the equation found in part (a), letting dv/dt = (dv/dr)(dr/dt) and knowing that dv gR2 v = 2 . dr r (c) The differential equation found in part (b) is separable and can be written in the form v dv =  gR2 dr , r2 gR2 dr. r2 dr/dt = v, we get Dividing both sides by m, the mass of the rocket, and letting g = GM/R2 we get If the projectile leaves Earth with a velocity of v0 we have the initial value problem v dv =  Integrating we get v
r=R = v0 . gR2 v2 = + K, 2 r where K is an arbitrary constant. We can find the constant K by using the initial value
2 v0 gR2 v2  = 0  gR. 2 R 2 Substituting this formula for K and solving for the velocity we obtain as follows: K= v2 = 2gR2 2 + v0  2gR. r 2 (d) In order for the velocity of the projectile to always remain positive, (2gR2 /r) + v0 must be greater than 2gR as r approaches infinity. This means
r 2 Therefore, v0  2gR > 0. lim 2gR2 2 + v0 r > 2gR 2 v0 > 2gR. 136 Exercises 3.5
(e) Using the equation ve = we have ve = 2gR = 2 9.81 m/sec2 (1 km/1000 m)(6370 km) 11.18 km/sec. 2gR for the escape velocity and converting meters to kilometers (f) Similarly to (e), we find ve = EXERCISES 3.5: 2(g/6)R = 2(9.81/6)(1/1000)(1738) = 2.38 (km/sec). Electrical Circuits, page 122 1. In this problem, R = 5 , L = 0.05 H, and the voltage function is given by E(t) = 5 cos 120t V. Substituting these data into a general solution (3) to the Kirchhoff's equation (2) yields I(t) = eRt/L = e5t/0.05 eRt/L E(t) dt + K L 5 cos 120t e5t/0.05 dt + K 0.05 = e100t 100 e100t cos 120t dt + K . Using the integral tables, we evaluate the integral in the righthand side and obtain I(t) = e100t 100 e100t (100 cos 120t + 120 sin 120t) cos 120t + 1.2 sin 120t + Ke100t . +K = 2 + (120)2 (100) 2.44 The initial condition, I(0) = 1, implies that 1 = I(0) = and so I(t) = 1 cos(120(0)) + 1.2 sin(120(0)) + Ke100(0) = +K 2.44 2.44 K = 1 1 1.44 = 2.44 2.44 1.44e100t + cos 120t + 1.2 sin 120t . 2.44 The subsequent inductor voltage is then determined by EL (t) = L dI dt d 1.44e100t + cos 120t + 1.2 sin 120t dt 2.44 100t 7.2e  6 sin 120t + 7.2 cos 120t = . 2.44 = 0.05 137 Chapter 3
3. In this RC circuit, R = 100 , C = 1012 F, the initial charge of the capacitor is Q = q(0) = 0 coulombs, and the applied constant voltage is V = 5 volts. Thus we can use a general equation for the charge q(t) of the capacitor derived in Example 2. Substitution of given data yields q(t) = CV + [Q  CV ]et/RC = 1012 (5) 1  et/(10010 and so EC (t) = Solving the equation EC (t) = 3, we get 5 1  e10
10 t 12 ) = 5 1012 1  e1010 t q(t) 10 = 5 1  e10 t . C ln 0.4 9.2 1011 (sec). 1010 =3 e10 10 t = 0.4 t= Therefore, it will take about 9.21011 seconds for the voltage to reach 3 volts at the receiving gate. 5. Let V (t) denote the voltage across an element, and let I(t) be the current through this element. Then for the power, say P = P (t), generated or dissipated by the element we have P = I(t)V (t). (3.13) We use formulas given in (a), (b), and (c) on page 119120 of the text to find P for a resistor, an inductor, and a capacitor. (a) Resistor. In this case, V (t) = ER (t) = RI(t), and substitution into (3.13) yields PR = I(t) [RI(t)] = I(t)2 R. (b) Inductor. We have V (t) = EL (t) = L 138 dI(t) dt d [LI(t)2 /2] dI(t) L dI(t) L d [I(t)2 ] = . = 2I(t) = PL = I(t) L dt 2 dt 2 dt dt Exercises 3.6
(c) Capacitor. Here, with q(t) denoting the electrical charge on the capacitor, V (t) = EC (t) = and so PC = d [CEC (t)2 /2] d [CEC (t)] C dEC (t) C d [EC (t)2 ] EC (t) = = . 2EC (t) = dt 2 dt 2 dt dt 1 q(t) C q(t) = CEC (t) I(t) = d [CEC (t)] dq(t) = dt dt 7. First, we find a formula for the current I(t). Given that R = 3 , L = 10 H, and the voltage function E(t) is a constant, say, V , the formula (3) on page 121 (which describes currents in RL circuits) becomes I(t) = e3t/10 e3t/10 V dt + K 10 = e3t/10 V 3t/10 +K e 3 = V + Ke3t/10 . 3 The initial condition, I(0) = 0 (there were no current in the electromagnet before the voltage source was applied), yields 0= V + Ke3(0)/10 3 K= V 3 I(t) = V 1  e3t/10 . 3 Next, we find the limiting value I of I(t), that is, I = lim V 1  e3t/10 3 = V V (1  0) = . 3 3 t Therefore, we are looking for the moment t when I(t) = (0.9)I = (0.9)V /3. Solving yields V 0.9V = 1  e3t/10 3 3 e3t/10 = 0.1 t= 10 ln 0.1 7.68 . 3 Thus it takes approximately 7.68 seconds for the electromagnet to reach 90% of its final value. EXERCISES 3.6: Improved Euler's Method, page 132 1. Given the step size h and considering equally spaced points we have xn+1 = xn + nh, n = 0, 1, 2, . . . . 139 Chapter 3
Euler's method is defined by equation (4) on page 125 of the text to be yn+1 = yn + hf (xn , yn ), n = 0, 1, 2, . . . , where f (x, y) = 5y. Starting with the given value of y0 = 1, we compute y1 = y0 + h(5y0 ) = 1 + 5h. We can then use this value to compute y2 to be y2 = y1 + h(5y1 ) = (1 + 5h)y1 = (1 + 5h)2 . Proceeding in this manner, we can generalize to yn : yn = (1 + 5h)n . Referring back to our equation for xn and using the given values of x0 = 0 and x1 = 1 we find 1 = nh n= 1 . h Substituting this back into the formula for yn we find the approximation to the initial value problem y = 5y, at x = 1 to be (1 + 5h)1/h . 3. In this initial value problem, f (x, y) = y, x0 = 0, and y0 = 1. Formula (8) on page 127 of the text then becomes yn+1 = yn + Solving this equation for yn+1 yields 1 h 2 yn+1 = 1+ h 2 yn yn+1 = 1 + h/2 yn , 1  h/2 n = 0, 1, . . . . (3.14) h (yn + yn+1) . 2 y(0) = 1 If n 1, we can use (3.14) to express yn in terms of yn1 and substitute this expression into the righthand side of (3.14). Continuing this process, we get yn+1 = 140 1 + h/2 1  h/2 1 + h/2 1  h/2 yn1 = 1 + h/2 1  h/2
2 n+1 yn1 = = 1 + h/2 1  h/2 y0 . Exercises 3.6
In order to approximate the solution (x) = ex at the point x = 1 with N steps, we take h = (x  x0 )/N = 1/N, and so N = 1/h. Then the above formula becomes yN = and hence e = (1) yN = 1 + h/2 1  h/2 1 + h/2 1  h/2
N y0 = 1 + h/2 1  h/2 N = 1 + h/2 1  h/2
1/h 1/h . Substituting h = 10k , k = 0, 1, 2, 3, and 4, we fill in Table 3A. 1 + h/2 1 + h/2
1/h Table 3A: Approximations
h 1 101 102 103 104 to e 2.718281828 . . . .
Error 0.281718172 0.002269586 0.000022653 0.000000227 0.000000003 Approximation 3 2.720551414 2.718304481 2.718282055 2.718281831 These approximations are better then those in Tables 3.4 and 3.5 of the text. 5. In this problem, we have f (x, y) = 4y. Thus, we have f (xn , yn ) = 4yn and f (xn + h, yn + hf (xn , yn )) = 4 [yn + h(4yn )] = 4yn + 16hyn . By equation (9) on page 128 of the text, we have yn+1 = yn + h (4yn + 4yn + 16hyn ) = 1 + 4h + 8h2 yn . 2 (3.15) Since the initial condition y(0) = 1/3 implies that x0 = 0 and y0 = 1/3, equation (3.15) above yields y1 = 1 + 4h + 8h2 y0 = 1 1 + 4h + 8h2 , 3 141 Chapter 3
y2 = 1 + 4h + 8h2 y1 = 1 + 4h + 8h2 y3 = 1 + 4h + 8h2 y2 = 1 + 4h + 8h2 Continuing this way we see that yn = 1 1 + 4h + 8h2 3
n 1 3 1 3 1 + 4h + 8h2 = 1 + 4h + 8h2
2 1 2 1 + 4h + 8h2 , 3 1 3 1 + 4h + 8h2 . = 3 . (3.16) (This can be proved by induction using equation (3.15) above.) We are looking for an approximation to our solution at the point x = 1/2. Therefore, we have h= 1/2  x0 1/2  0 1 = = n n 2n n= 1 . 2h Substituting this value for n into equation (3.16) yields yn = 1 1 + 4h + 8h2 3
1/(2h) . 7. For this problem, f (x, y) = xy 2 . We need to approximate the solution on the interval [1, 1.5] using a step size of h = 0.1. Thus the number of steps needed is N = 5. The inputs to the subroutine on page 129 are x0 = 1, y0 = 0, c = 1.5, and N = 5. For Step 3 of the subroutine we have F = f (x, y) = x  y 2 , G = f (x + h, y + hF ) = (x + h)  (y + hF )2 = (x + h)  y + h(x  y 2) Starting with x = x0 = 1 and y = y0 = 0 we get h = 0.1 (as specified) and F = 1  02 = 1, G = (1 + 0.1)  0 + 0.1(1  02 ) Hence in Step 4 we compute x = 1 + 0.1 = 1.1 , y = 0 + 0.05(1 + 1.09) = 0.1045 . 142
2 2 . = 1.1  (0.1)2 = 1.09 . Exercises 3.6
Thus the approximate value of the solution at 1.1 is 0.1045. Next we repeat Step 3 with x = 1.1 and y = 0.1045 to obtain F = 1.1 + (0.1045)2 1.0891, G = (1.1 + 0.1)  0.1045 + 0.1 1.1  (0.1045)2 Hence in Step 4 we compute x = 1.1 + 0.1 = 1.2 , y = 0.1045 + 0.05(1.0891 + 1.1545) 0.21668 . Thus the approximate value of the solution at 1.2 is 0.21668. By continuing in this way, we fill in Table 3B. (The reader can also use the software provided free with the text.) Table 3B: Improved Euler's method to approximate the solution of y = xy 2 , y(1) = 0, with h = 0.1 .
i 0 1 2 3 4 5 x 1 1.1 1.2 1.3 1.4 1.5 y 0 0.10450 0.21668 0.33382 0.45300 0.57135
2 1.1545 . 9. In this initial value problem, f (x, y) = x + 3 cos(xy), x0 = 0, and y0 = 0. To approximate the solution on [0, 2] with a step size h = 0.2, we need N = 10 steps. The functions F and G in the improved Euler's method subroutine are F = f (x, y) = x + 3 cos(xy); G = f (x + h, y + hF ) = x + h + 3 cos[(x + h)(y + hF )] = x + 0.2 + 3 cos[(x + 0.2)(y + 0.2 {x + 3 cos(xy)})]. 143 Chapter 3
Starting with x = x0 = 0 and y = y0 = 0, we compute F = 0 + 3 cos(0 0) = 3 ; G = 0 + 0.2 + 3 cos[(0 + 0.2)(0 + 0.2 {0 + 3 cos(0 0)})] 3.178426 . Using these values, we find on Step 4 that x = 0 + 0.2 = 0.2 , y = 0 + 0.1(3 + 3.178426) 0.617843 . With these new values of x and y, we repeat the Step 3 and obtain F = 0.2 + 3 cos(0.2 0.617843) 3.177125 ; G = 0.2 + 0.2 + 3 cos[(0.2 + 0.2)(0.617843 + 0.2 {0.2 + 3 cos(0.2 0.617843)})] 3.030865 . Step 4 then yields an approximation of the solution at x = 0.4: x = 0.2 + 0.2 = 0.4 , y = 0.617843 + 0.1(3.177125 + 3.030865) 1.238642 . By continuing in this way, we obtain Table 3C. Table 3C: Improved Euler's method approximations to the solution of y = x+ 3 cos(xy), y(0) = 0, on [0, 2] with h = 0.2 .
i 0 1 2 3 4 5 x 0 0.2 0.4 0.6 0.8 1.0 y 0 0.617843 1.238642 1.736531 1.981106 1.997052 i 6 7 8 9 10 x 1.2 1.4 1.6 1.8 2.0 y 1.884609 1.724472 1.561836 1.417318 1.297794 144 Exercises 3.6
y 2 1.5 1 0.5 0 x
0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Figure 3B: Polygonal line approximation to the solution of y = x + 3 cos(xy), y(0) = 0. A polygonal line, approximating the graph of the solution to the given initial value problem, which has vertices at points (x, y) from Table 3C, is sketched in Figure 3B. 13. We want to approximate the solution (x) to y = 1  y + y 3, y(0) = 0, at x = 1. (In other words, we want to find an approximate value for (1).) To do this, we will use the algorithm on page 130 of the text. (We assume that the reader has a programmable calculator or microcomputer available and can transform the stepbystep outline on page 130 into an executable program. Alternatively, the reader can use the software provided free with the text.) The inputs to the program are x0 = 0, y0 = 0, c = 1, = 0.003, and, say, M = 100. Notice that by Step 6 of the improved Euler's method with tolerance, the computations should terminate when two successive approximations differ by less that 0.003. The initial value for h in Step 1 of the improved Euler's method subroutine is h = (1  0)20 = 1. For the given equation, we have f (x, y) = 1  y + y 3, and so the numbers F and G in Step 3 145 Chapter 3
of the improved Euler's method subroutine are F = f (x, y) = 1  y + y 3 , G = f (x + h, y + hF ) = 1  (y + hF ) + (y + hF )3 . From Step 4 of the improved Euler's method subroutine with x = 0, y = 0, and h = 1, we get x = x + h = 0 + 1 = 1, h 1 y = y + (F + G) = 0 + 1 + (1  1 + 13 ) = 1. 2 2 Thus, (1) y(1; 1) = 1. The algorithm (Step 1 of the improved Euler's method subroutine) next sets h = 21 = 0.5. The inputs to the subroutine are x = 0, y = 0, c = 1, and N = 2. For Step 3 of the subroutine we have F = 1  0 + 0 = 1, G = 1  [0 + 0.5(1)] + [0 + 0.5(1)]3 = 0.625 . Hence in Step 4 we compute x = 0 + 0.5 = 0.5 , y = 0 + 0.25(1 + 0.625) = 0.40625 . Thus the approximate value of the solution at 0.5 is 0.40625. Next we repeat Step 3 with x = 0.5 and y = 0.40625 to obtain F = 1  0.40625 + (0.40625)3 = 0.6607971 , G = 1  [0.40625 + 0.5(0.6607971)] + [0.40625 + 0.5(0.6607971)]3 0.6630946 . In Step 4 we compute x = 0.5 + 0.5 = 1, 146 Exercises 3.6 Table 3D: Improved Euler's method approximations to (1), where (x) is the solution to y = 1  y + y 3, y(0) = 0.
h 1 21 22 23 y(1; h) (1) 1.0 0.7372229 0.7194115 0.7169839 y = 0.40625 + 0.25(0.6607971 + 0.6630946) 0.7372229 . Thus the approximate value of the solution at x = 1 is 0.7372229. Further outputs of the algorithm are given in Table 3D. Since y(1; 23)  y(1; 22) = 0.7169839  0.7194115 < 0.003 , the algorithm stops (see Step 6 of the improved Euler's method with tolerance) and prints out that (1) is approximately 0.71698. 15. For this problem, f (x, y) = (x + y + 2)2 . We want to approximate the solution, satisfying y(0) = 2, on the interval [0, 1.4] to find the point, with two decimal places of accuracy, where it crosses the xaxis, that is y = 0. Our approach is to use a step size of 0.005 and look for a change in the sign of y. This requires 280 steps. For this procedure inputs to the improved Euler's method subroutine are x0 = 0, y0 = 2, c = 1.4, and N = 280. We will stop the subroutine when we see a sign change in the value of y. (The subroutine is implemented on the software package provided free with the text.) For Step 3 of the subroutine we have F = f (x, y) = (x + y + 2)2 , G = f (x + h, y + hF ) = (x + h + y + hF + 2)2 = [x + y + 2 + h(1 + F )]2 . 147 Chapter 3
Starting with the inputs x = x0 = 0, y = y0 = 2, and h = 0.005 we obtain F = (0  2 + 2)2 = 0, G = [0  2 + 2 + 0.005(1 + 0)]2 = 0.000025 . Thus, in Step 4 we compute x = 0 + 0.005 = 0.005 , y = 2 + 0.005(0 + 0.000025)(1/2) 2. Thus the approximate value of the solution at x = 0.005 is 2. We continue with Steps 3 and 4 of the improved Euler's method subroutine until we arrive at x = 1.270 and y 0.04658269. The next iteration, with x = 1.275, yields y 0.006295411. This tells us that y = 0 is occurs somewhere between x = 1.270 and x = 1.275. Therefore, rounding off to two decimal places yields x = 1.27. 17. In this initial value problem, f (x, y) = 20y, x0 = 0, and y0 = 1. By applying formula (4) on page 125 of the text, we can find a general formula for yn in terms of h. Indeed, yn = yn1 + h(20yn1 ) = (1  20h)yn1 = = (1  20h)n y0 = (1  20h)n = [c(h)]n , where c(h) = 1  20h. For suggested values of h, we have h = 0.1 h = 0.2 c(0.1) = 1 c(0.2) = 3 xn = 0.1n, xn = 0.2n, yn = (1)n , n = 1, . . . , 10; yn = (3)n , n = 1, . . . , 5. h = 0.025 c(0.025) = 0.5 xn = 0.025n, yn = (0.5)n , n = 1, . . . , 40; These values are shown in Table 3E. Thus, for h = 0.1 we have alternating yn = 1; for h = 0.2, yn 's have an increasing magnitude and alternating sign; h = 0.025 is a good step size. From this example we conclude that, in Euler's method, one should be very careful in choosing a step size. Wrong choice can even lead to a diverging process. 148 Exercises 3.6
Table 3E: Euler's method approximations to the solution of y = 20y, y(0) = 1, on [0, 1] with h = 0.1, 0.2 , and 0.025.
xn 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 yn (h = 0.2) 3 9 27 81 243 yn (h = 0.1) 1 1 1 1 1 1 1 1 1 1 yn (h = 0.025) 0.062500 0.003906 0.000244 0.000015 0.000001 0.000000 0.000000 0.000000 0.000000 0.000000 19. In this problem, the variables are t and p. With suggested values of parameters, the initial value problem (13) becomes dp = 3p  pr , dt method subroutine have the form F = f (t, p) = 3p  pr ; G = f (t + 0.25, p + 0.25F ) = 3[p + 0.25F ]  [p + 0.25F ]r = 3 [p + 0.25 (3p  pr )]  [p + 0.25 (3p  pr )]r . The results of computations are shown in Table 3F. These results indicate that the limiting populations for r = 1.5, r = 2, and r = 3 are p = 9, p = 3, and p = 3, respectively. Since the righthand side of the given logistic equation, f (t, p) = 3p  pr , does not depend on t, we conclude that this equation is autonomous. Therefore, its equilibrium solutions (if any) 149 p(0) = 1. Therefore, f (t, p) = 3p  pr and, with h = 0.25, functions F and G in improved Euler's Chapter 3
Table 3F: Improved Euler's method approximations to the solution of p = 3p  pr , p(0) = 1, on [0, 5] with h = 0.25 for r = 1.5, 2, and 3.
xn 0.25 0.5 0.75 1.0 1.25 1.5 1.75 2.0 2.25 2.5 2.75 3.0 3.25 3.5 3.75 4.0 4.25 4.5 4.75 5.0 yn (r = 1.5) 1.582860 2.351441 3.267498 4.253156 5.216751 6.083402 6.811626 7.392146 7.837090 8.168507 8.410362 8.584317 8.708165 8.795710 8.857285 8.900443 8.930619 8.951682 8.966366 8.976596 yn (r = 2) 1.531250 2.049597 2.440027 2.686754 2.829199 2.908038 2.950802 2.973767 2.986037 2.992574 2.996053 2.997903 2.998886 2.999408 2.999685 2.999833 2.999911 2.999953 2.999975 2.999987 yn (r = 3) 1.390625 1.553472 1.628847 1.669992 1.694056 1.708578 1.717479 1.722980 1.726396 1.728522 1.729847 1.730674 1.731191 1.731513 1.731715 1.731841 1.731920 1.731969 1.732000 1.732019 can be found by solving f (p) = 3p  pr = 0 p 3  pr1 = 0 p = 0 or p = 31/(r1) . The condition r > 1 implies that f (p) > 0 on 0, 31/(r1) and f (p) < 0 on 31/(r1) , . Therefore, p = 31/(r1) is a sink and, regardless of the initial value p(0) = p0 > 0, there holds
t lim p(t) = 31/(r1) . 21. We will use the improved Euler's method with h = 2/3 to approximate the solution of the 150 Exercises 3.6
problem 75  20 cos t 12  T (t) + 0.1 + 1.5[70  T (t)], T (0) = 65, with K = 0.2 . Since h = 2/3, it will take 36 steps to go from t = 0 to t = 24. By simplifying the above expression, we obtain dT = (75K + 105.1)  20K cos dt t 12  (K + 1.5)T (t), T (0) = 65. (Note that here t takes the place of x and T takes the place of y.) Therefore, with K = 0.2 the inputs to the subroutine are t0 = 0, T0 = 65, c = 24, and N = 36. For Step 3 of the subroutine we have F = f (t, T ) = (75K + 105.1)  20K cos G = f (t + h, T + hF ) = (75K + 105.1)  20K cos For Step 4 in the subroutine we have t = t + h, h T = T + (F + G). 2 Now, starting with t = t0 = 0 and T = T0 = 65, and h = 2/3 (as specified) we have Step 3 of the subroutine to be F = [75(0.2) + 105.1]  20(0.2) cos 0  [(0.2) + 1.5](65) = 5.6 , (0.6667) G = [75(0.2) + 105.1]20(0.2) cos [(0.2) + 1.5][65 + (0.6667)(5.6)] 0.6862 . 12 Hence in Step 4 we compute t = 0 + 0.6667 = 0.6667 T = 65 + 0.3333(5.6  0.6862) 66.638 . 151 (t + h) 12  (K + 1.5){T + hF }. (3.18) t 12  (K + 1.5)T, (3.17) Chapter 3 Table 3G: Improved Euler's method to approximate the temperature in a building over a 24hour period (with K = 0.2).
Time Midnight 12:40 a.m. 1:20 a.m. 2:00 a.m. 2:40 a.m. 3:20 a.m. 4:00 a.m. 8:00 a.m. Noon 4:00 p.m. 8:00 p.m. Midnight tn 0 0.6667 1.3333 2.0000 2.6667 3.3333 4.0000 8.0000 12.000 16.000 20.000 24.000 Tn 65 66.63803 67.52906 68.07270 68.46956 68.81808 69.16392 71.48357 72.90891 72.07140 69.80953 68.38519 Recalling that t0 is midnight, we see that these results imply that at 0.6667 hours after midnight (or 12 : 40 a.m.) the temperature is approximately 66.638 . Continuing with this process for n = 1, 2, . . . , 35 gives us the approximate temperatures in a building with K = 0.2 over a 24 hr period. These results are given in Table 3G. (This is just a partial table.) The next step is to redo the above work with K = 0.4. That is, we substitute K = 0.4 and h = 2/3 0.6667 into equations (3.17) and (3.18) above. This yields t  1.9T, 12 (t + 0.6667) G = 135.1  8 cos  1.9(T + 0.6667F ), 12 F = 135.1  8 cos and T = T + (0.3333)(F + G). Then, using these equations, we go through the process of first finding F , then using this result to find G, and finally using both results to find T . (This process must be done for 152 Exercises 3.7
n = 0, 1, 2, . . . , 35.) Lastly, we redo this work with K = 0.6 and h = 2/3. By so doing, we obtain the results given in the table in the answers of the text. (Note that the values for T0 , T6 , T12 , T18 , T24 , T30 , and T36 are given in the answers.) EXERCISES 3.7: Higher Order Numerical Methods: Taylor and RungeKutta, page 142 1. In this problem, f (x, y) = cos(x + y). Applying formula (4) on page 135 of the text we compute f (x, y) = [cos(x + y)] =  sin(x + y) (x + y) =  sin(x + y); x x x f (x, y) = [cos(x + y)] =  sin(x + y) (x + y) =  sin(x + y); y y y f2 (x, y) = f (x, y) f (x, y) + f (x, y) =  sin(x + y) + [ sin(x + y)] cos(x + y) x y =  sin(x + y)[1 + cos(x + y)], and so, with p = 2, (5) and (6) on page 135 yield xn+1 = xn + h , yn+1 = yn + h cos (xn + yn )  3. Here we have f (x, y) = x  y and so f2 (x, y) = (x  y) (x  y) + (x  y) = 1 + (1)(x  y) = 1  x + y. x y h2 sin (xn + yn ) [1 + cos (xn + yn )] . 2 To obtain f3 (x, y) and then f4 (x, y), we differentiate the equation y = f2 (x, y) twice. This yields y (x) = [f2 (x, y)] = (1  x + y) = 1 + y = 1 + x  y =: f3 (x, y); y (4) (x) = [f3 (x, y)] = (1 + x  y) = 1  y = 1  x + y =: f4 (x, y). 153 Chapter 3
Therefore, the recursive formulas of order 4 for the Taylor method are xn+1 = xn + h, yn+1 = yn + h (xn  yn ) + h2 h3 h4 (1  xn + yn ) + (1 + xn  yn ) + (1  xn + yn ) 2 3! 4! h2 h3 h4 (1  xn + yn )  (1  xn + yn ) + (1  xn + yn ) = yn + h (xn  yn ) + 2 6 24 h2 h3 h4  + = yn + h + (1  xn + yn ) h + 2 6 24 2 3 h h4 h  + . = yn + h (xn  yn ) + (1  xn  yn ) 2 6 24 5. For the Taylor method of order 2, we need to find (see equation (4) on page 135 of the text) f2 (x, y) = f (x, y) f (x, y) + f (x, y) x y for f (x, y) = x + 1  y. Thus, we have f2 (x, y) = 1 + (1)(x + 1  y) = y  x. Therefore, by equations (5) and (6) on page 135 of the text, we see that the recursive formulas with h = 0.25 become xn+1 = xn + 0.25 , yn+1 = yn + 0.25 (xn + 1  yn ) + (0.25)2 (yn  xn ) . 2 By starting with x0 = 0 and y0 = 1 (the initial values for the problem), we find y1 = 1 + 0.0625 1.03125 . 2 Plugging this value into the recursive formulas yields y2 = 1.03125 + 0.25(0.25 + 1  1.03125) + 0.0625 (1.03125  0.25) 1.11035 . 2 By continuing in this way, we can fill in the first three columns in Table 3H. 154 Exercises 3.7
For the Taylor method of order 4, we need to find f3 and f4 . Thus, we have f2 (x, y) f2 (x, y) + f (x, y) = 1 + 1 (x + 1  y) = x  y, x y f3 (x, y) f3 (x, y) + f4 (x, y) = f (x, y) = 1 + (1) (x + 1  y) = y  x. x y f3 (x, y) = Hence, by equation (6) on page 135 of the text, we see that the recursive formula for yn+1 for the Taylor method of order 4 with h = 0.25 is given by yn+1 = yn + 0.25 (xn + 1  yn ) + (0.25)2 (0.25)3 (0.25)4 (yn  xn ) + (xn  yn ) + (yn  xn ) . 2 6 24 By starting with x0 = 0 and y0 = 1, we can fill in the fourth column of Table 3H. Table 3H: Taylor approximations of order 2 and 4 for the equation y = x + 1  y.
n 0 1 2 3 4 xn 0 0.25 0.50 0.75 1.00 yn (order 2) 1 1.03125 1.11035 1.22684 1.37253 yn (order 4) 1 1.02881 1.10654 1.22238 1.36789 Thus, the approximation (rounded to 4 decimal places) of the solution by the Taylor method at the point x = 1 is given by 2 (1) = 1.3725 if we use order 2 and by 4 (1) = 1.3679 if we use order 4. The actual solution is y = x + ex and so has the value y(1) = 1 + e1 1.3678794 at x = 1. Comparing these results, we see that y(1)  2 (1) = 0.00462 and y(1)  4 (1) = 0.00002 . 7. We will use the 4th order RungeKutta subroutine described on page 138 of the text. Since x0 = 0 and h = 0.25, we need N = 4 steps to approximate the solution at x = 1. With f (x, y) = 2y  6, we set x = x0 = 0, y = y0 = 1 and go to Step 3 to compute kj 's. k1 = hf (x, y) = 0.25[2(1)  6] = 1 ; 155 Chapter 3
k2 = hf (x + h/2, y + k1 /2) = 0.25[2(1 + (1)/2)  6] = 1.25 ; k3 = hf (x + h/2, y + k2 /2) = 0.25[2(1 + (1.25)/2)  6] = 1.3125 ; k4 = hf (x + h, y + k3 ) = 0.25[2(1 + (1.3125))  6] = 1.65625 . Step 4 then yields x = 0 + 0.25 = 0.25 , 1 1 y = 1 + (k1 + 2k2 + 2k3 + k4 ) = 1 + (1  2 1.25  2 1.3125  1.65625) 0.29688 . 6 6 Now we go back to Step 3 and recalculate kj 's for new values of x and y. k1 = 0.25[2(0.29688)  6] = 1.64844 ; k2 = 0.25[2(0.29688 + (1.64844)/2)  6] = 2.06055 ; k3 = 0.25[2(0.29688 + (2.06055)/2)  6] = 2.16358 ; k4 = 0.25[2(0.29688 + (2.16358))  6] = 2.73022 ; x = 0.25 + 0.25 = 0.5 , 1 y = 0.29688 + (1.64844  2 2.06055  2 2.16358  2.73022) 2.43470 . 6 We repeat the cycle two more times: k1 = 0.25[2(2.43470)  6] = 2.71735 ; k2 = 0.25[2(2.43470 + (2.71735)/2)  6] = 3.39670 ; k3 = 0.25[2(2.43470 + (3.39670)/2)  6] = 3.56652 ; k4 = 0.25[2(2.43470 + (3.56652))  6] = 4.50060 ; x = 0.5 + 0.25 = 0.75 , 1 y = 2.43470 + (2.71735  2 3.39670  2 3.56652  4.50060) 5.95876 6 and k1 = 0.25[2(5.95876)  6] = 4.47938 ; k2 = 0.25[2(5.95876 + (4.47938)/2)  6] = 5.59922 ; 156 Exercises 3.7
k3 = 0.25[2(5.95876 + (5.59922)/2)  6] = 5.87918 ; k4 = 0.25[2(5.95876 + (5.87918))  6] = 7.41895 ; x = 0.75 + 0.25 = 1.00 , 1 y = 5.95876 + (4.47938  2 5.59922  2 5.87918  7.41895) 11.7679 . 6 Thus (1) 11.7679 . The actual solution, (x) = 3  2e2x , evaluated at x = 1, gives (1) = 3  2e2(1) = 3  2e2 11.7781 . 9. For this problem we will use the 4th order RungeKutta subroutine with f (x, y) = x + 1  y. Using the step size of h = 0.25, the number of steps needed is N = 4 to approximate the solution at x = 1. For Step 3 we have k1 = hf (x, y) = 0.25(x + 1  y), k1 h k2 = hf x + , y + = 0.25(0.875x + 1  0.875y), 2 2 h k2 = 0.25(0.890625x + 1  0.890625y), k3 = hf x + , y + 2 2 k4 = hf (x + h, y + k3 ) = 0.25(0.77734375x + 1  0.77734375y). Hence, in Step 4 we have x = x + 0.25 , 1 y = y + (k1 + 2k2 + 2k3 + k4 ) . 6 Using the initial conditions x0 = 0 and y0 = 1, c = 1, and N = 4 for Step 3 we obtain k1 = 0.25(0 + 1  1) = 0, k2 = 0.25(0.875(0) + 1  0.875(1)) = 0.03125, k3 = 0.25(0.890625(0) + 1  0.890625(1)) 0.0273438, k4 = 0.25(0.77734375(0) + 1  0.77734375(1)) 0.0556641. Thus, Step 4 gives x = 0 + 0.25 = 0.25 , 157 Chapter 3
y 1+ 1 [0 + 2(0.03125) + 2(0.0273438) + 0.0556641] 1.02881 . 6 Thus the approximate value of the solution at 0.25 is 1.02881. By repeating Steps 3 and 4 of the algorithm we fill in the following Table 3I. Table 3I: 4th order RungeKutta subroutine approximations for y = x + 1  y at x = 1 with h = 0.25 .
x 0 0.25 0.50 0.75 1.0 y 1 1.02881 1.10654 1.22238 1.36789 Thus, our approximation at x = 1 is approximately 1.36789. Comparing this with Problem 5, we see we have obtained accuracy to four decimal places as we did with the Taylor method of order four, but without having to compute any partial derivatives. 11. In this problem, f (x, y) = 2x4  y 2 . To find the root of the solution within two decimal places of accuracy, we choose a step size h = 0.005 in 4th order RungeKutta subroutine. It will require (2  1)/0.005 = 200 steps to approximate the solution on [1, 2]. With the initial input x = x0 = 1, y = y0 = 0.414, we get k1 = hf (x, y) = 0.005[2(1)4  (0.414)2 ] = 0.009143; k2 = hf (x + h/2, y + k1 /2) = 0.005[2(1 + 0.005/2)4  (0.414 + 0.009143/2)2] = 0.009062; k3 = hf (x + h/2, y + k2 /2) = 0.005[2(1 + 0.005/2)4  (0.414 + 0.009062/2)2] = 0.009062; k4 = hf (x + h, y + k3 ) = 0.005[2(1 + 0.005)4  (0.414 + 0.009062)2] = 0.008983; x = 1 + 0.005 = 1.005, 1 y = 0.414 + (0.009143 + 2 0.009062 + 2 0.009062 + 0.008983) 0.404937; 6 . . . 158 Exercises 3.7
On the 82nd step we get x = 1.405 + 0.005 = 1.410 , 1 y = 0.004425 + (0.002566 + 2 0.002548 + 2 0.002548 + 0.002530) 0.001876 , 6 and the next step gives k1 = 0.005[2(1.410)4  (0.001876)2 ] = 0.002530 ; k2 = 0.005[2(1.410 + 0.005/2)4  (0.001876 + 0.002530/2)2] = 0.002512 ; k3 = 0.005[2(1.410 + 0.005/2)4  (0.001876 + 0.002512/2)2] = 0.002512 ; k4 = 0.005[2(1.410 + 0.005)4  (0.001876 + 0.002512)2] = 0.002494 ; x = 1.410 + 0.005 = 1.415 , 1 y = 0.414 + (0.002530 + 2 0.002512 + 2 0.002512 + 0.002494) 0.000636 . 6 Since y(1.41) < 0 and y(1.415) > 0 we conclude that the root of the solution is on the interval (1.41, 1.415). As a check, we apply the 4th order RungeKutta subroutine to approximate the solution to the given initial value problem on [1, 1.5] with a step size h = 0.001, which requires N = (1.5  1)/0.001 = 500 steps. This yields y(1.413) 0.000367, y(1.414) 0.000134, and so, within two decimal places of accuracy, x 1.41 . 13. For this problem f (x, y) = y 2 2ex y +e2x +ex . We want to find the vertical asymptote located in the interval [0, 2] within two decimal places of accuracy using the Forth Order RungeKutta subroutine. One approach is to use a step size of 0.005 and look for y to approach infinity. This would require 400 steps. We will stop the subroutine when the value of y ("blows up") becomes very large. For Step 3 we have k1 = hf (x, y) = 0.005 y 2  2ex y + e2x + ex , k2 = hf k1 h x + ,y + 2 2 = 0.005 y+ k1 2
2  2e(x+h/2) y + k1 2 + e2(x+h/2) + e(x+h/2) , 159 Chapter 3
k3 = hf k2 h x + ,y + 2 2 = 0.005 k2 y+ 2
2  2e(x+h/2) y + k2 2 + e2(x+h/2) + e(x+h/2) , k4 = hf (x + h, y + k3 ) = 0.005 (y + k3 )2  2e(x+h) (y + k3 ) + e2(x+h) + e(x+h) . Hence in Step 4 we have x = x + 0.005 , 1 y = y + (k1 + 2k2 + 2k3 + k4 ) . 6 Using the initial conditions x0 = 0, y0 = 3, c = 2, and N = 400 on Step 3 we obtain k1 = 0.005 32  2e0 (3) + e2(0) + e0 = 0.025 , k2 = 0.005 (3 + 0.0125)2  2e(0+0.0025) (3 + 0.0125) + e2(0+0.0025) + e(0+0.0025) 0.02522 , k3 = 0.005 (3 + 0.01261)2  2e(0+0.0025) (3 + 0.01261) + e2(0+0.0025) + e(0+0.0025) 0.02522 , k4 = 0.005 (3 + 0.02522)2  2e(0+0.0025) (3 + 0.02522) + e2(0+0.0025) + e(0+0.0025) 0.02543 . Thus, Step 4 yields x = 0 + 0.005 = 0.005 and y 3+ 1 (0.025 + 2(0.02522) + 2(0.02522) + 0.02543) 3.02522 . 6 Thus the approximate value at x = 0.005 is 3.02522. By repeating Steps 3 and 4 of the subroutine we find that, at x = 0.505, y = 2.0201 1013 . The next iteration gives a floating point overflow. This would lead one to think the asymptote occurs at x = 0.51 . As a check lets apply the 4th order RungeKutta subroutine with the initial conditions x0 = 0, y0 = 3, c = 1, and N = 400. This gives a finer step size of h = 0.0025. With these inputs, we find y(0.5025) 4.0402 1013 . Repeating the subroutine one more time with a step size of 0.00125, we obtain the value y(0.50125) 8.0804 1013 . Therefore we conclude that the vertical asymptote occurs at x = 0.50 and not at 0.51 as was earlier thought. 160 Exercises 3.7 y
0.5 0 1 1.5 2 2.5 3 x 1 2 3 Figure 3C: Polygonal line approximation to the solution of y = cos(5y)  x, y(0) = 0, on [0, 3]. 15. Here f (x, y) = cos(5y)  x, x0 = 0, and y0 = 0. With a step size h = 0.1 we take N = 30 in order to approximate the solution on [0, 3]. We set x = x0 = 0, y = y0 = 0 and compute k1 = hf (x, y) = 0.1[cos(5 0)  0] = 0.1 ; k2 = hf (x + h/2, y + k1 /2) = 0.1[cos(5(0 + 0.1/2))  (0 + 0.1/2)] = 0.091891 ; k3 = hf (x + h/2, y + k2 /2) = 0.1[cos(5(0 + 0.091891/2))  (0 + 0.1/2)] = 0.092373 ; k4 = hf (x + h, y + k3 ) = 0.1[cos(5(0 + 0.092373))  (0 + 0.1)] = 0.079522 ; x = 0 + 0.1 = 0.1 , 1 y = 0 + (0.1 + 2 0.091891 + 2 0.092373 + 0.079522) 0.091342 ; 6 . . . The results of computations are shown in Table 3J. Using these value, we sketch a polygonal line approximating the graph of the solution on [0, 3]. See Figure 3C. 161 Chapter 3
Table 3J: 4th order RungeKutta approximations to the solution of y = cos(5y)  x, y(0) = 0, on [0, 3] with h = 0.1 .
xn 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 yn 0 0.09134 0.15663 0.19458 0.21165 0.21462 0.20844 0.19629 0.18006 0.16079 0.13890 0.11439 0.08686 0.05544 0.01855 xn 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 yn 0.02668 0.85748 0.17029 0.30618 0.53517 0.81879 1.02887 1.17307 1.30020 1.45351 1.69491 2.03696 2.30917 2.50088 2.69767 2.99510 17. Taylor method of order 2 has recursive formulas given by equations (5) and (6) on page 135 of the text: that is xj+1 = xj + h With f (x, y) = y, we have f2 (x, y) = y = f (x, y) f (x, y) + f (x, y) = 0 + 1 (y) = y. x y and yj+1 = yj + hf (xj , yj ) + h2 f2 (xj , yj ) . 2! Therefore, since h = 1/n, the recursive formula for yj+1 is given by the equation yj+1 = yj + 1 1 yj + 2 yj = n 2n 1+ 1 1 + 2 n 2n yj . We are starting the process at x0 = 0 and y0 = 1, and we are taking steps of size 1/n until we reach x = 1. This means that we will take n steps. Thus, yn will be an approximation for the 162 Exercises 3.7
solution to the differential equation at x = 1. Since the actual solution is y = ex , this means that yn e. To find the equation we are looking for, we see that y1 = y2 = y3 = y4 = . . . yn = 1 1 1+ + 2 n 2n yn1 = 1 1 1+ + 2 n 2n
n 1+ 1 1 + 2 n 2n y0 = y1 = y2 = y3 = 1+ 1 1 + 2 n 2n ,
2 1 1 1+ + 2 n 2n 1 1 1+ + 2 n 2n 1+ 1 1 + 2 n 2n 1 1 1+ + 2 n 2n 1 1 1+ + 2 n 2n 1+ 1 1 + 2 n 2n ,
3 ,
4 , . (This can be proved rigorously by mathematical induction.) As we observed above, yn e, and so we have e 1 1 1+ + 2 n 2n
n . 19. In this initial value problem, the independent variable is u, the dependent variable is v, u0 = 2, v0 = 0.1, and u 5 + 1 v3 + u + v2 . 2 2 We will use the classical 4th order RungeKutta algorithm with tolerance given on page 139 of f (u, v) = u the text but, since the stopping criteria should be based on the relative error, we will replace the condition z  v < in Step 6 by (z  v)/v < (see Step 6 on page 138). We start with m = 0, N = 2m = 1, and a step size h = (3  2)/N = 1. Setting u = u0 = 2, v = v0 = 0.1, on Step 4 we compute 2 5 + 1 (0.1)3 + 2 + (0.1)2 = 0.049; 2 2 2 + 1/2 k2 = hf (u + h/2, v + k1 /2) = (1) (2 + 1/2) + 1 (0.1 + 0.049/2)3 2 5 + (2 + 1/2) + (0.1 + 0.049/2)2 = 0.088356 ; 2 k1 = hf (u, v) = (1) 2 163 Chapter 3
2 + 1/2 + 1 (0.1 + 0.088356/2)3 2 5 (0.1 + 0.088356/2)2 = 0.120795 ; + (2 + 1/2) + 2 2+1 k4 = hf (u + h, v + k3 ) = (1) (2 + 1) + 1 (0.1 + 0.120795)3 2 5 + (2 + 1) + (0.1 + 0.120795)2 = 0.348857 . 2 k3 = hf (u + h/2, v + k2 /2) = (1) (2 + 1/2) So, u = u + h = 2 + 1 = 3, 1 v = v + (0.049 + 2 0.088356 + 2 0.120795 + 0.348857) 0.236027 . 6 Because the relative error between two successive approximations, v(3; 20) = 0.236027 and v = 0.1 is = (0.236027  0.1)/0.236027 0.576320 > 0.0001, we go back to Step 2 and set m = 1, take N = 2m = 2 on Step 3, compute h = 1/N = 0.5, and use the 4th order RungeKutta subroutine on page 138 of the text to find v(3; 0.5). This takes two steps and yields k1 = (0.5) 2 2 5 + 1 (0.1)3 + 2 + (0.1)2 = 0.0245; 2 2 2 + 0.5/2 + 1 (0.1 + 0.0245/2)3 k2 = (0.5) (2 + 0.5/2) 2 5 + (2 + 0.5/2) + (0.1 + 0.0245/2)2 = 0.033306 ; 2 2 + 0.5/2 + 1 (0.1 + 0.033306/2)3 2 5 (0.1 + 0.033306/2)2 = 0.036114 ; + (2 + 0.5/2) + 2 2 + 0.5 + 1 (0.1 + 0.036114)3 2 5 + (2 + 0.5) + (0.1 + 0.036114)2 = 0.053410 . 2 k3 = (0.5) (2 + 0.5/2) k4 = (0.5) (2 + 0.5) 164 Exercises 3.7
This gives u = 2 + 0.5 = 2.5 , 1 v = 0.1 + (0.0245 + 2 0.033306 + 2 0.036114 + 0.053410) 0.136125 . 6 We compute kj 's again and find an approximate value of v(3). 2.5 5 + 1 (0.136125)3 + 2.5 + (0.136125)2 = 0.053419; 2 2 2.5 + 0.5/2 k2 = (0.5) (2.5 + 0.5/2) + 1 (0.136125 + 0.053419/2)3 2 5 (0.136125 + 0.053419/2)2 = 0.083702 ; + (2.5 + 0.5/2) + 2 2.5 + 0.5/2 k3 = (0.5) (2.5 + 0.5/2) + 1 (0.136125 + 0.083702/2)3 2 5 + (2.5 + 0.5/2) + (0.136125 + 0.083702/2)2 = 0.101558 ; 2 2.5 + 0.5 + 1 (0.136125 + 0.101558)3 k4 = (0.5) (2.5 + 0.5) 2 5 + (2.5 + 0.5) + (0.136125 + 0.101558)2 = 0.205709 . 2 k1 = (0.5) 2.5 Therefore, at u = 2.5 + 0.5 = 3.0, v = 0.136125 + 1 (0.053419 + 2 0.083702 + 2 0.101558 + 0.205709) 0.241066 . 6 This time the relative error is = 0.241066  0.236027 v(3; 21)  v(3; 20) = 0.020903 > 0.0001 . 1 ) v(3; 2 0.241066 Thus we set m = 2, N = 2m = 4, h = 1/N = 0.25, repeat computations with this new step, and find that v(3; 22) 0.241854 and = 0.241854  0.241066 v(3; 22)  v(3; 21) = 0.003258 > 0.0001 . 1 ) v(3; 2 0.241854 165 Chapter 3
We continue increasing m and get m = 3, h = 0.125 , v(3; 23) = 0.241924 , = 0.241924  0.241854 0.00029 > 104 ; 0.241924 0.241929  0.241924 0.00002 < 104 . 0.241929 m = 4, h = 0.0625 , v(3; 24) = 0.241929 , = Therefore, within an accuracy of 0.0001, v(3) 0.24193 . 166 CHAPTER 4: Linear Second Order Equations
EXERCISES 4.1: Introduction: The MassSpring Oscillator, page 159 1. With b = 0 and Fext = 0, equation (3) on page 155 becomes my + ky = 0. Substitution y = sin t, where = k/m, yields m(sin t) + k(sin t) = m 2 sin t + k sin t = sin t m 2 + k = sin t (m(k/m) + k) = 0. Thus y = sin t is indeed a solution. 3. Differentiating y(t), we find y = 2 sin 3t + cos 3t y = 6 cos 3t  3 sin 3t y = 18 sin 3t  9 cos 3t. Substituting y, y , and y into the given equation, we get 2y + 18y = 2(18 sin 3t  9 cos 3t) + 18(2 sin 3t + cos 3t) = [2(18) + 18(2)] sin 3t + [2(9) + 18(1)] cos 3t = 0. Next, we check that the initial conditions are satisfied. y(0) = (2 sin 3t + cos 3t)
t=0 = 2 sin 0 + cos 0 = 1, = 6 cos 0  3 sin 0 = 6. 167 y (0) = (6 cos 3t  3 sin 3t) t=0 Chapter 4
Writing y(t) in the form 1 2 5 sin 3t + cos 3t = 5 sin(3t + ), 5 5 where = arctan(1/2), we conclude that y(t) = 5 sin(3t + ), and so max y(t) = 5 y(t) = (since max  sin(3t + ) = 1). 5. We differentiate y(t) twice and obtain y(t) = e2t sin( 2t) y (t) = e2t [(2) sin( 2t) + 2 cos( 2t)] y (t) = e2t (2)2 sin( 2t) + (2) 2 cos( 2t) + (2) 2 cos( 2t)  ( 2)2 sin( 2t) = e2t 2 sin( 2t)  4 2 cos( 2t) . Substituting these functions into the differential equation, we get my + by + ky = y + 4y + 6y = e2t 2 sin( 2t)  4 2 cos( 2t) +4e2t [(2) sin( 2t) + 2 cos( 2t)] + 6e2t sin( 2t) = e2t (2  8 + 6) sin( 2t) + (4 2 + 4 2) cos( 2t) = 0. Therefore, y = e2t sin( 2t) is a solution. As t +, e2t 0 while sin( 2t) remains bounded. Therefore, lim y(t) = 0.
t+ 7. For y = A cos 5t + B sin 5t, y = 5A sin 5t + 5B cos 5t, y = 25A cos 5t  25B sin 5t. Inserting y, y , and y into the given equation and matching coefficients yield y + 2y + 4y = 3 sin 5t (25A cos 5t  25B sin 5t) + 2(5A sin 5t + 5B cos 5t) + 4(A cos 5t + B sin 5t) = (21A + 10B) cos 5t + (10A  21B) sin 5t = 3 sin 5t 21A + 10B = 0, 10A  21B = 3 A = 30/541, B = 63/541. Thus, y = (30/541) cos 5t(63/541) sin 5t is a synchronous solution to y +2y +4y = 3 sin 5t. 168 Exercises 4.2
9. We differentiate y = A cos 2t + B sin 2t twice to get y = 2A sin 2t + 2B cos 2t and y = 4A cos 2t  4B sin 2t, substitute y, y , and y into the given equation, and compare coefficients. This yields y + 2y + 4y = (4A cos 2t  4B sin 2t) + 2(2A sin 2t + 2B cos 2t) + 4(A cos 2t + B sin 2t) = 4B cos 2t  4A sin 2t = 3 cos 2t + 4 sin 2t 4B = 3, 4A = 4 A = 1, B = 3/4 y =  cos 2t + (3/4) sin 2t. EXERCISES 4.2: Homogeneous Linear Equations; The General Solution, page 167 1. The auxiliary equation for this problem is r 2 + 5r + 6 = (r + 2)(r + 3) = 0, which has the roots r = 2 and r = 3. Thus {e2t , e3t } is a set of two linearly independent solutions for this differential equation. Therefore, a general solution is given by y(t) = c1 e2t + c2 e3t , where c1 and c2 are arbitrary constants. 3. The auxiliary equation, r 2 + 8r + 16 = (r + 4)2 = 0, has a double root r = 4. Therefore, e4t and te4t are two linearly independent solutions for this differential equation, and a general solution is given by y(t) = c1 e4t + c2 te4t , where c1 and c2 are arbitrary constants. 5. The auxiliary equation for this problem is r 2 + r  1 = 0. By the quadratic formula, we have 1 5 1 1 + 4 = . r= 2 2 Therefore, a general solution is z(t) = c1 e(1 5)t/2 + c2 e(1+ 5)t/2 . 169 Chapter 4
7. Solving the auxiliary equation, 2r 2 + 7r  4 = 0, yields r = 1/2, 4. Thus a general solution is given by u(t) = c1 et/2 + c2 e4t , where c1 and c2 are arbitrary constants. 9. The auxiliary equation for this problem is r 2  r  11 = 0, which has roots 13 5 1 1 + 4 11 = . r= 2 2 Thus, a general solution to the given equation is y(t) = c1 e(1+3 5)t/2 + c2 e(13 5)t/2 . 11. Solving the auxiliary equation, 4r 2 + 20r + 25 = (2r + 5)2 = 0, we conclude that r = 5/2 is its double root. Therefore, a general solution to the given differential equation is w(t) = c1 e5t/2 + c2 te5t/2 . 13. The auxiliary equation for this problem is r 2 + 2r  8 = 0, which has roots r = 4, 2. Thus, a general solution is given by y(t) = c1 e4t + c2 e2t , where c1 , c2 are arbitrary constants. To satisfy the initial conditions, y(0) = 3, y (0) = 12, we find the derivative y (t) = 4c1 e4t + 2c2 e2t and solve the system y(0) = c1 e40 + c2 e20 = c1 + c2 = 3, y (0) = 4c1 e
40 20 + 2c2 e = 4c1 + 2c2 = 12 c1 = 3, c2 = 0. Therefore, the solution to the given initial value problem is y(t) = (3)e4t + (0)e2t = 3e4t . 15. The auxiliary equation for this equation is r 2 + 2r + 1 = (r + 1)2 = 0. We see that r = 1 is a repeated root. Thus, two linearly independent solutions are y1 (t) = et and y2 (t) = tet . This means that a general solution is given by y(t) = c1 et + c2 tet . 170 Exercises 4.2
To find the constants c1 and c2 , we substitute the initial conditions into the general solution and its derivative, y (t) = c1 et + c2 (et  tet ), and obtain y(0) = 1 = c1 e0 + c2 0 = c1 , y (0) = 3 = c1 e0 + c2 (e0  0) = c1 + c2 . So, c1 = 1 and c2 = 2. Therefore, the solution that satisfies the initial conditions is given by y(t) = et  2tet . 17. The auxiliary equation for this problem, r 2  2r  2 = 0, has roots r = 1 3. Thus, a general solution is given by z(t) = c1 e(1+ 3)t + c2 e(1 3)t . Differentiating, we find that z (t) = c1 (1 + 3)e(1+ 3)t + c2 (1  3)e(1 3)t . Substitution of z(t) and z (t) into the initial conditions yields the system z(0) = c1 + c2 = 0, z (0) = c1 (1 + 3) + c2 (1  3) = 3(c1  c2 ) = 3 3/2, c2 =  3/2. c1 = Thus, the solution satisfying the given initial conditions is 3 (1+3)t 3 (13)t 3 (1+3)t e e e  =  e(1 3)t . z(t) = 2 2 2 19. Here, the auxiliary equation is r 2  4r  5 = (r  5)(r + 1) = 0, which has roots r = 5, 1. Consequently, a general solution to the differential equation is y(t) = c1 e5t + c2 et , where c1 and c2 are arbitrary constants. To find the solution that satisfies the initial conditions, y(1) = 3 and y (1) = 9, we first differentiate the solution found above, then plug in y and y into the initial conditions. This gives y(1) = 3 = c1 e5 + c2 e y (1) = 9 = 5c1 e5  c2 e. Solving this system yields c1 = 2e5 , c2 = e1 . Thus y(t) = 2e5(t+1) + e(t+1) is the desired solution. 171 Chapter 4
21. (a) With y(t) = ert , y (t) = rert , the equation becomes arert + bert = (ar + b)ert = 0. Since the function ert is never zero on (, ), to satisfy the above equation we must have ar + b = 0. (b) Solving the characteristic equation, ar + b = 0, obtained in part (a), we get r = b/a. So y(t) = ert = ebt/a , and a general solution is given by y = cebt/a , where c is an arbitrary constant. 23. We form the characteristic equation, 5r + 4 = 0, and find its root r = 4/5. Therefore, y(t) = ce4t/5 is a general solution to the given equation. 25. The characteristic equation, 6r  13 = 0, has the root r = 13/6. Therefore, a general solution is given by w(t) = ce13t/6 . 27. Assuming that y1 (t) = et cos 2t and y2 (t) = et sin 2t are linearly dependent on (0, 1), we conclude that, for some constant c and all t (0, 1), y1 (t) = cy2 (t) et cos 2t = cet sin 2t cos 2t = c sin 2t. Choosing, say, t = /4, we get cos(/2) = c sin(/2) or c = 0. This implies that cos 2t 0 sin 2t 0, t (0, 1), which is a contradiction. Thus, y1 (t) and y2 (t) are linearly independent on (0, 1) (and so on (, ); see Problem 33(a) below). 29. These functions are linearly independent, because the equality y1 (t) cy2 (t) would imply that, for some constant c, te2t ce2t on (0, 1). 172 tc Exercises 4.2
31. Using the trigonometric identity 1 + tan2 t sec2 t, we conclude that y1 (t) = tan2 t  sec2 t 1 y2 (t) 3 (3)y1 (t), and so y1 (t) and y2 (t) are linearly dependent on (0, 1) (even on (, )). 33. (a) True. Since y1 (t) and y2 (t) are linearly dependent on [a, b], there exists a constant c such that y1 (t) = cy2 (t) (or y2 (t) = cy1(t)) for all t in [a, b]. In particular, this equality is satisfied on any smaller interval [c, d], and so y1 (t) and y2 (t) are linearly dependent on [c, d]. (b) False. As an example, consider y1 (t) = t and y2 (t) = t on [1, 1]. For t in [0, 1], y2 (t) = t = y1 (t), and so y2 (t) c1 y1 (t) with constant c1 = 1. For t in [1, 0], we have y2 (t) = t = y1 (t), and so y2 (t) c2 y1 (t) with constant c2 = 1. Therefore, these two functions are linearly dependent on [0, 1] and on [1, 0]. Since c1 = c2 , there is no such a constant c that y1 (t) cy2 (t) on [1, 1]. So, y1 (t) and y2 (t) are linearly independent on [1, 1]. 35. (a) No, because, for t 0, y2 (t) = t3  = t3 = y1 (t). (b) No, because, for t 0, y2 (t) = t3  = t3 = y1 (t). (c) Yes, because there is no constant c such that y2 (t) = cy1 (t) is satisfied for all t (for positive t we have c = 1, and c = 1 for negative t). (d) While y1 (t) = 3t2 on (, ), for the derivative of y2 (t) we consider three different cases: t < 0, t = 0, and t > 0. For t < 0, y2 (t) = t3 , y2 (t) = 3t2 , and so W [y1 , y2 ](t) = t3 3t
2 t3 3t
2 = t3 (3t2 )  3t2 (t3 ) = 0. Similarly, for t > 0, y2 (t) = t3 , y2 (t) = 3t2 , and W [y1, y2 ](t) = t3 3t
2 t3 3t
2 = t3 3t2  3t2 t3 = 0. 173 Chapter 4
For t = 0, y1 (0) = 302 = 0 and y2 (0) = 0. The latter follows from the fact that onesided derivatives of y2 (t), 3t2 and 3t2 , are both zero at t = 0. Also, y1 (0) = y2 (0) = 0. Hence W [y1 , y2](0) = 0 0 0 0 = 0, and so W [y1 , y2 ](t) 0 on (, ). This result does not contradict part (b) in Problem 34 because these functions are not a pair of solutions to a homogeneous linear equation with constant coefficients. 37. If y1 (t) and y2 (t) are solutions to the equation ay + by + c = 0, then, by Abel's formula, W [y1 , y2](t) = Cebt/a , where C is a constant depending on y1 and y2 . Thus, if C = 0, then W [y1 , y2](t) = 0 for any t in (, ), because the exponential function, ebt/a , is never zero. For C = 0, W [y1 , y2 ](t) 0 on (, ). 39. (a) A linear combination of y1 (t) = 1, y2 (t) = t, and y3 (t) = t2 , C1 1 + C2 t + C3 t2 = C1 + C2 t + C3 t2 , is a polynomial of degree at most two and so can have at most two real roots, unless it is a zero polynomial, i.e., has all zero coefficients. Therefore, the above linear combination vanishes on (, ) if and only if C1 = C2 = C3 = 0, and y1 (t), y2 (t), and y3 (t) are linearly independent on (, ). (b) Since 5y1(t) + 3y2(t) + 15y3(t) = 15 + 15 sin2 t + 15 cos2 t = 15(1 + sin2 t + cos2 t) 0 on (, ) (the Pythagorean identity), given functions are linearly dependent. (c) These functions are linearly independent. Indeed, since the function et does not vanish on (, ), C1 y1 + C2 y2 + C3 y3 = C1 et + C2 tet + C3 t2 et = C1 + C2 t + C3 t2 et = 0 if and only if C1 + C2 t + C3 t2 = 0. But functions 1, t, and t2 are linearly independent on (, ) (see (a)) and so their linear combination is identically zero if and only if C1 = C2 = C3 = 0. 174 Exercises 4.2
(d) By the definition of cosh t, 1 et + et 1 1 1 = et + et = y1 (t) + y2 (t) , y3 (t) = cosh t = 2 2 2 2 2 and given functions are linearly dependent on (, ). 41. The auxiliary equation for this problem is r 3 + r 2  6r + 4 = 0. Factoring yields r 3 + r 2  6r + 4 = r 3  r 2 + 2r 2  2r + (4r + 4) = r 2 (r  1) + 2r(r  1)  4(r  1) = (r  1)(r 2 + 2r  4). Thus the roots of the auxiliary equation are r = 1 and r = Therefore, the functions et , e(1 2 (2)2  4(1)(4) = 1 5 . 2 5)t 5)t , and e(1+ are solutions to the given equation, and they are linearly independent on (, ) (see Problem 40). Hence, a general solution to y + y  6y + 4y = 0 is given by y(t) = c1 et + c2 e(1 5)t + c3 e(1+ 5)t . 42. The auxiliary equation associated with this differential equation is r 3  6r 2  r + 6 = 0. We see, by inspection, that r = 1 is a root. Dividing the cubic polynomial r 3  6r 2  r + 6 by r  1, we find that r 3  6r 2  r + 6 = (r  1)(r 2  5r  6) = (r  1)(r + 1)(r  6). Hence r = 1, 1, 6 are the roots to the auxiliary equation, and a general solution is y(t) = c1 et + c2 et + c3 e6t . 43. Factoring the auxiliary polynomial yields r 3 + 2r 2  4r  8 = (r 3 + 2r 2 )  (4r + 8) = r 2 (r + 2)  4(r + 2) = (r + 2) (r 2  4) = (r + 2)(r + 2)(r  2). 175 Chapter 4
Therefore, the auxiliary equation has a double root 2 and a root 2. The functions e2t , te2t , and e2t form a linearly independent solution set. Therefore, a general solution in this problem is z(t) = c1 e2t + c2 te2t + c3 e2t . 45. By inspection, we see that r = 2 is a root of the auxiliary equation, r 3 + 3r 2  4r  12 = 0. Dividing the polynomial r 3 + 3r 2  4r  12 by r  2 yields r 3 + 3r 2  4r  12 = (r  2) r 2 + 5r + 6 = (r  2)(r + 2)(r + 3). Hence, two other roots of the auxiliary equation are r = 2 and r = 3. The functions e3t , e2t , and e2t are three linearly independent solutions to the given equation, and a general solution is given by y(t) = c1 e3t + c2 e2t + c3 e2t . 47. First we find a general solution to the equation y  y = 0. Its characteristic equation, r 3  r = 0, has roots r = 0, 1, and 1, and so a general solution is given by y(t) = c1 e(0)t + c2 e(1)t + c3 e(1)t = c1 + c2 et + c3 et . Differentiating y(t) twice yields y (t) = c2 et + c3 et , y (t) = c2 et + c3 et . Now we substitute y, y , and y into the initial conditions and find c1 , c2 , and c3 . y(0) = c1 + c2 + c3 = y (0) = c2 + c3 y (0) = c2 + c3 = 2, 3, c1 = 3, c2 = 2, c3 = 1. = 1 Therefore, the solution to the given initial value problem is y(t) = 3  2et + et . 176 Exercises 4.3
49. (a) To find the roots of the auxiliary equation, p(r) := 3r 3 + 18r 2 + 13r  19 = 0, one can use Newton's method or intermediate value theorem. We note that p(5) = 9 < 0, p(2) = 3 > 0, p(0) = 19 < 0, p(4) = 25 > 0, p(1) = 17 < 0, p(1) = 15 > 0. Therefore, the roots of p(r) belong to the intervals [5, 4], [2, 1], and [0, 1], and we can take r = 5, r = 2, and r = 0 as initial quesses. Approximation yields r1 4.832, r2 1.869, and r3 0.701. So, a general solution is given by y(t) = c1 er1 t + c2 er2 t + c3 er3 t = c1 e4.832t + c2 e1.869t + c3 e0.701t . (b) The auxiliary equation, r 4  5r 2 + 5 = 0, is of quadratic type. The substitution s = r 2 yields s  5s + 5 = 0 Therefore, r1 = 5 5 1.176 , 2 r2 = 5+ 5 1.902 , 2 r3 = r1 , and r4 = r2
2 5 5 s= 2 r= s= 5 5 . 2 are the roots of the auxiliary equation, and a general solution to y (iv)  5y + 5y = 0 is given by y(t) = c1 er1 t + c2 er1 t + c3 er2 t + c4 er2 t . (c) We can use numerical tools to find the roots of the auxiliary fifth degree polynomial equation r 5  3r 4  5r 3 + 15r 2 + 4r  12 = 0. Alternatively, one can involve the rational root theorem and examine the divisors of the free coefficient, 12. These divisors are 1, 2, 3, 4, 6, and 12. By inspection, r = 1, 2, and 3 satisfy the equation. Thus, a general solution is y(t) = c1 et + c2 et + c3 e2t + c4 e2t + c5 e3t . EXERCISES 4.3: Auxiliary Equations with Complex Roots, page 177 1. The auxiliary equation in this problem is r 2 + 9 = 0, which has roots r = 3i. We see that = 0 and = 3. Thus, a general solution to the differential equation is given by y(t) = c1 e(0)t cos 3t + c2 e(0)t sin 3t = c1 cos 3t + c2 sin 3t. 177 Chapter 4
3. The auxiliary equation, r 2  6r + 10 = 0, has roots r = 6 = 1, and z(t) = c1 e3t cos t + c2 e3t sin t is a general solution. 5. This differential equation has the auxiliary equation r 2 +4r +6 = 0. The roots of this auxiliary equation are r = 4 16  24 /2 = 2 2 i. We see that = 2 and = 2. Thus, a general solution to the differential equation is given by w(t) = c1 e2t cos 2t + c2 e2t sin 2t. 7. The auxiliary equation for this problem is given by 4r  4r + 26 = 0
2 62  40 /2 = 3 i. So = 3, 2r  2r + 13 = 0 2 r= 2 4  104 1 5 = i. 4 2 2 Therefore, = 1/2 and = 5/2. Thus, a general solution is given by y(t) = c1 et/2 cos 5t 2 + c2 et/2 sin 5t 2 . 9. The associated auxiliary equation, r 2  8r + 7 = 0, has two real roots, r = 1, 7. Thus the answer is y(t) = c1 et + c2 e7t . 11. The auxiliary equation for this problem is r 2 + 10r + 25 = (r + 5)2 = 0. We see that r = 5 is a repeated root. Thus two linearly independent solutions are z1 (t) = e5t and z2 (t) = te5t . This means that a general solution is given by z(t) = c1 e5t + c2 te5t , where c1 and c2 are arbitrary constants. 13. Solving the auxiliary equation yields complex roots r 2 + 2r + 5 = 0 178 r= 2 22  4(1)(5) = 1 2i. 2 Exercises 4.3
So, = 1, = 2, and a general solution is given by y(t) = c1 et cos 2t + c2 et sin 2t. 15. First, we find the roots of the auxiliary equation. r 2 + 10r + 41 = 0 r= 10 102  4(1)(41) = 5 4i. 2 These are complex numbers with = 5 and = 4. Hence, a general solution to the given differential equation is y(t) = c1 e5t cos 4t + c2 e5t sin 4t. 17. The auxiliary equation in this problem, r 2  r + 7 = 0, has the roots 1 12  4(1)(7) 1 27 1 3 3 = = i. r= 2 2 2 2 Therefore, a general solution is y(t) = c1 et/2 cos 3 3 t 2 + c2 et/2 sin 3 3 t . 2 19. The auxiliary equation, r 3 + r 2 + 3r  5 = 0, is a cubic equation. Since any cubic equation has a real root, first we examine the divisors of the free coefficient, 5, to find integer real roots (if any). By inspection, r = 1 satisfies the equation. Dividing r 3 + r 2 + 3r  5 by r  1 yields r 3 + r 2 + 3r  5 = (r  1)(r 2 + 2r + 5). Therefore, the other two roots of the auxiliary equation are the roots of the quadratic equation r 2 + 2r + 5 = 0, which are r = 1 2i. A general solution to the given equation is then given by y(t) = c1 et + c2 et cos 2t + c3 et sin 2t. 21. The auxiliary equation for this problem is r 2 + 2r + 2 = 0, which has the roots 2 4  8 = 1 i. r= 2 179 Chapter 4
So, a general solution is given by y(t) = c1 et cos t + c2 et sin t , where c1 and c2 are arbitrary constants. To find the solution that satisfies the initial conditions, y(0) = 2 and y (0) = 1, we first differentiate the solution found above, then plug in given initial conditions. This yields y (t) = c1 et ( cos t  sin t) + c2 et (cos t  sin t) and y(0) = c1 = 2, y (0) = c1 + c2 = 1 . Thus c1 = 2, c2 = 3, and the solution is given by y(t) = 2et cos t + 3et sin t . 23. The auxiliary equation for this problem is r 2  4r + 2 = 0. The roots of this equation are r= 4 16  8 = 2 2, 2 2)t which are real numbers. A general solution is given by w(t) = c1 e(2+ + c2 e(2 2)t , where c1 and c2 are arbitrary constants. To find the solution that satisfies the initial conditions, w(0) = 0 and w (0) = 1, we first differentiate the solution found above, then plug in our initial conditions. This gives w(0) = c1 + c2 = 0, w (0) = 2 + 2 c1 + 2  2 c2 = 1 . Solving this system of equations yields c1 = 1/(2 2) and c2 = 1/(2 2). Thus 1 (2+2)t 2 (2+2)t 1 (22)t w(t) = e  e =  e(2 2)t e 4 2 2 2 2 is the desired solution. 180 Exercises 4.3
25. The auxiliary equation, r 2  2r + 2 = 0, has the roots r = 1 i. Thus, a general solution is y(t) = c1 et cos t + c2 et sin t , where c1 and c2 are arbitrary constants. To find the solution that satisfies the initial conditions, y() = e and y () = 0, we find y (t) = c1 et (cos t  sin t) + c2 et (sin t + cos t) and solve the system e = y() = c1 e , 0 = y () = c1 e  c2 e . This yields c1 = 1, c2 = c1 = 1. So, the answer is y(t) = et cos t + et sin t = et (sin t  cos t) . 27. To solve the auxiliary equation, r 3  4r 2 + 7r  6 = 0, which is of the third order, we find its real root first. Examining the divisors of 6, that is, 1, 2, 3, and 6, we find that r = 2 satisfies the equation. Next, we divide r 3  4r 2 + 7r  6 by r  2 and obtain r 3  4r 2 + 7r  6 = (r  2) r 2  2r + 3 . Therefore, the other two roots of the auxiliary equation are 2 4  12 = 1 2i , r= 2 and a general solution to the given differential equation is given by y(t) = c1 e2t + c2 et cos 2t + c3 et sin 2t . Next, we find the derivatives, y (t) = 2c1 e2t + c2 et cos 2t  2 sin 2t + c3 et sin 2t + 2 cos 2t , y (t) = 4c1 e2t + c2 et  cos 2t  2 2 sin 2t + c3 et  sin 2t + 2 2 cos 2t , 181 Chapter 4
and substitute y, y , and y into the initial conditions. This yields c1 + c2 = 1, 2c1 + c2 + 2c3 = 0, 4c1  c2 + 2 2c3 = 0 c1 = 1, c2 = 0, c3 =  2 . With these values of the constants c1 , c2 , and c3 , the solution becomes y(t) = e2t  2et sin 2t . 29. (a) As it was stated in Section 4.2, third order linear homogeneous differential equations with constant coefficients can be handled in the same way as second order equations. Therefore, we look for the roots of the auxiliary equation r 3  r 2 + r + 3 = 0. By the rational root theorem, the only possible rational roots are r = 1 and 3. By checking these values, we find that one of the roots of the auxiliary equation is r = 1. Factorization yields r 3  r 2 + r + 3 = (r + 1)(r 2  2r + 3). Using the quadratic formula, we find that the other two roots are 2 4  12 r= = 1 2 i. 2 A general solution is, therefore, y(t) = c1 et + c2 et cos 2t + c3 et sin 2t . (b) By inspection, r = 2 is a root of the auxiliary equation, r 3 + 2r 2 + 5r  26 = 0. Since r 3 + 2r 2 + 5r  26 = (r  2) r 2 + 4r + 13 , the other two roots are the roots of r 2 + 4r + 13 = 0, that is, r = 2 3i. Therefore, a general solution to the given equation is y(t) = c1 e2t + c2 e2t cos 3t + c3 e2t sin 3t . 182 Exercises 4.3
(c) The fourth order auxiliary equation r 4 + 13r 2 + 36 = 0 can be reduced to a quadratic equation by making a substitution s = r 2 . This yields 13 5 13 169  144 = . s= s2 + 13r + 36 = 0 2 2 Thus, s = (13 + 5)/2 = 4 or s = (13  5)/2 = 9, and the solutions to the auxiliary equation are r = 4 = 2i and r = 9 = 3i. A general solution, therefore, has the form y(t) = c1 cos 2t + c2 sin 2t + c3 cos 3t + c4 sin 3t . 31. (a) Comparing the equation y + 16y = 0 with the massspring model (16) in Example 4, we conclude that the damping coefficient b = 0 and the stiffness constant k = 16 > 0. Thus, solutions should have an oscillatory behavior. Indeed, the auxiliary equation, r 2 + 16 = 0, has roots r = 4i, and a general solution is given by y(t) = c1 cos 4t + c2 sin 4t . Evaluating y (t) and substituting the initial conditions, we get y(0) = c1 = 2, y (0) = 4c2 = 0 c1 = 2, c2 = 0 y(t) = 2 cos 4t . (b) Positive damping b = 100 and stiffness k = 1 imply that the displacement y(t) tends to zero, as t . To confirm this prediction, we solve the given initial value problem explicitly. The roots of the associated equation are r= Thus the roots r1 = 50  solution is given by y(t) = c1 er1 t + c2 er2 t y (t) = c1 r1 er1 t + c2 r2 er2 t . 183 100 2 1002  4 = 50 2499 . 2499 and r2 = 50 + 2499 are both negative. A general Chapter 4
Solving the initial value problem yields y(0) = 1 = c1 + c2 , y (0) = 0 = c1 r1 + c2 r2 c1 = r2 /(r2  r1 ), c2 = r1 /(r1  r2 ), and so the desired solution is 50 + 2499 (502499)t 50 + 2499 (50+2499)t e e + . y(t) = 2 2499 2 2499 Since both powers in exponential functions tend to  as t , y(t) 0. (c) The corresponding massspring model has negative damping b = 6 and positive stiffness k = 8. Thus the magnitude y(t) of the displacement y(t) will increase without bound, as t . Moreover, because of the positive initial displacement and initial zero velocity, the mass will move in the negative direction. Thus, our guess is that y(t)  as t . Now we find the actual solution. Since the roots of the auxiliary equation are r = 2 and r = 4, a general solution to the given equation is y(t) = c1 e2t + c2 e4t . Next, we find c1 and c2 satisfying the initial conditions. y(0) = 1 = c1 + c2 , y (0) = 0 = 2c1 + 4c2 Thus, the desired solution is y(t) = 2e2t  e4t , and it approaches  as t . (d) In this problem, the stiffness k = 3 is negative. In the massspring model, this means that the spring forces the mass to move in the same direction as the sign of the displacement is. Initially, the displacement y(0) = 2 is negative, and the mass has no initial velocity. Thus the mass, when released, will move in the negative direction, and the spring will enforce this movement. So, we expect that y(t)  as t . To find the actual solution, we solve the auxiliary equation r 2 + 2r  3 = 0 and obtain r = 3, 1. Therefore, a general solution is given by y(t) = c1 e3t + c2 et . We find c1 and 184 c1 = 2, c2 = 1. Exercises 4.3
c2 from the initial conditions. y(0) = 2 = c1 + c2 , y (0) = 0 = 3c1 + c2 c1 = 1/2, c2 = 3/2. Thus, the solution to the initial value problem is e3t 3et  , y(t) =  2 2 and, as t , it approaches . (e) As in the previous problem, we have negative stiffness k = 6. But this time the initial displacement, y(0) = 1, as well as the initial velocity, y (0) = 1, is positive. So, the mass will start moving in the positive direction, and will continue doing this (due to the negative stiffness) with increasing velocity. Thus our prediction is that y(t) when t . Indeed, the roots of the characteristic equation in this problem are r = 2 and 3, and so a general solution has the form y(t) = c1 e2t + c2 e3t . To satisfy the initial conditions, we solve the system y(0) = 1 = c1 + c2 , y (0) = 1 = 2c1 + 3c2 Thus, the solution to the initial value problem is 2e2t 3e3t y(t) = + , 5 5 and it approaches as t . 33. From Example 3 we see that, in the study of a vibrating spring with damping, we have the initial value problem my (t) + by (t) + ky(t) = 0; y(0) = y0 , y (0) = v0 , c1 = 2/5, c2 = 3/5. where m is the mass of the spring system, b is the damping constant, k is the spring constant, y(0) is the initial displacement, y (0) is the initial velocity, and y(t) is the displacement of the mass from the equilibrium at time t. 185 Chapter 4
(a) We want to determine the equation of motion for a spring system with m = 10 kg, b = 60 kg/sec, k = 250 kg/sec2 , y(0) = 0.3 m, and y (0) = 0.1 m/sec. That is, we seek the solution to the initial value problem 10y (t) + 60y (t) + 250y(t) = 0; y(0) = 0.3 , y (0) = 0.1 . The auxiliary equation for the above differential equation is 10r 2 + 60r + 250 = 0 which has the roots r= 6 r 2 + 6r + 25 = 0, 6 8i 36  100 = = 3 4i. 2 2 Hence = 3 and = 4, and the displacement y(t) has the form y(t) = c1 e3t cos 4t + c2 e3t sin 4t. We find c1 and c2 by using the initial conditions. We first differentiate y(t) to get y (t) = (3c1 + 4c2 )e3t cos 4t + (4c1  3c2 )e3t sin 4t. Substituting y and y into the initial conditions, we obtain the system y(0) = 0.3 = c1 , y (0) = 0.1 = 3c1 + 4c2 . Solving, we find that c1 = 0.3 and c2 = 0.2. Therefore the equation of motion is given by y(t) = 0.3e3t cos 4t + 0.2e3t sin 4t (m). (b) From Problem 32 we know that the frequency of oscillation is given by /(2). In part (a) we found that = 4. Therefore the frequency of oscillation is 4/(2) = 2/. (c) We see a decrease in the frequency of oscillation. We also have the introduction of the factor e3t , which causes the solution to decay to zero. This is a result of energy loss due to the damping. 186 Exercises 4.3
35. The equation of the motion of a swinging door is similar to that for massspring model (with the mass m replaced by the moment of inertia I and the displacement y(t) replaced by the angle that the door is open). So, from the discussion following Example 3 we conclude that the door will not continually swing back and forth (that is, the solution (t) will not oscillate) if b 4Ik = 2 Ik. 37. (a) The auxiliary equation for this problem is r 4 + 2r 2 + 1 = (r 2 + 1)2 = 0. This equation has the roots r1 = r2 = i, r3 = r4 = i. Thus, cos t and sin t are solutions and, since the roots are repeated, we get two more solutions by multiplying cos t and sin t by t, that is, t cos t and t sin t are also solutions. This gives a general solution y(t) = c1 cos t + c2 sin t + c3 t cos t + c4 t sin t. (b) The auxiliary equation in this problem is r 4 + 4r 3 + 12r 2 + 16r + 16 = (r 2 + 2r + 4)2 = 0. The roots of the quadratic equation r 2 + 2r + 4 = 0 are 2 4  16 = 1 3i. r= 2 Hence the roots of the auxiliary equation are r1 = r2 = 1 3i and r3 = r4 = 1+ 3i. Thus two linearly independent solutions are et cos( 3t) and et sin( 3t), and we get two more linearly independent by multiplying them by t. This gives a general solution of the form y(t) = (c1 + c2 t)et cos( 3t) + (c3 + c4 t)et sin( 3t). 39. (a) Comparing given equation with the CauchyEuler equation (21) in general form, we conclude that a = 3, b = 11, and c = 3. Thus, the substitution x = et leads to the equation (22) in Problem 38 with these values of parameters. That is, a d2 y dy + (b  a) + cy = 0 dt2 dt 3 d2 y dy + 8  3y = 0. dt2 dt 187 Chapter 4
(b) The auxiliary equation to the differential equation obtained in (a) is 3r 2 + 8r  3 = 0, which has the roots r= 8 64  4(3)(3) 8 10 = 6 6 r = 3, 1 . 3 This yields a general solution y(t) = c1 et/3 + c2 e3t . (c) Since x = et , we can express y(t) as a function of x by writing y = c1 et/3 + c2 e3t = c1 et
1/3 + c2 et 3 = c1 x1/3 + c2 x3 . 41. This equation is a CauchyEuler equation. The substitution x = et leads to the equation (22) with a = 1, b = 2, and c = 6. Thus we have a dy d2 y + (b  a) + cy = 0 2 dt dt d2 y dy +  6y = 0. dt2 dt The auxiliary equation, r 2 + r  6 = 0, has the roots r = 3 and r = 2. Therefore, a general solution can be written as y = c1 e3t + c2 e2t = c1 et 43. The substitution x = et yields the equation dy d2 y + (9  1) + 17y = 0 dt2 dt dy d2 y + 8 + 17y = 0. dt2 dt
3 + c2 et 2 = c1 x3 + c2 x2 . Solving the characteristic equation, r 2 + 8r + 17 = 0, we get 8 64  68 = 4 i. r= 2 Thus, the roots are complex with = 4, = 1, and a general solution, as a function of t, is given by y(t) = c1 e4t cos t + c2 e4t sin t. Now we make the back substitution. Since x = et , we have t = ln x and so y = et 188
4 (c1 cos t + c2 sin t) = x4 [c1 cos(ln x) + c2 sin(ln x)] . Exercises 4.4
EXERCISES 4.4: Nonhomogeneous Equations: The Method of Undetermined Coefficients, page 186 1. We cannot use the method of undetermined coefficients to find a particular solution because of the t1 term, which is not a polynomial. 3. Rewriting the righthand side in the form 3t = e(ln 3)t = ert , where r = ln 3, we conclude that the method of undetermined coefficients can be applied. 5. Since sec = 1/ cos , we cannot use the method of undetermined coefficients. 7. Given equation is not an equation with constant coefficients. Thus the method of undetermined coefficients cannot be applied. 9. The roots of the auxiliary equation, r 2 + 3 = 0, are r = 3i. Since they are different from zero, we look for a particular solution of the form yp (t) A. Substitution into the original equation yields (A) + 3A = 9 3A = 9 A = 3. Thus, yp (t) 3 is a particular solution to the given nonhomogeneous equation. 11. The auxiliary equation in this problem, 2r 2 + 1 = 0, has complex roots. Therefore, e2t is not a solution to the corresponding homogeneous equation, and a particular solution to the original nonhomogeneous equation has the form zp (t) = Ae2t . Substituting this expression into the equation, we find the constant A. 2 Ae2t Hence, zp (t) = e2t . 12. This equation is a linear first order differential equation with constant coefficients. The corresponding homogeneous equation, 2x +x = 0, can be solved by the methods of Chapter 2. Alternatively, one can use the result of Problem 21 in Section 4.2. Either approach yields 189 + Ae2t = 2 4Ae2t + Ae2t = 9Ae2t = 9e2t A = 1. Chapter 4
xh (t) = Cet/2 . So, the homogeneous equation does not have a polynomial solution (other than x(t) 0), and we look for a particular solution to the nonhomogeneous equation of the form xp (t) = A2 t2 + A1 t + A0 . Substitution into the original differential equation yields 2xp (t) + xp (t) = 2 (2A2 t + A1 ) + A2 t2 + A1 t + A0 = A2 t2 + (4A2 + A1 ) t + (2A1 + A0 ) = 3t2 . By equating coefficients we obtain A2 = 3, 4A2 + A1 = 0 2A1 + A0 = 0 A1 = 12, A0 = 24. Therefore, a particular solution is xp (t) = 3t2  12t + 24. 13. The righthand side of the original nonhomogeneous equation suggest us the form yp (t) = ts (A cos 3t + B sin 3t) for a particular solution. Since the roots of the auxiliary equation, r 2  r + 9 = 0, are different from 3i, neither cos 3t nor sin 3t is a solution to the corresponding homogeneous equation. Therefore, we can choose s = 0, and so yp (t) = A cos 3t + B sin 3t, yp (t) = 3A sin 3t + 3B cos 3t, yp (t) = 9A cos 3t  9B sin 3t. Substituting these expressions into the original equation and equating the corresponding coefficients, we conclude that (9A cos 3t  9B sin 3t)  (3A sin 3t + 3B cos 3t) + 9 (A cos 3t + B sin 3t) = 3 sin 3t 3B cos 3t + 3A sin 3t = 3 sin 3t A = 1, B = 0. Hence, the answer is yp (t) = cos 3t. 190 Exercises 4.4
15. For this problem, the corresponding homogeneous equation is y  5y + 6y = 0, which has the associated auxiliary equation r 2  5r + 6 = 0. The roots of this equation are r = 3 and r = 2. Therefore, neither y = ex nor y = xex satisfies the homogeneous equation, and in the expression yp (x) = xs (Ax + B)ex for a particular solution we can take s = 0. So yp (x) = (Ax + B)ex yp (x) = (Ax + B + A)ex yp (x) = (Ax + B + 2A)ex (Ax + B + 2A)ex  5(Ax + B + A)ex + 6(Ax + B)ex = xex (2Ax  3A + 2B)ex = xex 2A = 1, 3A + 2B = 0 A = 1/2, B = 3/4 , and yp (x) = (x/2 + 3/4)ex . 16. The corresponding homogeneous equation has the auxiliary equation r 2  1 = 0, whose roots are r = 1. Thus, in the expression p (t) = (A1 t + A0 ) cos t + (B1 t + B0 ) sin t none of the terms is a solution to the homogeneous equation. We find p (t) = (A1 t + A0 ) cos t + (B1 t + B0 ) sin t p (t) = A1 cos t  (A1 t + A0 ) sin t + B1 sin t + (B1 t + B0 ) cos t = (B1 t + A1 + B0 ) cos t + (A1 t  A0 + B1 ) sin t p (t) = B1 cos t  (B1 t + B0 + A1 ) sin t  A1 sin t + (A1 t  A0 + B1 ) cos t = (A1 t  A0 + B1 ) cos t + (B1 t  B0  2A1 ) sin t. Substituting these expressions into the original differential equation, we get p  p = (A1 t  A0 + 2B1 ) cos t + (B1 t  B0  2A1 ) sin t  (A1 t + A0 ) cos t  (B1 t + B0 ) sin t = 2A1 t cos t + (2A0 + 2B1 ) cos t  2B1 t sin t + (2A1  2B0 ) sin t = t sin t. 191 Chapter 4
Equating the coefficients, we see that 2A1 = 0 2A0 + 2B1 = 0 2B1 = 1 2A1  2B0 = 0 A1 = 0, B1 = A0 , 1 B1 =  2 B0 = 0. 1 and so A0 =  , 2 Therefore, a particular solution of the nonhomogeneous equation  = t sin t is given by p (t) =  t sin t + cos t . 2 17. The righthand side of the original equation suggests that a particular solution should be of the form yp (t) = Ats et . Since r = 1 is a double root of the corresponding auxiliary equation, r 2  2r + 1 = (r  1)2 = 0, we take s = 2. Hence yp (t) = At2 et yp (t) = A t2 + 2t et yp (t) = A t2 + 4t + 2 et . Substituting these expressions into the original equation, we find the constant A. A t2 + 4t + 2 et  2A t2 + 2t et + At2 et = 8et Thus, yp (t) = 4t2 et . 19. According to the righthand side of the given equation, a particular solution has the form yp (t) = ts (A1 t + A0 )e3t . To choose s, we solve the auxiliary equation, 4r 2 + 11r  3 = 0, and find that r = 3 is its simple root. Therefore, we take s = 1, and so yp (t) = t (A1 t + A0 ) e3t = A1 t2 + A0 t e3t . Differentiating yields yp (t) = 3A1 t2 + (2A1  3A0 ) t + A0 e3t , yp (t) = 9A1 t2 + (9A0  12A1 ) t + 2A1  6A0 e3t . 192 2Aet = 8et A = 4. Exercises 4.4
Substituting y, y , and y into the original equation, after some algebra we get [26A1 t + (8A1  13A0 )]e3t = 2te3t Therefore, yp (t) = 8 t + te3t . 13 169 26A1 = 2, 8A1  13A0 = 0 A1 = 1/13, A0 = 8/169. 21. The nonhomogeneous term of the original equation is te2t . Therefore, a particular solution has the form xp (t) = ts (A1 t + A0 ) e2t . The corresponding homogeneous differential equation has the auxiliary equation r 2  4r + 4 = (r  2)2 = 0. Since r = 2 is its double root, s is chosen to be 2. Thus a particular solution to the nonhomogeneous equation has the form xp (t) = t2 (A1 t + A0 ) e2t = A1 t3 + A0 t2 e2t . We compute xp = 3A1 t2 + 2A0 t e2t + 2 A1 t3 + A0 t2 e2t , xp = (6A1 t + 2A0 ) e2t + 4 3A1 t2 + 2A0 t e2t + 4 A1 t3 + A0 t2 e2t . Substituting these expressions into the original differential equation yields xp  4xp + 4xp = (6A1 t + 2A0 ) e2t + 4 3A1 t2 + 2A0 t e2t + 4 A1 t3 + A0 t2 e2t 4 3A1 t2 + 2A0 t e2t  8 A1 t3 + A0 t2 e2t + 4 A1 t3 + A0 t2 e2t = (6A1 t + 2A0 ) e2t = te2t . Equating coefficients yields A0 = 0 and A1 = 1/6. Therefore xp (t) = t3 e2t /6 is a particular solution to the given nonhomogeneous equation. 23. The righthand side of this equation suggests that yp () = s (A2 2 + A1 + A0 ). We choose s = 1 because r = 0 is a simple root of the auxiliary equation, r 2  7r = 0. Therefore, yp () = (A2 2 + A1 + A0 ) = A2 3 + A1 2 + A0 yp () = 3A2 2 + 2A1 + A0 yp () = 6A2 + 2A1 . 193 Chapter 4
So, yp 7yp = (6A2 + 2A1 )7 3A2 2 + 2A1 + A0 = 21A2 2 +(6A2 14A1 )+2A1 7A0 = 2 . Comparing the corresponding coefficients, we find A2 , A1 , and A0 . 21A2 = 1, 6A2  14A1 = 0, 2A1  7A0 = 0 Hence yp () =  1 3 1 2 2   . 21 49 343 A2 = 1/21, A1 = 3A2 /7 = 1/49, A0 = 2A1 /7 = 2/343. 25. We look for a particular solution of the form yp (t) = ts (A cos 3t + B sin 3t)e2t . Since r = 2 + 3i is not a root of the auxiliary equation, which is r 2 + 2r + 4 = 0, we can take s = 0. Thus, yp (t) = (A cos 3t + B sin 3t)e2t yp (t) = [(2A + 3B) cos 3t + (3A + 2B) sin 3t)]e2t yp (t) = [(5A + 12B) cos 3t + (12A  5B) sin 3t)]e2t . Next, we substitute yp , yp , and yp into the original equation and compare the corresponding coefficients. yp + 2yp + 4yp = [(3A + 18B) cos 3t + (18A + 3B) sin 3t]e2t = 111e2t cos 3t 3A + 18B = 111, 18A + 3B = 0. This system has the solution A = 1, B = 6. So, yp (t) = (cos 3t + 6 sin 3t)e2t . 27. The righthand side of this equation suggests that yp (t) = ts A3 t3 + A2 t2 + A1 t + A0 cos 3t + ts B3 t3 + B2 t2 + B1 t + B0 sin 3t. 194 Exercises 4.4
To choose s, we find the roots of the characteristic equation, which is r 2 + 9 = 0. Since r = 3i are its simple roots, we take s = 1. Thus yp (t) = t A3 t3 + A2 t2 + A1 t + A0 cos 3t + t B3 t3 + B2 t2 + B1 t + B0 sin 3t. 29. The characteristic equation r 2  6r + 9 = (r  3)2 = 0 has a double root r = 3. Therefore, a particular solution is of the form yp (t) = t2 A6 t6 + A5 t5 + A4 t4 + A3 t3 + A2 t2 + A1 t + A0 e3t . 31. From the form of the righthand side, we conclude that a particular solution should be of the form yp (t) = ts A3 t3 + A2 t2 + A1 t + A0 cos t + B3 t3 + B2 t2 + B1 t + B0 sin t et . Since r = 1 i are simple roots of the characteristic equation, r 2 + 2r + 2 = 0, we should take s = 1. Therefore, yp (t) = t A3 t3 + A2 t2 + A1 t + A0 cos t + B3 t3 + B2 t2 + B1 t + B0 sin t et . 33. The righthand side of the equation suggests that yp (t) = ts (A cos t + B sin t). By inspection, we see that r = i is not a root of the corresponding auxiliary equation, r 3  r 2 + 1 = 0. Thus, with s = 0, yp (t) = A cos t + B sin t, yp (t) = A sin t + B cos t, yp (t) = A cos t  B sin t, yp (t) = A sin t  B cos t, and substitution into the original equation yields (A sin t  B cos t)  (A cos t  B sin t) + (A cos t + B sin t) = sin t (2A  B) cos t + (A + 2B) sin t = sin t 2A  B = 0, A + 2B = 1 A = 1/5, B = 2/5 yp (t) = 2 1 cos t + sin t. 5 5 195 Chapter 4
35. We look for a particular solution of the form yp (t) = ts (A1 t + A0 )et , and choose s = 1 because the auxiliary equation, r 3 + r 2  2 = (r  1)(r 2 + 2r + 2) = 0 has r = 1 as a simple root. Hence, yp (t) = t(A1 t + A0 )et = (A1 t2 + A0 t)et yp (t) = A1 t2 + (2A1 + A0 )t + A0 et yp (t) = A1 t2 + (4A1 + A0 )t + (2A1 + 2A0 ) et yp (t) = A1 t2 + (6A1 + A0 )t + (6A1 + 3A0 ) et y + y  2y = [10A1 t + (8A1 + 5A0 )] et = tet . Equating the corresponding coefficients, we find that 10A1 = 1, 8A1 + 5A0 = 0 EXERCISES 4.5: A1 = 1/10, A0 = 8A1 /5 = 4/25 yp (t) = 1 2 4 t  t et . 10 25 The Superposition Principle and Undetermined Coefficients Revisited, page 192 1. Let g1 (t) := sin t and g2 (t) := e2t . Then y1 (t) = cos t is a solution to y  y + y = g1 (t) and y2 (t) = e2t /3 is a solution to y  y + y = g2 (t). (a) The righthand side of the given equation is 5 sin t = 5g1 (t). Therefore, the function y(t) = 5y1 (t) = 5 cos t is a solution to y  y + y = 5 sin t. (b) We can express sin t3e2t = g1 (t) 3g2 (t). So, by the superposition principle the desired solution is y(t) = y1 (t)  3y2 (t) = cos t  e2t . (c) Since 4 sin t + 18e2t = 4g1 (t) + 18g2 (t), the function y(t) = 4y1(t) + 18y2(t) = 4 cos t + 6e2t is a solution to the given equation. 196 Exercises 4.5
3. The corresponding homogeneous equation, y  y = 0, has the associated auxiliary equation r 2  1 = (r  1)(r + 1) = 0. This gives r = 1 as the roots of this equation, and a general solution to the homogeneous equation is yh (t) = c1 et + c2 et . Combining this solution with the particular solution, yp (t) = t, we find that a general solution is given by y(t) = yp (t) + yh (t) = t + c1 et + c2 et . 5. The corresponding auxiliary equation, r 2  r  2 = 0, has the roots r = 1, 2. Hence, a general solution to the corresponding homogeneous equation is h (t) = c1 e2t + c2 et . By the superposition principle, a general solution to the original nonhomogeneous equation is (t) = p (t) + h (t) = t  1 + c1 e2t + c2 et . 7. First, we rewrite the equation in standard form, that is, y  2y + y = 2ex . The corresponding homogeneous equation, y  2y + y = 0, has the associated auxiliary equation r 2  2r + 1 = (r  1)2 = 0. Thus r = 1 is its double root, and a general solution to the homogeneous equation is yh (x) = c1 xex + c2 ex . Combining this with the particular solution, yp (x) = x2 ex , we find that a general solution is given by y(x) = yp (x) + yh (x) = x2 ex + c1 xex + c2 ex . 9. We can write the nonhomogeneous term as a difference t2 + 4t  t2 et sin t = (t2 + 4t)  (t2 et sin t) = g1 (t)  g2 (t). Both, g1 (t) and g2 (t), have a form suitable for the method of undetermined coefficients. Therefore, we can apply this method to find particular solutions yp,1(t) and yp,2 (t) to 3y + 2y + 8y = g1 (t) and 3y + 2y + 8y = g2 (t), respectively. Then, by the superposition principle, yp (t) = yp,1(t)  yp,2(t) is a particular solution to the given equation. 197 Chapter 4
11. The answer is "no", because the method of undetermined coefficients cannot be applied to y  6y  4y = 1 . t 13. In the original form, the function sin2 t does not fit any of the cases in the method of undetermined coefficients. But it can be written as sin2 t = (1  cos 2t)/2, and so 2t + sin2 t + 3 = 2t + 1  cos 2t +3= 2 2t + 7 2  1 cos 2t . 2 Now, the method of undetermined coefficients can be applied to each term in the above difference to find a particular solution to the corresponding nonhomogeneous equation, and the difference of these particular solutions, by the superposition principle, is a particular solution to the original equation. Thus, the answer is "yes". 15. "No", because the given equation is not an equation with constant coefficients. 17. The auxiliary equation in this problem is r 2  1 = 0 with roots r = 1. Hence, yh (t) = c1 et + c2 et is a general solution to the corresponding homogeneous equation. Next, we find a particular solution yp (t) to the original nonhomogeneous equation. The method of undetermined coefficients yields yp (t) = At + B yp (t) A yp (t) 0; A = 11, B = 1 yp  yp = 0  (At + B) = At  B = 11t + 1 yp (t) = 11t  1. By the superposition principle, a general solution is given by y(t) = yp (t) + yh (t) = 11t  1 + c1 et + c2 et . 19. Solving the auxiliary equation, r 2  3r + 2 = 0, we find that r = 1, 2. Therefore, a general solution to the homogeneous equation, y  3y + 2y = 0, is yh (x) = c1 ex + c2 e2x . 198 Exercises 4.5
By the method of undetermined coefficients, a particular solution yp (x) to the original equation has the form yp (x) = xs (A cos x + B sin x)ex . We choose s = 0 because r = 1 + i is not a root of the auxiliary equation. So, yp (x) = (A cos x + B sin x)ex yp (x) = [(A + B) cos x + (B  A) sin x]ex yp (x) = (2B cos x  2A sin x)ex . Substituting these expressions into the equation, we compare the corresponding coefficients and find A and B. {(2B cos x  2A sin x)  3[(A + B) cos x + (B  A) sin x] + 2(A cos x + B sin x)} ex = ex sin x (A + B) cos x + (A  B) sin x = sin x A + B = 0, AB =1 A = 1/2, B = 1/2. Therefore, yp (x) = and y(x) = (cos x  sin x)ex 2 (cos x  sin x)ex + c1 ex + c2 e2x 2 is a general solution to the given nonhomogeneous equation. 21. Since the roots of the auxiliary equation, which is r 2 + 2r + 2 = 0, are r = 1 i, we have a general solution to the corresponding homogeneous equation yh () = c1 e cos + c2 e sin = (c1 cos + c2 sin ) e , and look for a particular solution of the form yp () = s (A cos + B sin )e Differentiating yp (), we get yp () = (A cos + B sin )e + (A cos + B sin )e , 199 with s = 1. Chapter 4
yp () = 2 (A cos + B sin )e + (A cos + B sin )e = 2 [(B  A) cos  (B + A) sin ] e + (A cos + B sin )e . (Note that we did not evaluate the terms containing the factor because they give zero result when substituted into the original equation.) Therefore, yp + 2yp + 2yp = 2 [(B  A) cos  (B + A) sin ] e + 2(A cos + B sin )e = 2 (B cos  A sin ) e = e cos . Hence A = 0, B = 1/2, yp () = (1/2)e sin , and a general solution is given by y() = 1  e sin + (c1 cos + c2 sin ) e . 2 23. The corresponding homogeneous equation, y  y = 0, is separable. Solving yields dy =y dt dy = dt y ln y = t + c y = ec et = Cet , where C = 0 is an arbitrary constant. By inspection, y 0 is also a solution. Therefore, yh (t) = Cet , where C is an arbitrary constant, is a general solution to the homogeneous equation. (Alternatively, one can apply the method of solving first order linear equations in Section 2.3 or the method discussed in Problem 21, Section 4.2.) A particular solution has the form yp (t) = A. Substitution into the original equation yields (A)  A = 1 A = 1. Thus y(t) = Cet  1 is a general solution. To satisfy the initial condition, y(0) = 0, we find 0 = y(0) = Ce0  1 = C  1 So, the answer is y(t) = et  1. 25. The auxiliary equation, r 2 + 1 = 0, has roots r = i. Therefore, a general solution to the corresponding homogeneous equation is zh (x) = c1 cos x + c2 sin x, and a particular solution 200 C = 1. Exercises 4.5
to the original equation has the form zp (x) = Aex . Substituting this function into the given equation, we find the constant A. z + z = Aex + Aex = 2Aex = 2ex A = 1, and a general solution to the given nonhomogeneous equation is z(x) = ex + c1 cos x + c2 sin x . Next, since z (x) = ex  c1 sin x + c2 cos x, from the initial conditions we get a system for determining constants c1 and c2 . 0 = z(0) = 1 + c1 , 0 = z (0) = 1 + c2 c1 = 1, c2 = 1. Hence, z = (x) = ex  cos x + sin x is the solution to the given initial value problem. 27. The roots of the auxiliary equation, r 2  r  2 = 0, are r = 1 and r = 2. This gives a general solution to the corresponding homogeneous equation of the form yh (x) = c1 ex + c2 e2x . We use the superposition principle to find a particular solution to the nonhomogeneous equation. (i) For the equation y  y  2y = cos x, a particular solution has the form yp,1(x) = A cos x + B sin x. Substitution into the above equation yields (A cos x  B sin x)  (A sin x + B cos x)  2(A cos x + B sin x) = (3A  B) cos x + (A  3B) sin x = cos x 3A  B = 1, A  3B = 0 A = 3/10, B = 1/10. So, yp,1(x) = (3/10) cos x  (1/10) sin x. 201 Chapter 4
(ii) For the equation y  y  2y = sin 2x, a particular solution has the form yp,2(x) = A cos 2x + B sin 2x. Substitution yields (4A cos 2x  4B sin 2x)  (2A sin 2x + 2B cos 2x)  2(A cos 2x + B sin 2x) = (6A  2B) cos 2x + (2A  6B) sin 2x = sin 2x So, yp,2(x) = 3 1 cos 2x  sin 2x. 20 20 6A  2B = 0, 2A  6B = 1 A = 1/20, B = 3/20. Therefore, a general solution to the original equation is y(x) = yp,1(x)  yp,2(x) + yh (x) 3 1 1 3 =  cos x  sin x  cos 2x + sin 2x + c1 ex + c2 e2x . 10 10 20 20 Next, we find c1 and c2 such that the initial conditions are satisfied. 7/20 = y(0) = 3/10  1/20 + c1 + c2 , 1/5 = y (0) = 1/10 + 2(3/20)  c1 + 2c2 With these constants, the solution becomes y(x) =  1 1 3 3 cos x  sin x  cos 2x + sin 2x . 10 10 20 20 c1 + c2 = 0, c1 + 2c2 = 0 c1 = 0, c2 = 0. 29. The roots of the auxiliary equation, r 2  1 = 0, are r = 1. Therefore, a general solution to the corresponding homogeneous equation is yh () = c1 e + c2 e . (i) For the equation y  y = sin , 202 Exercises 4.5
a particular solution has the form yp,1(x) = A cos + B sin . Substitution into the equation yields (A cos  B sin )  (A cos + B sin ) = 2A cos  2B sin = sin 2A = 0, 2B = 1 A = 0, B = 1/2. So, yp,1() = (1/2) sin . (ii) For the equation y  y = e2 , a particular solution has the form yp,2() = Ae2 . Substitution yields Ae2 and yp,2() = (1/3)e2 . By the superposition principle, a particular solution to the original nonhomogeneous equation is given by yp () = yp,1()  yp,2() = (1/2) sin  (1/3)e2 , and a general solution is y() = yp () + yh () = (1/2) sin  (1/3)e2 + c1 e + c2 e . Next, we satisfy the initial conditions. 1 = y(0) = 1/3 + c1 + c2 , 1 = y (0) = 1/2  2/3 + c1  c2 c1 + c2 = 4/3, c1  c2 = 1/6 c1 = 3/4, c2 = 7/12.  Ae2 = 3Ae2 = e2 A = 1/3, Therefore, the solution to the given initial value problem is y() =  1 3 7  1 sin  e2 + e + e . 2 3 4 12 203 Chapter 4
31. For the nonhomogeneous term sin t + t cos t, a particular solution has the form yp,1(t) = (A1 t + A0 )ts cos t + (B1 t + B0 )ts sin t. For 10t = et ln 10 , a particular solution should be of the form yp,2(t) = Ctp et ln 10 = Ctp 10t . Since the roots of the auxiliary equation, r 2 + 1 = 0, are r = i, we choose s = 1 and p = 0. Thus, by the superposition principle, yp (t) = yp,1(t) + yp,2(t) = (A1 t + A0 )t cos t + (B1 t + B0 )t sin t + C 10t . 33. The roots of the auxiliary equation, which is r 2  r  2 = 0, are r = 1, 2. The righthand side of the given equation is a sum of two terms, et cos t and t2 + t + 1. Corresponding particular solutions have the forms yp,1(t) = (A cos t + B sin t)ts et and yp,2(t) = (C2 t2 + C1 t + C0 )tp , and we can take s = p = 0 since neither r = 1 + i nor r = 0 is a root of the auxiliary equation. By the superposition principle, yp (t) = (A cos t + B sin t)et + C2 t2 + C1 t + C0 . 35. Since the roots of the auxiliary equation are 4 16  20 = 2 i, r= 2 which are different from 5 and 3i, a particular solution has the form yp (t) = (A1 t + A0 ) cos 3t + (B1 t + B0 ) sin 3t + Ce5t . (The last term corresponds to e5t in the righthand side of the original equation, and the first two come from t sin 3t  cos 3t.) 204 Exercises 4.5
37. Clearly, r = 0 is not a root of the auxiliary equation, r 3  2r 2  r + 2 = 0. (One can find the roots, say, using the factorization r 3  2r 2  r + 2 = (r  2)(r  1)(r + 1), but they are not needed for the form of a particular solution: the only important thing is that they are different from zero.) Therefore, a particular solution has the form yp (t) = A2 t2 + A1 t + A0 . Substitution into the original equation yields yp  2yp  yp + 2yp = (0)  2(2A2 )  (2A2 t + A1 ) + 2(A2 t2 + A1 t + A0 ) = 2A2 t2 + (A1  2A2 )t + (A0  A1  4A2 ) = 2t2 + 4t  9. Equating the coefficients, we obtain 2A2 = 2, 2A1  2A2 = 4, 2A0  A1  4A2 = 9 Therefore, yp (t) = t2 + 3t  1. 39. The auxiliary equation in this problem is r 3 + r 2  2 = 0. By inspection, we see that r = 0 is not a root. Next, we find that r = 1 is a simple root because r3 + r2  2 =0
r=1 A2 = 1, A1 = 3, A0 = 1. and r3 + r2  2 = 3r 2 + 2r
r=1 r=1 = 0. Therefore, by the superposition principle, a particular solution has the form yp (t) = t(A1 t + A0 )et + B = (A1 t2 + A0 t)et + B. Differentiating, we get yp (t) = A1 t2 + (A0 + 2A1 )t + A0 et , yp (t) = A1 t2 + (A0 + 4A1 )t + 2A0 + 2A1 et , yp (t) = A1 t2 + (A0 + 6A1 )t + 3A0 + 6A1 et . 205 Chapter 4
We substitute yp and its derivatives into the original equation and equate the corresponding coefficients. This yields A1 t2 + (A0 + 6A1 )t + 3A0 + 6A1 + A1 t2 + (A0 + 4A1 )t + 2A0 + 2A1 2 A1 t2 + A0 t [10A1 t + 8A1 + 5A0 ] et  2B = tet + 1 10A1 = 1, 8A1 + 5A0 = 0, 2B = 1 Hence, a particular solution is yp (t) = 1 4 t 10 25 tet  1 . 2 A1 = 1/10, A0 = 4/25, B = 1/2. et  2B = tet + 1 41. The characteristic equation in this problem is r 2 + 2r + 5 = 0, which has roots r = 1 2i. Therefore, a general solution to the corresponding homogeneous equation is given by yh (t) = (c1 cos 2t + c2 sin 2t) et . (a) For 0 t 3/2, g(t) 10, and so the equation becomes y + 2y + 5y = 10. Hence a particular solution has the form yp (t) A. Substitution into the equation yields (A) + 2(A) + 5(A) = 10 5A = 10 A = 2, (4.1) and so, on [0, 3/2], a general solution to the original equation is y1 (t) = (c1 cos 2t + c2 sin 2t) et + 2. We find c1 and c2 by substituting this function into the initial conditions. 0 = y1 (0) = c1 + 2, 0 = y1 (0) = c1 + 2c2 206 c1 = 2, c2 = 1
t y1 (t) =  (2 cos 2t + sin 2t) e + 2. Exercises 4.5
(b) For t > 3/2, g(t) 0, and so the given equation becomes homogeneous. Thus, a general solution, y2 (t), is given by (4.1), i.e., y2 (t) = yh (t) = (c1 cos 2t + c2 sin 2t) et . (c) We want to satisfy the conditions y1 (3/2) = y2 (3/2), y1 (3/2) = y2 (3/2). Evaluating y1 , y2 , and their derivatives at t = 3/2, we solve the system 2e3/2 + 2 = c1 e3/2 , 0 = (c1  2c2 )e3/2 c1 = 2 e3/2 + 1 , c2 =  e3/2 + 1 . 43. Recall that the motion of a massspring system is governed by the equation my + by + ky = g(t), where m is the mass, b is the damping coefficient, k is the spring constant, and g(t) is the external force. Thus, we have an initial value problem y + 4y + 3y = 5 sin t, y(0) = 1 , 2 y (0) = 0. The roots of the auxiliary equation, r 2 + 4r + 3 = 0, are r = 3, 1, and a general solution to the corresponding homogeneous equation is yh (t) = c1 e3t + c2 et . We look for a particular solution to the original equation of the form yp (t) = A cos t + B sin t. Substituting this function into the equation, we get yp + 4yp + 3yp = (A cos t  B sin t) + 4(A sin t + B cos t) + 3(A cos t + B sin t) = (2A + 4B) cos t + (2B  4A) sin t = 5 sin t 2A + 4B = 0, 2B  4A = 5 A = 1, B = 1/2. 207 Chapter 4
Thus, a general solution to the equation describing the motion is y(t) =  cos t + 1 sin t + c1 e3t + c2 et . 2 Differentiating, we find y (t) = sin t + (1/2) cos t  3c1 e3t  c2 et . Initial conditions give y(0) = 1 + c1 + c2 = 1/2, y (0) = 1/2  3c1  c2 = 0 Hence, the equation of motion is y(t) =  cos t + 1 1 sin t  e3t + 2et . 2 2 c1 = 1/2, c2 = 2. 45. (a) With m = k = 1 and L = given initial value problem becomes y(t) = 0, y +y = , 2V cos V t, /(2V ) < t < /(2V ), t 0, t /(2V ) . The corresponding homogeneous equation y + y = 0 is the simple harmonic equation whose general solution is yh (t) = C1 cos t + C2 sin t . (4.2) First, we find the solution to the given problem for /(2V ) < t < /(2V ). The nonhomogeneous term, cos V t, suggests a particular solution of the form yp (t) = A cos V t + B sin V t. Substituting yp (t) into the equation yields (A cos V t + B sin V t) + (A cos V t + B sin V t) = cos V t V 2 A cos V t  V 2 B sin V t + (A cos V t + B sin V t) = cos V t 1  V 2 A cos V t + 1  V 2 B sin V t = cos V t . Equating coefficients, we get A= 208 1 , 1V2 B = 0, Exercises 4.5
Thus a general solution on (/(2V ), /(2V )) is y1 (t) = yh (t) + yp (t) = C1 cos t + C2 sin t + 1 cos V t . 1V2 (4.3) Since y(t) 0 for t /(2V ), the initial conditions for the above solution are y1  From (4.3) we obtain y1  y1 2V  2V + C2 sin  =0 2V 2V V = C1 sin  + C2 cos  + = 0. 2V 2V 1V2 = C1 cos  2V = y1  2V = 0. Solving the system yields C1 = and y1 (t) = V 1 V sin cos t + 2 cos sin t + cos V t V 1 2V V 1 2V 1V2 V 1 sin t + cos V t,  <t< . =  2 2 1 V 2V V 1 2V 2V
2 V sin , V 1 2V
2 C2 = V cos , V 1 2V
2 For t > /(2V ) given equation is homogeneous, and its general solution, y2 (t), is given by (4.2). That is, y2 (t) = C3 cos t + C4 sin t. From the initial conditions y2 y2 we conclude that C3 cos V + C4 sin = 2 sin , 2V 2V V 1 V 209 2V 2V = y1 = y1 2V 2V , , Chapter 4
A(V )
1.5 1 0.5 0 0.73
0.8 1 V 0.2 0.4 0.6 Figure 4A: The graph of the function A(V ). C3 sin V V 2V + C4 cos = 2 cos + 2 = 2 cos2 . 2V 2V V 1 V V 1 V 1 2V 2V cos . V 1 2V
2 The solution of this system is C3 = 0, So, y2 (t) = (b) The graph of the function 2V cos V 1 2V is given in Figure 4A. From this graph, we find that the most violent shaking of the A(V ) =
2 C4 = 2V cos sin t. V 1 2V
2 vehicle (the maximum of A(V )) happens when the speed V 0.73 . 47. The auxiliary equation in this problem is r 2 + 9 = 0 with roots r = 3i. So, a general solution to the corresponding homogeneous equation is yh = c1 cos 3t + c2 sin 3t. The form of a particular solution, corresponding to the righthand side, is yp (t) = A cos 6t + B sin 6t. 210 Exercises 4.6
Substitution into the original equation yields 27(A cos 6t + B sin 6t) = 27 cos 6t A = 1, B = 0 yp (t) =  cos 6t. Therefore, a general solution has the form y(t) = c1 cos 3t + c2 sin 3t  cos 6t. In (a)(c), we have the same boundary condition at t = 0, that is, y(0) = 1. This yields 1 = y(0) = c1  1 c1 = 0. Hence, all the solutions satisfying this condition are given by y(t) = c2 sin 3t  cos 6t. (a) The second boundary condition gives 3 = y (/6) = c2 + 1 is y = 2 sin 3t  cos 6t. (b) This time we have 5 = y (/3) = c2 0  1 (c) Now we have 1 = y (/3) = c2 0  1 5 = 1, and so there is no solution of 1 = 1, which is a true identity. This (4.4) c2 = 2, and the answer the form (4.4) satisfying this second boundary condition. means that any function in (4.4) satisfies both boundary conditions. EXERCISES 4.6: Variation of Parameters, page 197 1. The auxiliary equation in this problem is r 2 + 4 = 0, which has the roots r = 2i. Therefore, y1 (t) = cos 2t and y2 (t) = sin 2t are two linearly independent solutions, and a general solution to the corresponding homogeneous equation is given by yh (t) = c1 cos 2t + c2 sin 2t. Using the variation of parameters method, we look for a particular solution to the original nonhomogeneous equation of the form yp (t) = v1 (t)y1 (t) + v2 (t)y2 (t) = v1 (t) cos 2t + v2 (t) sin 2t. 211 Chapter 4
The system (9) on page 195 in the text becomes v1 (t) cos 2t + v2 (t) sin 2t = 0 2v1 (t) sin 2t + 2v2 (t) cos 2t = tan 2t. (4.5) Multiplying the first equation in (4.5) by sin 2t, the second equation by (1/2) cos 2t, and adding the resulting equations together, we get v2 (t) = 1 sin 2t 2 v2 = 1 2 sin 2t dt =  1 cos 2t + c3 . 4 From the first equation in (4.5) we also obtain v1 (t) = v2 (t) tan 2t =  v1 (t) = 1 2 1 1  cos2 2t 1 1 sin2 2t = = (cos 2t  sec 2t) 2 cos 2t 2 cos 2t 2 1 (cos 2t  sec 2t) dt = (sin 2t  ln  sec 2t + tan 2t) + c4 . 4 We take c3 = c4 = 0 since we need just one particular solution. Thus yp (t) = 1 1 (sin 2t  ln  sec 2t + tan 2t) cos 2t  cos 2t sin 2t 4 4 1 =  cos 2t ln  sec 2t + tan 2t 4 and a general solution to the given equation is y(t) = yh (t) + yp (t) = c1 cos 2t + c2 sin 2t  1 cos 2t ln  sec 2t + tan 2t. 4 2. From Example 1 on page 196 in the text, we know that functions y1 (t) = cos t and y2 (t) = sin t are two linearly independent solutions to the corresponding homogeneous equation, and so its general solution is given by yh (t) = c1 cos t + c2 sin t. Now we apply the method of variation of parameters to find a particular solution to the original equation. By the formula (3) on page 194 in the text, yp (t) has the form yp (t) = v1 (t)y1 (t) + v2 (t)y2 (t). 212 Exercises 4.6
Since y1 (t) = (cos t) =  sin t, the system (9) on page 195 becomes v1 (t) cos t + v2 (t) sin t = 0, v1 (t) sin t + v2 (t) cos t = sec t. Multiplying the first equation by sin t and the second equation by cos t yields v1 (t) sin t cos t + v2 (t) sin2 t = 0, v1 (t) sin t cos t + v2 (t) cos2 t = 1. Adding these equations together, we obtain v2 (t) cos2 t + sin2 t = 1 or v2 (t) = 1. (4.6) y2 (t) = (sin t) = cos t, From the first equation in (4.6), we can now find v1 (t): v1 (t) = v2 (t) So, v1 (t) =  tan t, v2 (t) = 1 v1 (t) =  tan t dt = ln  cos t + c3 , v2 (t) = dt = t + c4 . sin t =  tan t. cos t Since we are looking for a particular solution, we can take c3 = c4 = 0 and get yp (t) = cos t ln  cos t + t sin t. Thus a general solution to the given equation is y(t) = yp (t) + yh (t) = cos t ln  cos t + t sin t + c1 cos t + c2 sin t. 3. First, we can simplify the equation by dividing both sides by 2. This yields x  x  2x = e3t . 213 Chapter 4
This equation has associated homogeneous equation x  x  2x = 0. The roots of the associated auxiliary equation, r 2  r  2 = 0, are r = 2 and r = 1. Therefore, a general solution to this equation is xh (t) = c1 e2t + c2 et . For the variation of parameters method, we let xp (t) = v1 (t)x1 (t) + v2 (t)x2 (t) , where x1 (t) = e2t and x2 (t) = et . Thus, x1 (t) = 2e2t and x2 (t) = et . This means that we have to solve the system e2t v1 + et v2 = 0, 2e2t v1  et v2 = e3t . Adding these two equations yields 3e2t v1 = e3t v1 = 1 t e 3 v1 (t) = 1 t e . 3 Substututing v1 into the first equation, we get 1 3t e + et v2 = 0 3 Therefore, xp (t) = and a general solution is x(t) = c1 e2t + c2 et + 1 3t e . 4 1 t 2t 1 4t t 1 3t ee  e e = e , 3 12 4 1 v2 =  e4t 3 v2 (t) =  1 4t e . 12 5. This equation has associated homogeneous equation y  2y + y = 0. Its auxiliary equation, r 2  2r + 1 = 0, has a double root r = 1. Thus a general solution to the homogeneous equation is yh (t) = c1 et + c2 tet . For the variation of parameters method, we let yp (t) = v1 (t)y1 (t) + v2 (t)y2 (t) , 214 where y1 (t) = et and y2 (t) = tet . Exercises 4.6
Thus, y1 (t) = et and y2 (t) = tet + et . This means that we want to solve the system (see system (9) on page 195 of text) et v1 + tet v2 = 0, et v1 + tet + et v2 = t1 et . Subtracting these two equations yields et v2 = t1 et So v2 (t) = t1 dt = ln t + c3 . v2 = t1 . Also, we have from the first equation of the system et v1 = tet v2 = tet t1 = et So, v1 (t) = t + c4 . By letting c3 and c4 equal to zero, and plugging the expressions found above for v1 (t) and v2 (t) into the equation defining yp (t) , we obtain a particular solution yp (t) = tet + tet ln t. We obtain a general solution of the nonhomogeneous equation by adding this expression for yp (t) to the expression for yh (t). Thus, we obtain y(t) = c1 et + c2 tet  tet + tet ln t = c1 et + (c2  1)tet + tet ln t. If we let C1 = c1 and C2 = c2  1, we can express this general solution in the form y(t) = C1 et + C2 tet + tet ln t. 215 v1 = 1. Chapter 4
7. The auxiliary equation in this problem is r 2 + 16 = 0, which has the roots r = 4i. Therefore, y1 () = cos 4 and y2 () = sin 4 are two linearly independent solutions, and a general solution to the corresponding homogeneous equation is given by yh () = c1 cos 4 + c2 sin 4. Using the variation of parameters method, we look for a particular solution to the original nonhomogeneous equation of the form yp () = v1 ()y1 () + v2 ()y2 () = v1 () cos 4 + v2 () sin 4. The system (9) on page 195 in the text becomes v1 () cos 4 + v2 () sin 4 = 0, 4v1 () sin 4 + 4v2 () cos 4 = sec 4. (4.7) Multiplying the first equation in (4.7) by sin 4 and the second equation by (1/4) cos 4, and adding the resulting equations together, we get v2 () = 1 4 v2 = 1 + c3 . 4 From the first equation in (4.7) we also obtain 1 v1 () =  tan 4 4 Taking c3 = c4 = 0, we obtain yp () = 1 cos 4 ln  cos 4 + sin 4 16 4 cos 4 y() = c1 cos 4 + c2 sin 4 + sin 4 + ln  cos 4. 4 16 v1 () =  1 4 tan 4 d = 1 ln  cos 4 + c4 . 16 9. In this problem, the corresponding homogeneous equation is the same as that in Problem 1. Hence y1 (t) = cos 2t and y2 (t) = sin 2t are two linearly independent solutions, and a general solution to the homogeneous equation is given by yh (t) = c1 cos 2t + c2 sin 2t, 216 Exercises 4.6
and, in the variation of parameters method, a particular solution has the form yp (t) = v1 (t) cos 2t + v2 (t) sin 2t, where v1 (t), v2 (t) satisfy the system v1 (t) cos 2t + v2 (t) sin 2t = 0, 2v1 (t) sin 2t + 2v2 (t) cos 2t = csc2 2t. Multiplying the first equation by sin 2t and the second equation by (1/2) cos 2t, and adding the resulting equations, we get v2 (t) = 1 csc2 2t cos 2t 2 v2 = 1 2 csc2 2t cos 2t dt =  1 csc 2t + c3 . 4 From the first equation in the system above we also find v1 (t) = v2 (t) tan 2t =  With c3 = c4 = 0, yp (t) = 1 1 1 cos 2t ln  csc 2t + cot 2t  csc 2t sin 2t = (cos 2t ln  csc 2t + cot 2t  1) 4 4 4 1 y(t) = c1 cos 2t + c2 sin 2t + (cos 2t ln  csc 2t + cot 2t  1) . 4 v1 (t) =  1 2 1 1 csc2 2t cos 2t tan 2t =  csc 2t 2 2 1 csc 2t dt = ln  csc 2t + cot 2t + c4 . 4 11. This equation is similar to that in Example 1 on page 196 in the text. Only the nonhomogeneous term is different. Thus we will follow steps in Example 1. Two independent solutions to the corresponding homogeneous equation, y + y = 0, are y1 (t) = cos t and y2 (t) = sin t. A particular solution to the original equation is of the form yp (t) = v1 (t) cos t + v2 (t) sin t, where v1 (t) and v2 (t) satisfy v1 (t) cos t + v2 (t) sin t = 0, v1 (t) sin t + v2 (t) cos t = tan2 t. 217 Chapter 4
Multiplying the first equation by sin t and the second equation by cos t, and adding them together yield v2 (t) = tan2 t cos t = (sec2 t  1) cos t = sec t  cos t. We find v1 (t) from the first equation in the system. v1 (t) = v2 (t) tan t = (sec t  cos t) tan t = sin t  Integrating, we get v1 (t) = v2 (t) = sin t  sin t cos2 t dt =  cos t  sec t, sin t . cos2 t (sec t  cos t) dt = ln  sec t + tan t  sin t, where we have taken zero integration constants. Therefore, yp (t) = (cos t + sec t) cos t + (ln  sec t + tan t  sin t) sin t = sin t ln  sec t + tan t  2, and a general solution is given by y(t) = c1 cos t + c2 sin t + sin t ln  sec t + tan t  2. 13. The corresponding homogeneous equation in this problem is the same as that in Problem 1 (with y replaced by v). Similarly to the solution of Problem 1, we conclude that v1 (t) = cos 2t and v2 (t) = sin 2t are two linearly independent solutions of the corresponding homogeneous equation, and a particular solution to the original equation can be found as vp (t) = u1 (t) cos 2t + u2 (t) sin 2t , where u1 (t) and u2 (t) satisfy u1 (t) cos 2t + u2 (t) sin 2t = 0, 2u1 (t) sin 2t + 2u2(t) cos 2t = sec4 2t. Multiplying the first equation by sin 2t and the second equation by (1/2) cos 2t, and adding the results together, we get u2 (t) = 218 1 sec3 2t. 2 Exercises 4.6
From the first equation in the above system we also obtain u1 (t) = u2 (t) tan 2t =  Integrating yields u1 (t) =  u2 (t) = Thus, vp (t) =  1 1 sec3 2t cos 2t + (sec 2t tan 2t + ln  sec 2t + tan 2t) sin 2t 12 8 1 1 1 2 2 =  sec 2t + tan 2t + sin 2t ln  sec 2t + tan 2t 12 8 8 1 1 1 2 sec 2t  + sin 2t ln  sec 2t + tan 2t, = 24 8 8 1 1 1 sec2 2t  + sin 2t ln  sec 2t + tan 2t. 24 8 8 1 2 1 2 sec4 2t sin 2t dt =  sec3 2t dt = 1 2 cos4 2t sin 2t dt =  1 sec3 2t, 12 1 sec4 2t sin 2t . 2 1 (sec 2t tan 2t + ln  sec 2t + tan 2t). 8 and a general solution to the given equation is v(t) = c1 cos 2t + c2 sin 2t + 15. The corresponding homogeneous equation is y + y = 0. Its auxiliary equation has the roots r = i. Hence, a general solution to the homogeneous problem is given by yh (t) = c1 cos t + c2 sin t. We will find a particular solution to the original equation by first finding a particular solution for each of two problems, one with the nonhomogeneous term g1 (t) = 3 sec t and the other one with the nonhomogeneous term g2 (t) = t2 + 1. Then we will use the superposition principle to obtain a particular solution for the original equation. The term 3 sec t is not in a form that allows us to use the method of undetermined coefficients. Therefore, we will use the method of variation of parameters. To this end, let y1 (t) = cos t and y2 (t) = sin t (linearly independent solutions to the corresponding homogeneous problem). Then a particular solution yp,1 to y + y = 3 sec t has the form yp,1(t) = v1 (t)y1 (t) + v2 (t)y2 (t) = v1 (t) cos t + v2 (t) sin t, 219 Chapter 4
where v1 (t) and v2 (t) are determined by the system v1 cos t + v2 sin t = 0, v1 sin t + v2 cos t = 3 sec t. Multiplying the first equation by cos t and the second equation by sin t and subtracting the results, we get v1 = 3 sec t sin t = 3 tan t. Hence v1 (t) = 3 tan t dt = 3 ln  cos t + C1 . To find v2 (t), we multiply the first equation of the above system by sin t, the second by cos t, and add the results to obtain v2 = 3 sec t cos t = 3 v2 (t) = 3t + C2 . Therefore, for this first equation (with g1 (t) = 3 sec t), by letting C1 = C2 = 0, we have a particular solution given by yp,1(t) = 3 cos t ln  cos t + 3t sin t. The nonhomogeneous term g2 (t) = t2 + 1 is of a form that allows us to use the method of undetermined coefficients. Thus, a particular solution to this nonhomogeneous equation will have the form yp,2(t) = A2 t2 + A1 t + A0 yp,2 (t) = 2A2 t + A1 yp,2 (t) = 2A2 . Plugging these expressions into the equation y + y = t2 + 1 yields yp,2 + yp,2 = 2A2 + A2 t2 + A1 t + A0 = A2 t2 + A1 t + (2A2 + A0 ) = t2 + 1. By equating coefficients, we obtain A2 = 1, 220 A1 = 0, 2A2 + A0 = 1 A0 = 3. Exercises 4.6
Therefore, we have yp,2(t) = t2 + 3. By the superposition principle, we see that a particular solution to the original problem is given by yp (t) = yp,1(t) + yp,2 (t) = 3 cos t ln  cos t + 3t sin t  t2 + 3. Combining this solution with the general solution to the homogeneous equation yields a general solution to the original differential equation, y(t) = c1 cos t + c2 sin t  t2 + 3 + 3t sin t + 3 cos t ln  cos t. 17. Multiplying the given equation by 2, we get y + 4y = 2 tan 2t  et . The nonhomogeneous term, 2 tan 2t  et , can be written as a linear combination 2g1 (t)  g2 (t), where g1 (t) = tan 2t and g2 (t) = et . A particular solution to the equation y + 4y = tan 2t is found in Problem 1, that is, 1 yp,1(t) =  cos 2t ln  sec 2t + tan 2t. 4 A particular solution to y + 4y = et can be found using the method of undetermined coefficients. We look for yp,2 of the form yp,2(t) = Aet . Substitution yields Aet + 4 Aet = et 5Aet = et A= 1 , 5 and so yp,2 = (1/5)et . By the superposition principle, a particular solution to the original equation is 1 1 yp (t) = 2yp,1  yp,2 =  cos 2t ln  sec 2t + tan 2t  et . 2 5 221 Chapter 4
Adding a general solution to the homogeneous equation, we get y(t) = c1 cos 2t + c2 sin 2t  1 1 cos 2t ln  sec 2t + tan 2t  et . 2 5 19. A general solution of the corresponding homogeneous equation is given by yh (t) = c1 et + c2 et . We will try to find a particular solution to the original nonhomogeneous equation of the form yp (t) = v1 (t)y1 (t) + v2 (t)y2 (t), where y1 (t) = et and y2 (t) = et . We apply formulas (10) on page 195 in the text, but replace indefinite integrals by definite integrals. Note that y1 (t)y2 (t)  y1 (t)y2 (t) = ex ex  ex ex = 2. With g(t) = 1/t and integration from 1 to t, formulas (10) yield
t v1 (t) =
1 t g(x)y2 (x) 1 dx =  2 2
t t 1 ex dx , x v2 (t) =
1 g(x)y1 (x) 1 dx = 2 2 1 ex dx . x (Notice that we have chosen the lower limit of integration to be equal to 1 because the initial conditions are given at 1. We could have chosen any other value for the lower limit, but the choice of 1 will make the determination of the constants c1 and c2 easier.) Thus et yp (t) = 2
t t 1 et ex dx  x 2 1 ex dx , x and so a general solution to the original differential equation is y(t) = c1 e
t et + c2 e + 2
t t 1 et ex dx  x 2 t 1 ex dx . x By plugging in the first initial condition (and using the fact that the integral of a function from a to a is zero which is why we have chosen the lower limit of integration to be the initial point, t = 1), we find that y(1) = c1 e1 + c2 e1 = 0. 222 Exercises 4.6
Differentiating y(t) yields y (t) = c1 e
t et + c2 e + 2
t t 1 ex dx + x et 2 et t et + 2 t 1 ex dx  x et 2 et t , where we have used the product rule and the fundamental theorem of calculus to differentiate the last two terms of y(t). We now plug in the second initial condition into the equation we just found for y (t) to obtain y (1) = c1 e1 + c2 e1 + Solving the system c1 e1 + c2 e1 = 0, c1 e1 + c2 e1 = 2 yields c2 = e1 and c1 = e1 . Therefore, the solution to our problem is given by y(t) = e
1t e1 2 e1 1 + e1 2 e1 1 = c1 e1 + c2 e1  1 1 + = 2. 2 2 e t1 et + 2 t 1 et ex dx  x 2 t 1 ex dx . x (4.8) Simpson's rule is implemented on the software package provided free with the text (see also the discussion of the solution to Problem 25 in Exercises 2.3). Simpson's rule requires an even number of intervals, but we don't know how many are required to obtain the 2place accuracy desired. We will compute the approximate value of y(t) at t = 2 using 2, 4, 6, . . . intervals for Simpson's rule until the approximate value changes by less than five in the third place. For n = 2, we divide [1, 2] into 4 equal subintervals. Thus each interval will be of length (2  1)/4 = 1/4. Therefore the integrals are approximated by
2 1 2 1 e1 e1.25 e1.5 e1.75 e2 ex dx +4 +2 +4 + 3.0592 , x 12 1 1.25 1.5 1.75 2 1 e1 e1.25 e1.5 e1.75 e2 ex dx +4 +2 +4 + 0.1706 . x 12 1 1.25 1.5 1.75 2 223 1 Chapter 4
Substituting these values into equation (4.8) we obtain y(2) e
12 e 21 e2 e2 (3.0592) + (0.1706) = 1.9271 .  2 2 Repeating these calculations for n = 3, 4, and 5 yields the approximations in Table 4A. Table 4A: Successive approximations for y(2) using Simpson's rule.
Intervals 6 8 10 y(2) 1.9275 1.9275 1.9275 Since these values do not change in the third place, we can expect that the first three places are accurate and we obtained an approximate solution of y(2) = 1.93 . 21. A particular solution to the given equation has the form yp (t) = v1 (t)y1 (t) + v2 (t)y2 (t) = v1 (t)et + v2 (t)(t + 1). Since y1 (t) = et , y2 (t) 1, the system (9), with a = a(t) = t and g(t) = t2 , becomes v1 (t)et + v2 (t)(t + 1) = 0, t2 t = t. v1 (t)e + v2 (t) = t Subtracting the second equation from the first one, we get tv2 (t) = t v2 (t) = 1 v2 (t) = t. Substituting v2 (t) into the first equation yields v1 (t)et  (t + 1) = 0 224 v1 (t) = v1 (t) = (t + 1)et et dt = (t + 2)et . (t + 1)et dt = (t + 1)et + Exercises 4.6
Thus yp (t) = (t + 2)et et  t(t + 1) = t2  2t  2. (Note that 2t  2 = 2(t + 1) = 2y2 (t) is a solution to the corresponding homogeneous equation. Thus, t2 = yp (t) + 2y2 (t) is another particular solution.) 23. We are seeking for a particular solution to the given equation of the form yp (t) = v1 (t)y1 (t) + v2 (t)y2 (t) = v1 (t)(5t  1) + v2 (t)e5t . Since y1 (t) 5, y2 (t) = 5e5t , the system (9), with a = a(t) = t and g(t) = t2 e5t , becomes v1 (t)(5t  1) + v2 (t)e5t = 0, t2 e5t 5v1 (t)  5v2 (t)e5t = = te5t . t Dividing the second equation by 5 and adding to the first equation yields 5tv1 (t) = 1 5t te 5 v1 (t) = 1 5t e 25 v1 (t) =  1 5t e . 125 Substituting v1 (t) into the first equation, we get 1 5t e (5t  1) + v2 (t)e5t = 0 25 Thus yp (t) =  5t  1 v2 (t) =  25 e5t = 1 t2  125 10 t2 t v2 (t) =  + . 10 25 e5t . t 1 5t t2 e (5t  1) +  + 125 10 25 (Since (1/125)e5t = (1/125)y2(t) is a solution to the corresponding homogeneous equation, the function (t2 /10)e5t is also a particular solution.) 25. A general solution to the corresponding homogeneous equation is yh (x) = c1 y1 (x) + c2 y2 (x) = x1/2 (c1 cos x + c2 sin x) . To find a particular solution to the original equation, we apply the method of variation of parameters. To form the system (9) on page 195, we need y1 and y2 . Applying the product rule, we get 1 y1 (x) =  x3/2 cos x  x1/2 sin x, 2 225 Chapter 4
1 y2 (x) =  x3/2 sin x + x1/2 cos x. 2 Thus, functions v1 (x) and v2 (x) in a particular solution, yp (x) = v1 (x)y1 (x) + v2 (x)y2 (x), satisfy the system v1 x1/2 cos x + v2 x1/2 sin x = 0, 1 1 v1  x3/2 cos x  x1/2 sin x + v2  x3/2 sin x + x1/2 cos x 2 2 = x5/2 = x1/2 . x2 From the first equation, we express v1 = v2 tan x and substitute this expression into the second equation. After some algebra, the result simplifies to v2 = x cos x Integrating, we get v1 (x) =  v2 (x) = With C1 = C2 = 0, yp (x) = (x cos x  sin x)x1/2 cos x + (x sin x + cos x)x1/2 sin x = x1/2 . Therefore, a general solution to the given nonhomogeneous Bessel equation is y(t) = x1/2 + x1/2 (c1 cos x + c2 sin x) . EXERCISES 4.7: Qualitative Considerations for VariableCoefficient and Nonlinear Equations, page 208 1. Let Y (t) := y(t). Then, using the chain rule, we get dY d(t) = y (t) = y (t), dt dt 226 x sin x dx = x cos x  sin x + C1 , x cos x dx = x sin x + cos x + C2 . v1 = v2 tan x = x sin x. Exercises 4.7
d2 Y d[y (t)] d(t) = y (t) = y (t). = 2 dt dt dt Therefore, denoting t = s, we obtain Y (t) + tY (t) = y (t) + ty(t) = y (s)  sy(s) = 0. 2. Comparing the given equation with (13) on page 202 in the text, we conclude that inertia m = 1, damping b = 0, stiffness "k" = 6y. For y > 0, the stiffness "k" is negative, and it tends to reinforce the displacement. So, we should expect that the solutions y(t) grow without bound. 3. As in Problem 2, this equation describes the motion of the massspring system with unit mass, no damping, and stiffness "k" = 6y. The initial displacement y(0) = 1 is negative as well as the initial velocity y (0) = 1. So, starting from t = 0, y(t) will decrease for a while. This will result increasing positive stiffness, 6y, i.e., "the spring will become stiffer and stiffer". Eventually, the spring will become so strong that the mass will stop and then go in the positive direction. While y(t) is negative, the positive stiffness will force the mass to approach zero displacement point, y = 0. Thereafter, with y(t) > 0, the stiffness becomes negative, which means that the spring itself will push the mass further away from y = 0 in the positive direction with force, which increases with y. Thus, the curve y(t) will increase unboundedly. Figure 4.23 confirms our prediction. 5. (a) Comparing the equation y = 2y 3 with equation (7) in Lemma 3, we conclude that f (y) = 2y 3 , and so 1 4 y + C, 2 where C is a constant. We can choose any particular value for C, say, C = 0. Thus F (y) = 2y 3 dy = F (y) = (1/2)y 4. Next, with constant K = 0 and sign "" in front of the integral, equation (11) on page 201, becomes t= dy = 2(1/2)y 4 y 2 dy = y 1 + c, 227 Chapter 4
or, equivalently, y= where c is an arbitrary constant. (b) A linear combination of y1 (t) := 1/(t  c1 ) and y2 (t) := 1/(t  c2 ), C1 y1 (t) + C2 y2 (t) = C1 C2 (C1 + C2 )t  (C1 c2 + C2 c1 ) , + = t  c1 t  c2 (t  c1 )(t  c2 ) 1 , tc is identically zero in a neighborhood of t = 0 if and only if (C1 + C2 )t(C1 c2 + C2 c1 ) 0. Thus the numerator must be the zero polynomial, i.e., C1 and C2 must satisfy C1 + C2 = 0, C 1 c2 + C 2 c1 = 0 C2 = C1 , C1 (c2  c1 ) = 0. Since c1 = c2 , the second equation implies that C1 = 0, and then C2 = 0 from the first equation. Thus, only the trivial linear combination of y1 (t) and y2 (t) vanishes identically around the origin, and so these functions are linearly independent. (c) For any function of the form yc (t) := 1/(t  c), the equality yc (t) =  1 =  [yc (t)]2 (t  c)2 holds for all t = c. In particular, at t = 0, yc (0) =  [yc (0)]2 . (We assume that c = 0; otherwise, t = 0 is not in the domain.) Obviously, this equality fails for any positive initial velocity y (0), in particular, it is false for given data, y(0) = 1 and y (0) = 2. 6. Rewriting given equation in the equivalent form y = (k/m)y, we see that the function f (y) in the energy integral lemma is (k/m)y. So, F (y) = 228  k 2 k y dy =  y + C. m 2m Exercises 4.7
With C = 0, F (y) = [k/(2m)]y 2 , and the energy E(t) = 1 1 k 2 2 2 [y (t)]  F [y(t)] = [y (t)]   y 2 2 2m = k 2 1 2 [y (t)] + y . 2 2m By the energy integral lemma, k 2 1 2 [y (t)] + y = const. 2 2m Multiplying both sides by 2m, we get the stated equation. 7. (a) Since, for a point moving along a circle of radius , the magnitude v of its linear velocity v and the angular velocity = d/dt are connected by v = = (d/dt) , and the vector v is tangent to the circle (and so, perpendicular to the radius), we have angular momentum = mv = m d =m dt
2 d . dt (b) From Figure 4.18, we see that the component of the gravitational force, mg, which is perpendicular to the level arm, has the magnitude mg sin  and is directed towards decreasing . Thus, torque = (mg sin ) =  mg sin . (c) According to the Newton's law of rotational motion, torque = d d (angular momentum)  mg sin = dt dt d2 d2 g + sin = 0.  mg sin = m 2 2 dt dt2 = g, the function (t) satisfies the identity ( )2  cos = C = const. 2 (4.9) m
2 d dt 9. According to Problem 8, with Our first purpose is to determine the constant C. Let ta denote the moment when pendulum is in the apex point, i.e., (ta ) = . Since it doesn't cross the apex over, we also have (ta ) = 0. Substituting these two values into (4.9), we obtain 02  cos = C 2 C = 1. 229 Chapter 4
( )2  cos = 1. 2 In particular, at the initial moment, t = 0, [ (0)]2  cos[(0)] = 1. 2 Since (0) = 0, we get [ (0)]2 2  cos 0 = 1 [ (0)] = 4 2 or (0) = 2. (0) = 2 11. The "damping coefficient" in the Rayleigh equation is b = (y )2  1. Thus, for low velocities y , we have b < 0, and b > 0 for high velocities. Therefore, the low velocities are boosted, while high velocities are slowed, and so one should expect a limit cycle. 13. Qualitative features of solutions to Airy, Duffing, and van der Pol equations, are discussed after Example 3, in Examples 6 and 7, respectively. Comparing curves in Figure 4.26 with graphs depicted in Figures 4.13, 4.16, and 4.17, we conclude that the answers are (a) Airy; (b) Duffing; (c) van der Pol. 15. (a) Yes, because the "stiffness" t2 is positive and no damping. (b) No, because of the negative "stiffness" t2 . (c) Writing y + y 5 = y + (y 4 )y, we conclude that the massspring model, corresponding to this equation, has positive "stiffness" y 4 and no damping. Thus the answer is "yes". (d) Here, the "stiffness" is y 5 , which is negative for y < 0. So, "no". (e) Yes, because the "stiffness" 4 + 2 cos t 2 > 0 and no damping. (f) Since both the "damping" t and the stiffness 1 are positive, all solutions are bounded. 230 Thus (4.9) becomes Exercises 4.7
(g) No, because the "stiffness", 1, is negative. 17. For the radius, r(t), we have the initial value problem r (t) = GMr 2 , r(0) = a, r (0) = 0. Thus, in the energy integral lemma, f (r) = GMr 2 . Since f (r)dr = GMr 2 dr = GMR1 + C, we can take F (r) = GMR1 , and the energy integral lemma yields 1 GM 2 [r (t)]  = C1 = const. 2 r(t) To find the constant C1 , we use the initial conditions. C1 = Therefore, r(t) satisfies GM GM 1 2 [r (t)]  = 2 r(t) a 1 GM GM 2 (r ) =  2 r a r = 2GM a ar . r 1 GM 1 GM GM 2 [r (0)]  = 02  = . 2 r(0) 2 a a (Remember, r(t) is decreasing, and so r (t) < 0.) Separating variables and integrating, we get r dr = ar  2GM a dt a arctan r  ar r(a  r) a = 2GM t+C2 . a We apply the initial condition, r(0) = a, once again to find the constant C2 . But this time we have to be careful because the argument of "arctan" function becomes infinite at r = a. So, we take the limit of both sides rather than making simple substitution. lim a arctan r(t)  a  r(t) lim arctan r(t)[a  r(t)] a r(t)  lim a  r(t) t+0 r(t)[a  r(t)] a =a 0 =a , 2 2 231 t+0 =a t+0 Chapter 4
and, in the righthand side, lim  2GM t + C2 a = 2GM 0 + C2 = C2 . a t+0 Thus C2 = a/2 and r(t) satisfies a arctan r(t)  a  r(t) r(t)[a  r(t)] a = a 2GM t+ . a 2 At the moment t = T0 , when Earth splashes into the sun, we have r(T0 ) = 0. Substituting this condition into the last equation yields a arctan Then the required ratio is T0 = T 2 2 EXERCISES 4.8: a3 GM 2 1 a3 = . GM 4 2 0  a0 0(a  0) a = 2GM a T0 + a 2 2GM a T0 + a 2 a a T0 = = 2 2GM 2 2 0= a3 . GM A Closer Look at Free Mechanical Vibrations, page 219 1. In this problem, we have undamped free vibration case governed by equation (2) on page 210 in the text. With m = 3 and k = 48, the equation becomes 3y + 48y = 0 with the initial conditions y(0) = 0.5, y (0) = 2. The angular velocity of the motion is = 232 k = m 48 = 4. 3 (4.10) Exercises 4.8
It follows that 2 2 = = , 4 2 2 natural frequency = = . 2 period = A general solution to (4.10), given in (4) on page 211 in the text, becomes y(t) = C1 cos t + C2 sin t = C1 cos 4t + C2 sin 4t. We find C1 and C2 from the initial conditions. y(0) = (C1 cos 4t + C2 sin 4t)
t=0 = C1 = 1/2 ,
t=0 y (0) = (4C1 sin 4t + 4C2 cos 4t) = 4C2 = 2 C1 = 1/2, C2 = 1/2. Thus, the solution to the initial value problem is 1 2 1 sin 4t  , y(t) =  cos 4t + sin 4t = 2 2 2 4 where we have used formulas (6) rewriting the solution in form (5), page 211 in the text. The amplitude of the motion therefore is 2/2. Setting y = 0 in the above solution, we find values of t when the mass passes through the point of equilibrium. 2 sin 4t  2 4 =0 4t  = n, 4 n = 0, 1, . . . . (Time t is nonnegative.) The first moment when this happens, i.e., the smallest value of t, corresponds to n = 0. So, 4t  =0 4 t= . 16 3. The characteristic equation in this problem, r 2 + br + 16 = 0, has the roots b b2  64 . r= 2 (4.11) Substituting given particular values of b into (4.11), we find roots of the characteristic equation and solutions to the initial value problems in each case. 233 Chapter 4
b=0 0. 64 = 4i. r= 2 A general solution has the form y = C1 cos 4t + C2 sin 4t. Constants C1 and C2 can be found from the initial conditions. y(0) = (C1 cos 4t + C2 sin 4t)
t=0 = C1 = 1 ,
t=0 y (0) = (4C1 sin 4t + 4C2 cos 4t) and so y(t) = cos 4t. b=6 6. = 4C2 = 0 C1 = 1, C2 = 0 36  64 = 3 7i. 2 A general solution has the form y = (C1 cos 7t + C2 sin 7t)e3t . For constants C1 and r= 6 y(0) = C1 cos 7t + C2 sin 7t e3t t=0 = C1 = 1 , y (0) = ( 7C2  3C1 ) cos 7t  ( 7C1 + 3C2 ) sin 7t e3t C1 = 1, C2 = 3/ 7 , C2 , we have the system = t=0 7C2  3C1 = 0 3 4 7t + , y(t) = cos 7t + sin 7t e3t = e3t sin 7 7 where = arctan( 7/3) 0.723 . b=8 8. r= 8 64  64 = 4. 2 and so Thus, r = 4 is a double root of the characteristic equation. So, a general solution has the form y = (C1 t + C0 )e4t . For constants C1 and C2 , we obtain the system y(0) = (C1 t + C0 ) e4t
t=0 = C0 = 1 ,
4t t=0 y (0) = (4C1 t  4C0 + C1 ) e and so y(t) = (4t + 1)e4t . 234 = C1  4C0 = 0 C0 = 1, C1 = 4, Exercises 4.8
b = 10 10. 100  64 r= = 5 3. 2 Thus, r = 2, 8, and a general solution is given by y = C1 e2t + C2 e8t . Initial 10 conditions yield y(0) = (C1 e2t + C2 e8t )
t=0 8t = C1 + C2 = 1 ,
t=0 y (0) = (2C1 e2t  8C2 e ) = 2C1  8C2 = 0 C1 = 4/3, C2 = 1/3, and, therefore, y(t) = (4/3)e2t  (1/3)e8t is the solution to the initial value problem. The graphs of the solutions are depicted in Figures B.19B.22 in the answers in the text. 5. The auxiliary equation associated with given differential equation is r 2 + 10r + k = 0, and its roots are r = 5 25  k. k = 20 In this case, r = 5 20. by y = C1 e(5+ 5)t + C2 e(5 25  20 = 5 5. Thus, a general solution is given 5)t . The initial conditions yield y(0) = C1 e(5+ 5)t + C2 e(5 5)t = C1 + C2 = 1 , t=0 y (0) = (5 + 5)C1 e(5+ 5)t + (5  5)C2 e(5 5)t t=0 = (5 + 5)C1 + (5  5)C2 = 0 C1 = 1 + 5 /2, C2 = 1  5 /2, and, therefore, y(t) = [ 1 + the initial value problem. k = 25 Then r = 5 25. 25  25 = 5. Thus, r = 5 is a double root of the 5 /2]e(5+ 5)t +[ 1 5 /2]e(5 5)t is the solution to characteristic equation. So, a general solution has the form y = (C1 t + C0 )e5t . For constants C1 and C2 , using the initial conditions, we obtain the system y(0) = (C1 t + C0 ) e5t and so y(t) = (5t + 1)e5t . 235
t=0 = C0 = 1 ,
t=0 y (0) = (5C1 t  5C0 + C1 ) e5t = C1  5C0 = 0 C0 = 1, C1 = 5, Chapter 4 k = 30 In this case, r = 5 25  30 = 5 5i. A general solution has the form 30. y = (C1 cos 5t + C2 sin 5t)e5t . For constants C1 and C2 , we have the system y(0) = C1 cos 5t + C2 sin 5t e5t t=0 = C1 = 1 , y (0) = ( 5C2  5C1 ) cos 5t  ( 5C1 + 5C2 ) sin 5t e5t t=0 = 5C2  5C1 = 0 and so C1 = 1, C2 = 5, y(t) = cos 5t + 5 sin 5t e5t = 6e5t sin 5t + , where = arctan(1/ 5) 0.421 . Graphs of the solutions for k = 20, 25, and 30 are shown in Figures B.23B.25 in the answers in the text. 7. The motion of this massspring system is governed by equation (12) on page 213 in the text. With m = 1/8, b = 2, and k = 16 this equation becomes 1 y + 2y + 16y = 0, 8 and the initial conditions are y(0) = 3/4, y (0) = 2. Since b2  4mk = 4  4(1/8)16 = 4 < 0, we have a case of underdamped motion. A general solution to (4.12) is given in (16), that is, with = b/(2m) = 8 and = (1/2m) 4mk  b2 = 8, we have y = (C1 cos 8t + C2 sin 8t) e8t . Using the initial conditions, we find the constants C1 and C2 . y(0) = (C1 cos 8t + C2 sin 8t) e8t C1 = 3/4, C2 = 1,
t=0 (4.12) = C1 = 3/4 ,
t=0 y (0) = 8 [(C2  C1 ) cos 8t  (C2 + C1 ) sin 8t] e8t 236 = 8 (C2  C1 ) = 2 Exercises 4.8
and so y(t) =  3 5 cos 8t  sin 8t e8t = e8t sin(8t + ), 4 4 where tan = (3/4)/(1) = 3/4 and cos = 1 < 0. Thus, = + arctan(3/4) 3.785 . The damping factor is (5/4)e8t , the quasiperiod is P = 2/8 = /4, and the quasifrequency is 1/P = 4/. 9. Substituting the values m = 2, k = 40, and b = 8 5 into equation (12) on page 213 in the text and using the initial conditions, we obtain the initial value problem 2 dy d2 y + 40y = 0, +8 5 dt2 dt y(0) = 0.1 (m), y (0) = 2 (m/sec). The initial conditions are positive to reflect the fact that we have taken down to be positive in our coordinate system. The auxiliary equation for this system is 2r 2 + 8 5r + 40 = 0 or r 2 + 4 5r + 20 = 0. This equation has a double root at r = 2 5. Therefore, this system is critically damped and the equation of motion has the form y(t) = (C1 + C2 t) e2 5t . To find the constants C1 and C2 , we use the initial conditions y(0) = 0.1 and y (0) = 2. Thus, we have y(0) = 0.1 = C1 , y (0) = 2 = C2  2 5C1 C2 = 2 + 0.2 5 . From this we obtain y(t) = 0.1 + 2 + 0.2 5 t e2 5t . 237 Chapter 4
The maximum displacement of the mass is found by determining the first time the velocity of the mass becomes zero. Therefore, we have y (t) = 0 = 2 + 0.2 5 e2 5t  2 5 0.1 + 2 + 0.2 5 t e2 5t , which gives 2 1 = . t= 2 5(2 + 0.2 5) 5(2 + 0.2 5) Thus the maximum displacement is y 1 = 0.1 + 2 + 0.2 5 5(2 + 0.2 5) 1 5(2 + 0.2 5) e2 5/[ 5(2+0.2 5)] 0.242 (m). 11. The equation of the motion of this massspring system is y + 0.2y + 100y = 0, Clearly, this is an underdamped motion because b2  4mk = (0.2)2  4(1)(100) = 399.96 < 0. So, we use use equation (16) on page 213 in the text for a general solution. With = b 0.2 1 1 = = 0.1 and = 4mk  b2 = 399.96 = 99.99 , 2m 2 2m 2 y(0) = 0, y (0) = 1. equation (16) becomes y(t) = C1 cos 99.99t + C2 sin 99.99t e0.1t . From the initial condiions, y(0) = C1 cos 99.99t + C2 sin 99.99t e0.1t t=0 = C1 = 0 , 99.99C2  0.1C1 cos 99.99t  0.1C2 + 99.99C1 sin 99.99t e0.1t y (0) = = 99.99C2  0.1C1 = 1 238 C2 = 1/ 99.99 . C1 = 0, t=0 Exercises 4.8
Therefore, the equation of motion is given by y(t) = 1 e0.1t sin 99.99t . 99.99 The maximum displacement to the right occurs at the first point of local maximum of y(t). The critical points of y(t) are solutions to e0.1t y (t) = 99.99 cos 99.99t  0.1 sin 99.99t = 0 99.99 99.99 cos 99.99t  0.1 sin 99.99t = 0 tan 99.99t = 10 99.99 = 9999 . Solving for t, we conclude that the first point of local maximum is at t = (1/ 99.99) arctan 9999 0.156 sec. 13. In Example 3, the solution was found to be y(t) = 7 2t e sin 2 3t + , 12 (4.13) where = + arctan( 3/2). Therefore, we have y (t) =  7 2t e sin 2 3t + + 7 e2t cos 2 3t + . 3 Thus, to find the relative extrema for y(t), we set y (t) =  Since tan = 7 2t e sin 2 3t + + 7 e2t cos 2 3t + = 0 3 sin 2 3t + 7 = = 3 cos 2 3t + 7/3 tan 2 3t + = 3 . 3 when = (/3) + n, where n is an integer, we see that the relative extrema 2 3tn + = + n 3 (/3) + n  . 2 3 239 will occur at the points tn , where tn = Chapter 4
By substituting + arctan 3/2 for in the last equation above and by requiring that t be greater than zero, we obtain (/3) + (n  1)  arctan tn = 2 3 3/2 , n = 1, 2, 3, . . . . We see that the solution curve given by equation (4.13) above will touch the exponential curves y(t) = 7/12 e2t when we have 7 2t e sin 2 3t + = 12 7 2t e , 12 where = + arctan 3/2 . This will occur when sin 2 3t + = 1. Since sin = 1 when = (/2) + m for any integer m, we see that the times Tm , when the solution touches the exponential curves, satisfy 2 3Tm + = + m 2 where = + arctan Tm = (/2) + m  , 2 3 3/2 and m is an integer. Again requiring that t be positive we see that y(t) touches the exponential curve when (/2) + (m  1)  arctan Tm = 2 3 3/2 , m = 1, 2, 3, . . . . From these facts it follows that, for y(t) to be an extremum and, at the same time, touch the curves y(t) = 7/12e2t , there must be integers m and n such that (/2) + m  arctan 3/2 (/3) + n  arctan 3/2 = 2 3 2 3 + n = + m 3 2 1 1 1 nm=  = . 2 3 6 But, since m and n are integers, their difference is an integer and never 1/6. Thus, the extrema of y(t) do not occur on the exponential curves. 240 Exercises 4.9
15. Since the exponential function is never zero, from the equation of motion (16) on page 213 in the text we conclude that the mass passes the equilibrium position, that is, y(t) = 0, if and only if sin(t + ) = 0. Therefore, the time between two successive crossings of the equilibrium position is /, which is a half of the quasiperiod P . So, we can find the quasiperiod P by multiplying the time between two successive crossings of the equilibrium position by two. Whenever P is computed, we can measure the displacement y(t) at any moment t (with y(t) = 0) and then at the moment t + P . Taking the quotient y(t + P ) Ae(b/2m)(t+P ) sin[(t + P ) + ] = = e(b/2m)P , y(t) Ae(b/2m)t sin(t + ) we can calculate the damping coefficient b as b= EXERCISES 4.9: 2m ln[y(t + P )/y(t)] . P A Closer Look at Forced Mechanical Vibrations, page 227 1. The frequency response curve (13) on page 223, with m = 4, k = 1, and b = 2, becomes M() = 1 (k  m 2 )2 + b2 2 = 1 (1  4 2 )2 + 4 2 . The graph of this function is shown in Figure B.26 in the answers in the text. 3. The auxiliary equation in this problem is r 2 + 9 = 0, which has roots r = 3i. Thus, a general solution to the corresponding homogeneous equation has the form yh (t) = C1 cos 3t + C2 sin 3t. We look for a particular solution to the original nonhomogeneous equation of the form yp (t) = ts (A cos 3t + B sin 3t), 241 Chapter 4
where we take s = 1 because r = 3i is a simple root of the auxiliary equation. Computing the derivatives y (t) = A cos 3t + B sin 3t + t(3A sin 3t + 3B cos 3t), y (t) = 6B cos 3t  6A sin 3t + t(9A cos 3t  9B sin 3t), and substituting y(t) and y (t) into the original equation, we get 6B cos 3t  6A sin 3t + t(9A cos 3t  9B sin 3t) + 9t(A cos 3t + B sin 3t) = 2 cos 3t 6B cos 3t  6A sin 3t = 2 cos 3t A = 0, B = 1/3. So, yp (t) = (1/3)t sin 3t, and y(t) = C1 cos 3t + C2 sin 3t + (1/3)t sin 3t is a general solution. To satisfy the initial conditions, we solve y(0) = C1 = 1, y (0) = 3C2 = 0 C1 = 1, C2 = 0. So, the solution to the given initial value problem is y(t) = cos 3t + 1 t sin 3t . 3 The graph of y(t) is depicted in Figure B.27 in the answers section in the text. 5. (a) The corresponding homogeneous equation, my + ky = 0, is the equation of a simple harmonic motion, and so its general solution is given by yh (t) = C1 cos t + C2 sin t, = k/m . Since = , we look for a particular solution of the form yp (t) = A cos t + B sin t 242 yp (t) = A sin t + B cos t yp (t) = A 2 cos t  B 2 sin t. Exercises 4.9
Substitution into the original equation yields m A 2 cos t  B 2 sin t + k (A cos t + B sin t) = F0 cos t A m 2 + k cos t + B m 2 + k sin t = F0 cos t A = F0 / (k  m 2 ) , B=0 yp (t) = F0 cos t. k  m 2 Therefore, a general solution to the original equation is y(t) = C1 cos t + C2 sin t + F0 cos t . k  m 2 With the initial conditions, y(0) = y (0) = 0, we get y(0) = C1 + F0 / (k  m 2 ) = 0, y (0) = C2 = 0 Therefore, F0 F0 cos t + cos t , 2 k  m k  m 2 which can also be written in the form y(t) =  y(t) = F0 F0 (cos t  cos t) . (cos t  cos t) = 2 2  2) k  m m( C1 = F0 / (k  m 2 ) , C2 = 0. (b) Here one can apply the "differencetoproduct" identity cos A  cos B = 2 sin with A = t and B = t to get y(t) = 2F0 sin m( 2  2 ) + t sin 2  t . 2 B+A 2 sin BA 2 (c) For F0 = 32, m = 2, = 9, and = 7, the solution in part (b) becomes y(t) = 2(32) sin 2(92  72 ) 9+7 t sin 2 97 t 2 = sin 8t sin t . The graph of this function is shown in Figure B.28. 243 Chapter 4
7. The auxiliary equation to equation (1) on page 220 in the text, mr 2 + br + k = 0, has roots b b2  4mk , r= 2m which are both real (b2 > 4mk) and negative because b2  4mk < b. Let b  b2  4mk r1 := , 2m b + b2  4mk r2 := . 2m Then a general solution to the homogeneous equation corresponding to (1) has the form yh (t) = c1 er1 t + c2 er2 t . A particular solution to (1) is still given by (7) on page 221 in the text. Thus, y(t) = c1 er1 t + c2 er2 t + F0 (k  m 2 )2 + b2 2 sin(t + ), tan = (k  m 2 )/(b), is a general solution to the forced overdamped equation. 9. If a mass of m = 8 kg stretches the spring by k= mg = = 1.96 m, then the spring stiffness must be 8 9.8 = 40 (N/m). 1.96 Substitution m = 8, b = 3, k = 40, and the external force F (t) = cos 2t into the equation (23) on page 226 in the text yields 8y + 3y + 40y = cos 2t. The steadystate (a particular) solution to this equation is given in (6) and (7), page 221, that is, yp (t) = F0 k  m 2 cos t + b sin t (k  m 2 )2 + b2 2 1 40  8(2)2 cos 2t + (3)(2) sin 2t = 2 ]2 + (3)2 (2)2 [40  (8)(2) 1 1 {8 cos 2t + 6 sin 2t} = sin(2t + ), = 100 10 where = arctan(8/6) 0.927 . 244 Exercises 4.9
11. First, we find the mass m= 8 lb 1 = slug. 32 ft/sec2 4 Thus the equation (23), describing the motion, with m = 1/4, b = 1, k = 10, and the external force F (t) = 2 cos 2t becomes 1 y + y + 10y = 2 cos 2t, 4 (4.14) with the initial conditions are y(0) = y (0) = 0. A general solution to the corresponding homogeneous equation is given in Section 4.8, formula (16). That is, yh (t) = et (C1 cos t + C2 sin t) . We compute = So, yh (t) = e2t (C1 cos 6t + C2 sin 6t) . For a particular solution, we use formula (7), page 221 in the text. yp (t) = = F0 sin(t + ) (k  m 2 )2 + b2 2 2 2 sin(2t + ) = sin(2t + ), 85 [10  (1/4)(2)2]2 + (1)2 (2)2 b 1 1 = = 2 and = 2m 2(1/4) 2(1/4) 4(1/4)(10)  12 = 6. where = arctan[(k  m 2 )/(b)] = arctan(9/2) 1.352 . A general solution to (4.14) is then given by 2 y(t) = e2t (C1 cos 6t + C2 sin 6t) + sin(2t + ) . 85 From the initial conditions, we find y(0) = C1 + (2/ 85) sin = 0, y (0) = 2C1 + 6C2 + (4/ 85) cos = 0 C1 = (2/ 85) sin = 18/85, C2 = C1  (2/ 85) cos /3 = 22/255. 245 Chapter 4 y(t) = e2t  22 2 18 cos 6t  sin 6t + sin(2t + ) . 85 255 85 2 2 40  8 = , 2 The resonance frequency for the system is r = 2 (k/m)  (b2 )/(2m2 ) = 2 where we have used formula (15) on page 223 in the text for r . 13. The mass attached to the spring is m= 32 lb = 1 slug. 32 ft/sec2 Thus the equation governing the motion, my + by + ky = Fext , with m = 1, b = 2, k = 5, and Fext (t) = 3 cos 4t becomes y + 2y + 5y = 3 cos 4t. This is an underdamped motion because b2  4mk = (2)2  4(1)(5) = 16 < 0. For the steadystate solution of this equation we use formula (6) on page 221 in the text. Since Fext (t) = 3 cos 4t, we have F0 = 3, and = 4. Substituting m, b, k, F0 , and into (6), we obtain yp (t) = 3 (1)(4)2 ]2 [5  + (2)2 (4)2 3 = (8 sin 4t  11 cos 4t) . 185 page 228 [5  (1)(4)2] cos 4t + (2)(4) sin 4t REVIEW PROBLEMS: 1. Solving the auxiliary equation, r 2 + 8r  9 = 0, we find r1 = 9, r2 = 1. Thus a general solution is given by y(t) = c1 er1 t + c2 er2 t = c1 e9t + c2 et . 3. The auxiliary equation, 4r 2  4r + 10 = 0, has roots r1,2 = (1 3i)/2. Therefore a general solution is y(t) = c1 cos 246 3t 2 + c2 sin 3t 2 et/2 . Review Problems
5. The roots of the auxiliary equation, 6r 2  11r + 3 = 0, are r1 = 3/2 and r2 = 1/3. Thus, y(t) = c1 er1 t + c2 er2 t = c1 e3t/2 + c2 et/3 is a general solution. 7. Solving the auxiliary equation, 36r 2 + 24r + 5 = 0, we find r= 24 242  4(36)(5) 1 1 =  i. 2(36) 3 6 Thus a general solution is given by y(t) = c1 cos t 6 + c2 sin t 6 et/3 . 9. The auxiliary equation, 16r 2  56r + 49 = (4r  7)2 = 0, has a double root r = 7/4. Therefore, e7t/4 and te7t/4 are two linearly independent solutions, and a general solution is given by y(t) = c1 e7t/4 + c2 te7t/4 = (c1 + c2 t) e7t/4 . 11. This equation is a CauchyEuler equation. Using the approach discussed in Problem 38, Exercises 4.3, we make the substitution t = es and obtain dx dt dx dx = =t , ds dt ds dt d2 x d dx dx d2 x + t2 2 , = = ds2 ds ds ds dt 2 2 d x dx dx  t2 2 + 5x = dt ds2 ds + 5x = d2 x dx + 5x = 0.  ds2 ds The axiliary equation to this constant coefficient linear equation is r 2  r + 5 = 0, which has roots r= Thus, y(s) = e
s/2 1 12  4(1)(5) 1 19 = . 2 2 19s 2 + c2 sin 19s 2 247 c1 cos Chapter 4
is a general solution as a function of s. The back substitution, s = ln t, yields 19 19 1/2 ln t + c2 sin ln t . c1 cos y(t) = t 2 2 13. The roots of the auxiliary equation, r 2 + 16 = 0, are r = 4i. Thus a general solution to the corresponding homogeneous equation is given by yh (t) = c1 cos 4t + c2 sin 4t . The method of undetermined coefficients suggests the form yp (t) = (A1 t+A0 )et for a particular solution to the original equation. We compute yp (t) = (A1 t + A0 + A1 )et , yp (t) = (A1 t + A0 + 2A1 )et and substitute yp (t) and yp (t) into the given equation. This yields yp + 16yp = (A1 t + A0 + 2A1 )et + 16 (A1 t + A0 )et = tet 1 2 , A0 =  . (17A1 t + 17A0 + 2A1 ) et = tet A1 = 17 289 Therefore, yp (t) = t 2  et 17 289 y(t) = yh (t) + yp (t) = c1 cos 4t + c2 sin 4t + 2 t  17 289 et . 15. This is a third order homogeneous linear differential equation with constant coefficients. Its auxiliary equation is 3r 3 + 10r 2 + 9r + 2 = 0. Factoring yields 3r 3 + 10r 2 + 9r + 2 = (3r 3 + 3r 2 ) + (7r 2 + 7r) + (2r + 2) = (3r 2 + 7r + 2)(r + 1). Thus the roots of the auxiliary equation are r = 1 and r = and a general solution is given by y(t) = c1 e2t + c2 et + c3 et/3 . 248 7 72  4(3)(2) 1 = 2,  , 6 3 Review Problems
17. To solve the auxiliary equation, r 3 + 10r  11 = 0, we note that r1 = 1 is a root. Dividing the polynomial r 3 + 10r  11 by r  1 we get r 3 + 10r  11 = (r  1)(r 2 + r + 11), and so the other two roots are r2,3 = 1 1  4(1)(11) 1 43 = i. 2 2 2 y(t) = c1 et + et/2 c2 cos + c3 sin A general solution is then given by 43t 2 43t 2 . 19. By inspection, we find that r = 3 as a root of the auxiliary equation, 4r 3 + 8r 2  11r + 3 = 0. Using, say, the long division, we get 4r 3 + 8r 2  11r + 3 = (r + 3)(4r 2  4r + 1) = (r + 3)(2r  1)2 . Thus, in addition, r = 1/2 is a double root of the auxiliary equation. A general solution then has the form y(t) = c1 e3t + c2 et/2 + c3 tet/2 . 21. First, we solve the corresponding homogeneous equation, y  3y + 7y = 0. Since the roots of the auxiliary equation, r 2  3r + 7 = 0, are 3 19i 3 9  28 = , r= 2 2 a general solution to the homogeneous equation is yh (t) = c1 cos 19t 2 + c2 sin 19t 2 e3t/2 . 249 Chapter 4
We use the superposition principle to find a particular solution to the original nonhomogeneous equation. A particular solution, yp,1 (t) to y  3y + 7y = 7t2 has the form yp,1(t) = A2 t2 + A1 t + A0 . Substitution yields yp,1  3yp,1 + 7yp,1 = 2A2  3(2A2 t + A1 ) + 7(A2 t2 + A1 t + A0 ) = 7t2 and so 4 6 t+ . 7 49 The other term in the righthand side of the original equation is et . A particular solution to yp,1 (t) = t2 + y  3y + 7y = et has the form yp,2(t) = Aet . Substitution yields yp,2  3yp,2 + 7yp,2 = 5Aet = et A= 1 5 yp,2 (t) = 1 t e . 5 (7A2 )t2 + (7A1  6A2 )t + (7A0  3A1 + 2A2 ) = 7t2 7A2 = 7, 7A1  6A2 = 0, 7A0  3A1 + 2A2 = 0 A2 = 1, A1 = 6/7, A0 = 4/49, By the superposition principle, a general solution to the original equation is y(t) = yh (t)  yp,2(t) + yp,1(t) 19t + c2 sin = c1 cos 2 19t 2 e3t/2  4 1 t 6 e + t2 + t + . 5 7 49 23. The corresponding homogeneous equation in this problem is similar to that in Problem 13. Thus, y1 (t) = cos 4 and y2 (t) = sin 4 are its two linearly independent solutions, and a general solution is given by yh () = c1 cos 4 + c2 sin 4 . For a particular solution to the original equation, we use the variation of parameters method. Letting yp () = v1 () cos 4 + v2 () sin 4, 250 Review Problems
we get the following system for v1 and v2 (see (9) on page 195 in the text): v1 () cos 4 + v2 () sin 4 = 0 4v1 () sin 4 + 4v2 () cos 4 = tan 4. Multiplying the first equation by sin 4 and the second equation by (1/4) cos 4, and adding the resulting equations together, we get v2 () = 1 sin 4 4 v2 =  1 cos 4 + c3 . 16 From the first equation in the above system we also obtain v1 () = v2 () tan 4 =  v1 () =  1 4 1 1 sin2 4 =  (sec 4  cos 4) 4 cos 4 4 1 1 ln  sec 4 + tan 4 + sin 4 + c4 . (sec 4  cos 4) d =  16 16 Taking c3 = c4 = 0, we obtain yp () =  1 1 1 ln  sec 4 + tan 4 + sin 4 cos 4 +  cos sin 4 16 16 16 =  1 (cos 4) ln  sec 4 + tan 4, 16 and a general solution to the original equation is y() = c1 cos 4 + c2 sin 4  1 (cos 4) ln  sec 4 + tan 4 . 16 25. Since the auxiliary equation, 4r 2  12r + 9 = (2r  3)2 = 0, has a double root r = 3/2, a general solution to the corresponding homogeneous equation is yh (t) = c1 e3t/2 + c2 te3t/2 . By the superposition principle, a particular solution to the original equation has the form yp (t) = Ae5t + Be3t . Substituting this expression into the given nonhomogeneous equation, we get 4yp  12yp + 9yp = 4 25Ae5t + 9Be3t  12 5Ae5t + 3Be3t + 9 Ae5t + Be3t 251 Chapter 4
= 49Ae5t + 9Be3t = e5t + e3t A = 1/49, B = 1/9. Therefore, yp (t) = (1/49)e5t + (1/9)e3t and a general solution to the original equation is y(t) = c1 e3t/2 + c2 te3t/2 + 1 5t 1 3t e + e . 49 9 27. This is a CauchyEuler equation. Thus we make the substitution x = et and get d2 y dy + 2x  2y = 6x2 + 3x 2 dx dx d2 y dy dy  + 2  2y = 6(et )2 + 3(et ) dt2 dt dt d2 y dy +  2y = 6e2t + 3et . 2 dt dt x2 (4.15) The auxiliary equation, r 2 + r  2 = 0, has the roots r = 2, 1. Therefore, a general solution to the corresponding homogeneous equation is yh (t) = c1 et + c2 e2t . A particular solution to (4.15) has the form yp (t) = Ate2t + Btet . (The factor t appeared in both terms because et and e2t are both solutions to the homogeneous equation.) Differentiating, we find yp (t) = Ate2t + Btet Substitution into (4.15) yields 3Ae2t + 3Bet = 6e2t + 3et Thus a general solution to (4.15) is given by y(t) = yh (t) + yp (t) = c1 et + c2 e2t  2te2t + tet . 252 A = 2, B = 1. yp (t) = A(1  2t)e2t + B(t + 1)et yp (t) = A(4t  4)e2t + B(t + 2)et . Review Problems
The back substitution et = x (or t = ln x) results y(x) = c1 x + c2 x2  2x2 ln x + x ln x . 29. The roots of the auxiliary equation in this problem are r= 4 42  4(1)(7) = 2 3i . 2 Therefore, a general solution is given by y(t) = c1 cos 3t + c2 sin 3t e2t . Substituting the initial conditions, we obtain y(0) = c1 cos 3t + c2 sin 3t e2t t=0 = c1 = 1, y (0) = (2c1 + 3c2 ) cos 3t  ( 3c1 + 2c2 ) sin 3t e2t = 2c1 + t=0 3c2 = 2. Solving this system yields c1 = 1, c2 = 0. The solution to the given initial value problem is y(t) = e2t cos 3t . 31. We solve the corresponding homogeneous equation. Its auxiliary equation, r 2  2r + 10 = 0, has the roots r = 1 3i. Thus yh (t) = (c1 cos 3t + c2 sin 3t) et is a general solution. Now, we apply the method of undetermined coefficients and look for a particular solution to the original nonhomogeneous equation of the form yp (t) = A cos 3t + B sin 3t. Differentiating yp (t) twice, we obtain yp (t) = 3A sin 3t+3B cos 3t, yp = 9A cos 3t9B sin 3t and substitute these expressions into the original equation. Thus we get (9A cos 3t  9B sin 3t)  2(3A sin 3t + 3B cos 3t) + 10(A cos 3t + B sin 3t) = 6 cos 3t  sin 3t 253 Chapter 4 (A  6B) cos 3t + (6A + B) sin 3t = 6 cos 3t  sin 3t A  6B = 6, 6A + B = 1 A = 0, B = 1. So, yp (t) =  sin 3t, and y(t) = (c1 cos 3t + c2 sin 3t) et  sin 3t is a general solution to the given equation. Next, we satisfy the initial conditions. y(0) = c1 = 2, y (0) = c1 + 3c2  3 = 8 Hence, the answer is y(t) = 2 cos 3t  7 sin 3t et  sin 3t. 3 c1 = 2, c2 = 7/3. 33. The associated characteristic equation in this problem is r 3  12r 2 + 27r + 40 = 0, which is a third order equation. Using the rational root theorem, we look for its integer roots among the divisors of 40, which are 1, 2, 4, 8, 10, 20, and 40. By inspection, r = 1 is a root. Dividing r 3  12r 2 + 27r + 40 by r + 1, we get r 3  12r 2 + 27r + 40 = (r 2  13r + 40)(r + 1), and so the other two roots of the auxiliary equation are the roots of r 2 13r+40 = 0, which are r = 5 and 8. Therefore, a general solution to the given equation is y(t) = c1 et + c2 e5t + c3 e8t . We find the values of c1 , c2 , and c3 from the initial conditions. y(0) = (c1 et + c2 e5t + c3 e8t ) y (0) = (c1 e
t 5t t=0 8t = c1 + c2 + c3 = 3,
t=0 c1 = 1, c2 = 3, c3 = 1. + 5c2 e + 8c3 e ) = c1 + 5c2 + 8c3 = 6, = c1 + 25c2 + 64c3 = 12 y (0) = (c1 et + 25c2 e5t + 64c3 e8t ) t=0 Therefore, y(t) = et  3e5t + e8t is the solution to the given initial value problem. 35. Since the roots of the auxiliary equation, r 2 + 1 = 0, are r = i, the functions y1 () = cos and y2 () = sin are two linearly independent solutions to the corresponding homogeneous equation, and its general solution is given by yh () = c1 cos + c2 sin . 254 Review Problems
We apply the method of variation of parameters to find a particular solution to the original equation. We look for a particular solution of the form yp () = v1 () cos + v2 () sin , where v1 () and v2 () satisfy the system (9), Section 4.6. That is, v1 cos + v2 sin = 0, v1 sin + v2 cos = sec . Multiplying the first equation by sin , the second equation by cos , and adding them together yield v2 sin2 + v2 cos2 = sec cos v2 = 1 v2 () = . From the first equation in the above system we also get v1 = v2 tan =  tan v1 () =  tan d = ln  cos , where we have taken the zero integration constant. So, yp () = cos ln  cos  + sin , and y() = c1 cos + c2 sin + cos ln  cos  + sin is a general solution to the original equation. Differentiating we find that y () = c1 sin + c2 cos  sin ln  cos  + cos . Substitution of y() and y () into the initial conditions yields y(0) = c1 = 1, y (0) = c2 = 2 c1 = 1, c2 = 2, and so the answer is y() = cos + 2 sin + cos ln  cos  + sin . 255 Chapter 4
37. Comparing the given homogeneous equations with massspring oscillator equation (13) in Section 4.7, [inertia] y + [damping] y + [stiffness] y = 0, we see that in equations (a) through (d) the damping coefficient is 0. So, the behavior, of solutions, as t +, depends on the sign of the stiffness coefficient "k". (a) "k"= t4 > 0. This implies that all the solutions remain bounded as t +. (b) "k"= t4 < 0. The stiffness of the system is negative and increases unboundedly as t +. It reinforces the displacement, y(t), with magnitude increasing with time. Hence some solutions grow rapidly with time. (c) "k"= y 6 > 0. Similarly to (a), we conclude that all the solutions are bounded. (d) "k"= y 7. The function f (y) = y 7 is positive for positive y and negative if y is negative. Hence, we can expect that some of the solutions (say, ones satisfying negative initial conditions) are unbounded. (e) "k"= 3 + sin t. Since  sin t 1 for any t, we conclude that "k" 3 + (1) = 2 > 0, and all the solutions are bounded as t +. (f) Here there is positive damping "b"= t2 increasing with time, which results an increasing drain of energy from the system, and positive constant stiffness k = 1. Thus all the solutions are bounded. (g) Negative damping "b"= t2 increases (in absolute value) with time, which imparts energy to the system instead of draining it. Note that the stiffness k = 1 is also negative. Thus we should expect that some of the solutions increase unboundedly as t +. 39. If a weight of w = 32 lb stretches the spring by = 6 in = 0.5 ft, then the spring stiffness must be k= 256 w = 32 = 64 (lb/ft). 0.5 Review Problems
Also, the mass m of the weight is m= 32 w = = 1 (slug), g 32 and the damping constant b = 2 lbsec/ft. The external force is given to be F (t) = F0 cos t with F0 = 4 and = 8. Clearly, we have an underdamped motion because b2  4mk = 4  256 < 0. So, we can use formula (6) in Section 4.9 for the steadystate solution. This yields yp (t) = F0 (k  m 2 ) cos t + b sin t 2 )2 + b2 2 (k  m 1 4 (64  82 ) cos 8t + (2)(8) sin 8t = sin 8t . = 2 )2 + 2 2 8 2 (64  8 4 The resonant frequency for the system is r /(2), where r is given in (15), Section 4.9. Applying this formula, we get 1 resonant frequency = 2 b2 k 1  = 2 m 2m 2 22 64 62  = . 2) 1 2(1 2 257 Chapter 4 258 CHAPTER 5: Introduction to Systems and Phase Plane Analysis
EXERCISES 5.2: Elimination Method for Systems, page 250 1. Subtracting the second equation in the system from the first one, we eliminate y and obtain x +y y = 2y, x = x. = x  2y This equation is separable (also, it is linear). Separation yields dx = dt x ln x = t + C x(t) = c2 et . Substituting this solution into the second equation, we obtain an equation for y: y + 2y = x = c2 et . This equation is a first order linear equation. Solving we obtain (t) = exp e2t y = (2)dt = e2t et dt = c2 et + c1 c2 et e2t dt = c2 y(t) = c1 e2t + c2 et . Therefore, a general solution is x(t) = c2 et , y(t) = c1 e2t + c2 et . 3. We eliminate x by subtracting the second equation from the first equation. This yields y + 2y = 0 dy = 2dt y ln y = 2t + c y(t) = c2 e2t . 259 Chapter 5
From the second equation we get x y = 0 (xy) = 0 x(t) y(t) = c1 x(t) = c1 + c2 e2t , and a general solution is given by x(t) = c1 + c2 e2t , y(t) = c2 e2t . 5. Writing this system in operator notation yields the system (D  1)[x] + D[y] = 5, D[x] + (D + 1)[y] = 1. (5.1) We will first eliminate the function x(t), although we could proceed just as easily by eliminating the function y(t). Thus, we apply the operator D to the first equation and the operator (D  1) to the second equation to obtain D(D  1)[x] + D 2 [y] = D[5] = 0, (D  1)D[x]  (D  1)(D + 1)[y] = (D  1)[1] = 1. Adding these two equations yields {D(D  1)  (D  1)D} [x] + D 2  (D 2  1) [y] = 1 0x+1y = 1 y(t) = 1. To find the function x(t), we will eliminate y from the system given in (5.1). Therefore, we multiply the first equation in (5.1) by (D + 1) and the second by D to obtain the system (D + 1)(D  1)[x] + (D + 1)D[y] = (D + 1)[5] = 5, D 2 [x]  D(D + 1)[y] = D[1] = 0. By adding these two equations we obtain (D 2  1)  D 2 [x] = 5 x = 5 x(t) = 5. Therefore, this system of linear differential equation is solved by the functions x(t) = 5 260 and y(t) = 1. Exercises 5.2
7. In order to eliminate u, we multiply the first equation by (D  1), the second equation by (D + 1), and subtract the results. (D  1) {(D + 1)[u]  (D + 1)[v]} = (D  1) [et ] = (et )  et = 0, (D + 1) {(D  1)[u] + (2D + 1)[v]} = (D + 1) [5] = (5) + 5 = 5 (D 2  1) [u]  (D 2  1) [v] = 0, (D 2  1) [u] + {(D + 1)(2D + 1)} [v] = 5 (D + 1)(2D + 1) + D2  1 [v] = 5 {D(D + 1)} [v] = 5 . 3 (5.2) The corresponding homogeneous equation, {D(D + 1)} [v] = 0, has the characteristic equation r(r + 1) = 0 and so its general solution is vh (t) = c1 + c2 et . Applying the method of undetermined coefficients, we look for a particular solution to (5.2) of the form vp (t) = cts , where we choose s = 1 (because the homogeneous equation has constant solutions and does not have solutions of the form ct). Substitution v = ct into (5.2) yields {D(D + 1)} [ct] = (D + 1)[c] = c = Therefore, a general solution to (5.2) is v(t) = vh (t) + vp (t) = c1 + c2 et + 5 t. 3 5 3 vp (t) = 5 t. 3 r = 0, 1, We now go back to the original system and subtract the second equation from the first one. 2u  (3D + 2)[v] = et  5 1 3 5 D + 1 [v] + et  u= 2 2 2 1 5 5 5 3 u= c1 + c2 et + t + c1 + c2 et + t + et  2 3 3 2 2 1 1 5 u(t) = c1  c2 et + et + t. 2 2 3 261 Chapter 5
Thus, a general solution to the given system is 1 1 5 c2 et + et + t, 2 2 3 5 v(t) = c1 + c2 et + t. 3 u(t) = c1  9. Expressed in operator notation, this system becomes (D + 2)[x] + D[y] = 0, (D  1)[x] + (D  1)[y] = sin t. In order to eliminate the function y(t), we will apply the operator (D  1) to the first equation above and the operator D to the second one. Thus, we have (D  1)(D + 2)[x] + (D  1)D[y] = (D  1)[0] = 0, D(D  1)[x]  D(D  1)[y] = D[sin t] =  cos t. Adding these two equations yields the differential equation involving the single function x(t) given by (D 2 + D  2)  (D 2  D) [x] =  cos t 2(D  1)[x] =  cos t. (5.3) This is a linear first order differential equation with constant coefficients and so can be solved by the methods of Chapter 2. (See Section 2.3.) However, we will use the methods of Chapter 4. We see that the auxiliary equation associated with the corresponding homogeneous equation is given by 2(r  1) = 0, which has the root r = 1. Thus, a general solution to the corresponding homogeneous equation is xh (t) = C1 et . We will use the method of undetermined coefficients to find a particular solution to the nonhomogeneous equation. To this end, we note that a particular solution to this differential equation will have the form xp (t) = A cos t + B sin t 262 xp (t) = A sin t + B cos t. Exercises 5.2
Substituting these expressions into the nonhomogeneous equation given in (5.3) yields 2xp  2xp = 2(A sin t + B cos t)  2(A cos t + B sin t) = (2B  2A) cos t + (2A  2B) sin t =  cos t. By equating coefficients we obtain 2B  2A = 1 and  2A  2B = 0. By solving these two equations simultaneously for A and B, we see that A= 1 4 and 1 B= . 4 Thus, a particular solution to the nonhomogeneous equation given in (5.3) will be xp (t) = 1 1 cos t  sin t 4 4 and a general solution to the nonhomogeneous equation (5.3) will be x(t) = xh (t) + xp (t) = C1 et + 1 1 cos t  sin t. 4 4 We now must find a function y(t). To do this, we subtract the second of the two differential equations in the system from the first to obtain 3x + y =  sin t Therefore, we see that y(t) = 3 C1 et + 1 1 cos t  sin t  sin t 4 4 1 3 y(t) = 3C1 et  cos t  sin t. 4 4 y = 3x  sin t. Hence this system of differential equations has the general solution x(t) = C1 et + 1 1 cos t  sin t 4 4 and y(t) = 3C1 et  1 3 cos t  sin t. 4 4 263 Chapter 5
11. From the second equation, we obtain u =  (D 2 + 2) [v]/2. Substitution into the first equation eliminates u and gives D2  1 1 D 2 + 2 [v] + 5v = et 2 2 D  1 D 2 + 2  10 [v] = 2et  D 4 + D 2  12 [v] = 2et . (5.4) Solving the characteristic equation, r 4 + r 2  12 = 0, r 4 + r 2  12 = 0 r2 + 4 r2  3 = 0 r = 2i, 3 , we conclude that a general solution to the corresponding homogeneous equation is vh (t) = c1 cos 2t + c2 sin 2t + c3 e 3t + c4 e 3t . A particular solution to (5.4) has the form vp (t) = cet . Substitution yields D 4 + D 2  12 cet = cet + cet  12cet = 10cet = 2et 3t c= 1 . 5 Therefore, v = vh + vp = c1 cos 2t + c2 sin 2t + c3 e u =  1 1 D 2 + 2 [v] =  2 2 + c4 e 3t 3t + et /5 and 3t c1 cos 2t + c2 sin 2t + c3 e + c4 e + 1 t e 5 + c4 e 3t  c1 cos 2t + c2 sin 2t + c3 e = c1 cos 2t + c2 sin 2t  5 5 3 t c3 e 3t  c4 e 3t  e . 2 2 10 3t + 1 t e 5 By replacing (5/2)c3 by c3 and (5/2)c4 by c4 we obtain the same answer as given in the text. 13. Expressing x from the second equation and substituting the result into the first equation, we obtain x=y y 264 d(y  y) = (y  y)  4y dt y  2y + 5y = 0. Exercises 5.2
This homogeneous linear equation with constant coefficients has the characteristic equation r 2  2r + 5 = 0 with roots r = 1 2i. Thus a general solution is y = c1 et cos 2t + c2 et sin 2t . Therefore, x = = c1 et cos 2t + c2 et sin 2t  c1 et cos 2t + c2 et sin 2t c1 et cos 2t  2c1 et sin 2t + c2 et sin 2t + 2c2 et cos 2t  c1 et cos 2t + c2 et sin 2t = 2c2 et cos 2t  2c1 et sin 2t. 15. In operator form, the system becomes 2z + (D  5)[w] = 5t, (D  4)[z]  3w = 17t. We multiply the first equation by 3, the second equation by (D  5), and add the resulting equations. {6 + (D  5)(D  4)} [z] = 3(5t) + (D  5)[17t] = 70t + 17 D 2  9D + 14 [z] = 70t + 17. Solving the characteristic equation, r 2  9r + 14 = 0, we obtain r = 2, 7. Hence, a general solution to the corresponding homogeneous equation is zh (t) = c1 e2t + c2 e7t . A particular solution has the form zp (t) = At + B. Substitution yields D 2  9D + 14 [At + B] = (At + B)  9(At + B) + 14(At + B) = 14At  9A + 14B = 70t + 17 17 + 9A 70 = 5, B = = 2 A= 14 14 z(t) = zh (t) + zp (t) = c1 e2t + c2 e7t  5t  2. We use now the second equation from the original system to find w. w= 2 1 (z  4z  17t) =  c1 e2t + c2 e7t + t + 1. 3 3 265 Chapter 5
17. Expressed in operator notation, this system becomes (D 2 + 5) [x]  4[y] = 0, [x] + (D 2 + 2) [y] = 0. In order to eliminate the function x(t), we apply the operator (D 2 + 5) to the second equation. Thus, we have D 2 + 5 [x]  4[y] = 0,  D 2 + 5 [x] + D 2 + 5 D 2 + 2 [y] = 0. Adding these two equations together yields the differential equation involving the single function y(t) given by (D 2 + 5)(D 2 + 2)  4 [y] = 0 D 4 + 7D 2 + 6 [y] = 0. The auxiliary equation for this homogeneous equation, r 4 + 7r 2 + 6 = (r 2 + 1)(r 2 + 6) = 0, has roots r = i, i 6. Thus, a general solution is y(t) = C1 sin t + C2 cos t + C3 sin 6t + C4 cos 6t. We must now find a function x(t) that satisfies the system of differential equations given in the problem. To do this we solve the second equation of the system of differential equations for x(t) to obtain x(t) = D 2 + 2 [y]. Substituting the expression we found for y(t), we see that x(t) = C1 sin t  C2 cos t  6C3 sin 6t  6C4 cos 6t 6t + C4 cos 6t +2 C1 sin t + C2 cos t + C3 sin x(t) = C1 sin t + C2 cos t  4C3 sin 6t  4C4 cos 6t. Hence this system of differential equations has the general solution x(t) = C1 sin t + C2 cos t  4C3 sin 6t  4C4 cos 6t y(t) = C1 sin t + C2 cos t + C3 sin 6t + C4 cos 6t. 266 Exercises 5.2
19. From the first equation, we conclude that y = x  4x. Substitution into the second equation yields (x  4x) = 2x + (x  4x) x  5x + 6x = 0. The characteristic equation, r 2  5r + 6 = 0, has roots r = 2, 3, and so a general solution is x(t) = c1 e2t + c2 e3t y(t) = c1 e2t + c2 e3t  4 c1 e2t + c2 e3t = 2c1 e2t  c2 e3t . We find constants c1 and c2 from the initial condition. 1 = x(0) = c1 e2(0) + c2 e3(0) = c1 + c2 , 0 = y(0) = 2c1 e
2(0)  c2 e 3(0) = 2c1  c2 c1 = 1, c2 = 2. Therefore, the answer to this problem is x(t) = e2t + 2e3t , y(t) = 2e2t  2e3t . 21. To apply the elimination method, we write the system using operator notation: D 2 [x]  y = 0, x + D 2 [y] = 0. (5.5) Eliminating y by applying D 2 to the first equation and adding to the second equation gives D 2 D 2  1 [x] = 0, which reduces to D 4  1 [x] = 0. (5.6) The corresponding auxiliary equation, r 4 1 = 0, has roots 1, i. Thus, the general solution to (5.6) is given by x(t) = C1 et + C2 et + C3 cos t + C4 sin t. Substituting x(t) into the first equation in (5.5) yields y(t) = x (t) = C1 et + C2 et  C3 cos t  C4 sin t. (5.8) 267 (5.7) Chapter 5
We use initial conditions to determine constants C1 , C2 , C3 , and C4 . Differentiating (5.7) and (5.8), we get 3 = x(0) = C1 e0 + C2 e0 + C3 cos 0 + C4 sin 0 = C1 + C2 + C3 , 1= x (0) = C1 e0  C2 e0  C3 sin 0 + C4 cos 0 = C1  C2 + C4 , 1 = y(0) = C1 e0 + C2 e0  C3 cos 0  C4 sin 0 = C1 + C2  C3 , 1 = y (0) = C1 e0  C2 e0 + C3 sin 0  C4 cos 0 = C1  C2  C4 C1 + C2 + C3 = 3, C1  C2 + C4 = 1, C1 + C2  C3 = 1, C1  C2  C4 = 1. Solving we obtain C1 = C2 = C3 = C4 = 1. So, the desired solution is x(t) = et + et + cos t + sin t, y(t) = et + et  cos t  sin t. 23. We will attempt to solve this system by first eliminating the function y(t). Thus, we multiply the first equation by (D + 2) and the second by (D  1). Therefore, we obtain (D + 2)(D  1)[x] + (D + 2)(D  1)D[y] = (D + 2) 3e2t = 6e2t  6e2t = 0, (D  1)(D + 2)[x]  (D  1)(D + 2)[y] = (D  1) 3et = 3et + 3et = 0. Adding these two equations yields 0 x + 0 y = 0, which will be true for any two functions x(t) and y(t). (But not every pair of functions will satisfy this system of differential equations.) Thus, this is a degenerate system, and has infinitely many linearly independent solutions. To see if we can find these solutions, we will examine the system more closely. Notice that we could write this system as (D  1)[x + y] = 3e2t , (D + 2)[x + y] = 3et . 268 Exercises 5.2
Therefore, let's try the substitution z(t) = x(t) + y(t). We want a function z(t) that satisfies the two equations z (t)  z(t) = 3e2t and z (t) + 2z(t) = 3et , (5.9) simultaneously. We start by solving the first equation given in (5.9). This is a linear differential equation with constant coefficients which has the associated auxiliary equation r  1 = 0. Hence, the solution to the corresponding homogeneous equation is zh (t) = Cet . By the method of undetermined coefficients, we see that a particular solution will have the form zp (t) = Ae2t zp = 2Ae2t . Substituting these expressions into the first differential equation given in (5.9) yields zp (t)  zp (t) = 2Ae2t  Ae2t = 3Ae2t = 3e2t Thus, the first equation given in (5.9) has the general solution z(t) = Cet + e2t . Now, substituting z(t) into the second equation in (5.9) gives Cet  2e2t + 2 Cet + e2t = 3et 3Cet = 3et . A = 1. Hence, C must be 1. Therefore, z(t) = et + e2t is the only solution that satisfies both differential equations given in (5.9) simultaneously. Thus, any two differentiable functions that satisfy the equation x(t) + y(t) = et + e2t will satisfy the original system. 25. Writing the system in operator form yields (D  1)[x]  2y + z = 0, x + D[y]  z = 0, 4x + 4y + (D  5)[z] = 0. 269 Chapter 5
We use the second equation to express z in terms of x and y. z = x + D[y]. Substituting this expression into the other two equations, we obtain (D  1)[x]  2y + (x + D[y]) = 0, 4x + 4y + (D  5)[x + D[y]] = 0 (D  2)[x] + (D  2)[y]) = 0, (D  1)[x] + (D2  5D + 4) [y] = 0. (5.11) (5.10) Now we eliminate x by multiplying the first equation by (D  1), the second equation by (D  2), and adding the results. This yields (D  1)(D  2) + (D  2)(D 2  5D + 4) [y] = 0 (D  2)(D 2  4D + 3) [y] = 0 {(D  2)(D  1)(D  3)} [y] = 0. The roots of the characteristic equation, (r  2)(r  1)(r  3) = 0, are r = 1, 2, and 3. Thus, a general solution for y is y = c1 et + c2 e2t + c3 e3t . With h := x + y, the first equation in (5.11) can be written in the form (D  2)[h] = 0 or h  2h = 0, which has a general solution h = Ke2t . Therefore, x = h  y = c1 et + (K  c2 )e2t  c3 e3t . To find K, we substitute the above solutions x(t) and y(t), with c1 = c3 = 0, into the second equation in (5.11). Thus we get (D  1) (K  c2 )e2t + D 2  5D + 4 270 c2 e2t = 0 (K  c2 )e2t + (4(c2 )  5(2c2 ) + 4(c2 )) e2t = 0 K  c2 = 0 K = c2 . Exercises 5.2
Hence, x = c1 et  2c2 e2t  c3 e3t . Finally, we find z using (5.10). z =  c1 et  2c2 e2t  c3 e3t + c1 et + c2 e2t + c3 e3t 27. We eliminate z by expressing z= 1 1 (x + 4x) =  (D  4)[x] 4 4 (5.12) = 2c1 et + 4c2 e2t + 4c3 e3t . from the first equation and substituting (5.12) into the second and third equations. We obtain 1 2  (D  4)[x] + (D  4)[y] = 0, 4 1 1 2x + 4y + D  (D  4)[x]  4  (D  4)[x] 4 4 After some algebra, the above system simplifies to (D  4)[x] + 2(D  4)[y] = 0, D2  8D + 8 [x]  16y = 0. We use the second equation to find that y= Then the first equation becomes (D  4)[x] + 2(D  4) (D  4) 1 + 1 D 2  8D + 8 [x] = 0 16 [x] = 0 (D  4)D(D  8)[x] = 0. 1 D 2  8D + 8 [x]. 16 (5.13) = 0. 1 D 2  8D + 8 8 Solving the characteristic equation, we get r = 0, 4, and 8; so x = c1 e8t + c2 e4t + c3 . 271 Chapter 5
Substitution of this solution into (5.12) and (5.13) yield 1 (x + 4x) = c1 e8t + c3 , 4 1 1 c1 e8t  c2 e4t + c3 . y= (x  8x + 8x) = 16 2 z= 29. We begin by expressing the system in operator notation (D  )[x] + y = 0, 3x + (D  1)[y] = 0. We eliminate y by applying (D  1) to the first equation and subtracting the second equation from it. This gives {(D  1)(D  )  (3)} [x] = 0 D 2  ( + 1)D + ( + 3) [x] = 0. (5.14) Note that since the given system is homogeneous, y(t) also satisfies this equation (compare (7) and (8) on page 247 of the text). So, we can investigate solutions x(t) only. The auxiliary equation, r 2  ( + 1)r + ( + 3) = 0, has roots ( + 1)  ( + 1) + , r2 = , r1 = 2 2 where the discriminant := ( + 1)2  4( + 3). We consider two cases: i) If + 3 < 0, i.e. < 3, then > ( + 1)2 and the root r2 > ( + 1) +  + 1 = 0. 2 Therefore, the solution x(t) = er2 t is unbounded as t +. ii) If + 3 0, i.e. 3, then ( + 1)2 . If < 0, then a fundamental solution set to (5.14) is e 272
(+1)t/2 cos t 2 ,e
(+1)t/2 sin t 2 . (5.15) Exercises 5.2
If 0, then <  + 1 and a fundamental solution set is {er1 t , er2 t } ,
r1 t r1 t if > 0, {e , te } , if = 0, (5.16) where both roots r1 , r2 are nonpositive if and only if 1. For = 1 we have = (1 + 1)2  4(1 + 3) < 0, and we have a particular case of the fundamental solution set (5.15) (without exponential term) consisting of bounded functions. Finally, if < 1, then r1 < 0, r2 0, and all the functions listed in (5.15), (5.16) are bounded. Any solution x(t) is a linear combination of fundamental solutions and, therefore, all solutions x(t) are bounded if and only if 3 1. 31. Solving this problem, we follow the arguments described in Section 5.1, page 242 of the text, i.e., x(t), the mass of salt in the tank A, and y(t), the mass of salt in the tank B, satisfy the system dx = inputA  outputA , dt (5.17) dy = inputB  outputB , dt with initial conditions x(0) = 0, y(0) = 20. It is important to notice that the volume of each tank stays at 100 L because the net flow rate into each tank is the same as the net outflow. Next we observe that "inputA " consists of the salt coming from outside, which is 0.2 kg/L 6 L/min = 1.2 kg/min, and the salt coming from the tank B, which is given by y(t) y(t) kg/L 1 L/min = kg/min. 100 100 Thus, inputA = 1.2 + y(t) kg/min. 100 273 Chapter 5
"outputA " consists of two flows: one is going out of the system and the other one is going to the tank B. So, outputA = 7x(t) x(t) kg/L (4 + 3) L/min = kg/min, 100 100 and the first equation in (5.17) becomes dx y 7x = 1.2 +  . dt 100 100 Similarly, the second equation in (5.17) can be written as 3x 3y dy =  . dt 100 100 Rewriting this system in the operator form, we obtain (D + 0.07)[x]  0.01y = 1.2 , 0.03x + (D + 0.03)[y] = 0 . Eliminating y yields {(D + 0.07)(D + 0.03)  (0.01)(0.03)} [x] = (D + 0.03)[1.2] = 0.036 , which simplifies to D2 + 0.1D + 0.0018 [x] = 0.036 . The auxiliary equation, r 2 + 0.1r + 0.0018 = 0, has roots 18 1 5  7 1 7 1  =  = 0.0765 , r1 =   20 400 10000 20 100 100 5 + 7 r2 = 0.0235 . 100 Therefore, the general solution the corresponding homogeneous equation is xh (t) = C1 er1 t + C2 er2 t . Since the nonhomogeneous term in (5.19) is a constant (0.036), we are looking for a particular solution of the form xp (t) = A =const. Substituting into (5.19) yields 0.0018A = 0.036 274 A = 20, (5.19) (5.18) Exercises 5.2
and the general solution, x(t), is x(t) = xh (t) + xp (t) = C1 er1 t + C2 er2 t + 20. From the first equation in (5.18) we find y(t) = 100 {(D + 0.07)[x]  1.2} = 100 dx + 7x(t)  120 dt = 100 r1 C1 er1 t + r2 C2 er2 t + 7 C1 er1 t + C2 er2 t + 20  120 = 2  7 C1 er1 t + 2 + 7 C2 er2 t + 20. The initial conditions imply 0 = x(0) = C1 + C2 + 20, 20 = y(0) = 2  7 C1 + 2 + 7 C2 + 20 C1 + C2 = 20, 2  7 C1 + 2 + 7 C2 = 0 20 C1 =  10 + , 7 20 C2 =  10  7 . Thus the solution to the problem is 20 20 x(t) =  10 + er1 t  10  er2 t + 20 (kg), 7 7 30 r2 t 30 r1 t y(t) = e  e + 20 (kg). 7 7 33. Since no solution flows in or out of the system from the tank B, we conclude that the solution flows from the tank B to the tank A with the same rate as it does from A to B, that is, 1 L/min. Furthermore, the solution flows in and out of the tank A with the same rate, 4 L/min, and so the volume of the solution in the tank A (as well as in the tank B) remains constant, 100 L. Thus, with x(t) and y(t) denoting the amount of salt in the tanks A and B, respectively, the law "rate of change = input rate  output rate" becomes Tank A: x = 4 L/min 0.2 kg/L + 1 L/min x y kg/L  kg/L (1 L/min + 4 L/min); 100 100 275 Chapter 5
Tank B: y = 1 L/min Hence, we obtain the system y x x = 0.8  + , 20 100 y x y =  . 100 100 From the second equation, we find that x = 100y + y. Substitution into the first equation yields (100y + y) = 0.8  100y + y y + 20 100 1 y = 0.8 100y + 6y + 25 y x kg/L  1 L/min kg/L. 100 100 y + 0.06y + 0.0004y = 0.008 . (5.20) The characteristic equation r 2 +0.06r+0.0004 = 0 of the corresponding homogeneous equation has roots r= and so yh (t) = c1 e(3 0.06 (0.06)2  4(1)(0.0004) 3 5 = , 2 100 5)t/100 + c2 e(3+ 5)t/100 is a general solution to the homogeneous equation. We now look for a particular solution of the form yp (t) = c. Substitution into (5.20) gives 0.0004c = 0.008 Thus y(t) = yp (t) + yh (t) = 20 + c1 e(3 is a general solution to (5.20). Then x(t) = y + 100y = 20 + (1  3  5)c1 e(3 5)t/100 + (1  3 + 5)c2 e(3+ 5)t/100 = 20  (2 + 5)c1 e(3 5)t/100 + (2 + 5)c2 e(3+ 5)t/100 . (5.22) c= 0.008 = 20. 0.0004 5)t/100 + c2 e(3+ 5)t/100 (5.21) 276 Exercises 5.2
Next, we use the initial condition, x(0) = 0, y(0) = 20, to find values of c1 and c2 . c1 = 10/ 5 , 20  (2 + 5)c1 + (2 + 5)c2 = 0, 20 + c1 + c2 = 20 c2 = 10/ 5. With these values, the solution given in (5.21), (5.22) becomes 20  10 5 20 + 10 5 (3 5)t/100 + x(t) = 20  e 5 5 y(t) = 20 + 10 e(3 5)t/100  5 e(3+ 5)t/100 , 10 5 e(3+ 5)t/100 . 35. Let x(t) and y(t) denote the temperatures at time t in zones A and B, respectively. Therefore, the rate of change of temperature in zone A will be x (t) and in zone B will be y (t). We can apply Newton's law of cooling to help us express these rates of change in an alternate manner. Thus, we observe that the rate of change of the temperature in zone A due to the outside temperature is k1 [100  x(t)] and due to the temperature in zone B is k2 [y(t)  x(t)]. Since the time constant for heat transfer between zone A and the outside is 2 hrs (= 1/k1 ), we see that k1 = 1/2. Similarly, we see that 1/k2 = 4 which implies that k2 = 1/4. Therefore, since there is no heating or cooling source in zone A, we can write the equation for the rate of change of the temperature in the attic as x (t) = 1 1 [100  x(t)] + [y(t)  x(t)]. 2 4 In the same way, we see that the rate of change of the temperature in zone B due to the temperature of the attic is k3 [x(t)  y(t)], where 1/k3 = 4; and the rate of change of the temperature in this zone due to the outside temperature is k4 [100  y(t)], where 1/k4 = 4. In this zone, however, we must consider the cooling due to the air conditioner. Since the heat capacity of zone B is (1/2) F per thousand Btu and the air conditioner has the cooling capacity of 24 thousand Btu per hr, we see that the air conditioner removes heat from this zone at the rate of (1/2) 24 = 12 F/hr. (Since heat is removed from the house, this rate will be negative.) By combining these observations, we see that the rate of change of the temperature in zone B is given by y (t) = 12 + 1 1 [x(t)  y(t)] + [100  y(t)]. 4 4 277 Chapter 5
By simplifying these equations, we observe that this cooling problem satisfies the system 4x (t) + 3x(t)  y(t) = 200, x (t) + 4y (t) + 2y(t) = 52. In operator notation, this system becomes (4D + 3)[x]  [y] = 200, [x] + (4D + 2)[y] = 52. Since we are interested in the temperature in the attic, x(t), we will eliminate the function y(t) from the system above by applying (4D + 2) to the first equation and adding the resulting equations to obtain {(4D + 2)(4D + 3)  1} [x] = (4D + 2)[200] + 52 = 452 16D 2 + 20D + 5 [x] = 452. (5.23) This last equation is a linear equation with constant coefficients whose corresponding homogeneous equation has the associated auxiliary equation 16r 2 + 20r + 5 = 0. By the quadratic formula, the roots to this auxiliary equation are 5 + 5 0.345 and r1 = 8 5 5  r2 = 8 0.905 . Therefore, the homogeneous equation associated with this equation has a general solution given by xh (t) = c1 er1 t + c2 er2 t , where r1 and r2 are given above. By the method of undetermined coefficients, we observe that a particular solution to equation (5.23) will have the form xp (t) = A xp (t) = 0 xp (t) = 0. Substituting these expressions into equation (5.23) yields 16xp + 20xp + 5xp = 5A = 452 278 A = 90.4 . Exercises 5.2
Thus, a particular solution to the differential equation given in (5.23) is xp (t) = 90.4 and the general solution to this equation will be x(t) = c1 er1 t + c2 er2 t + 90.4 , where r1 = (5 + 5)/8 and r2 = (5  5)/8. To determine the maximum temperature of the attic, we will assume that zones A and B have sufficiently cool initial temperatures. (So that, for example, c1 and c2 are negative.) Since r1 and r2 are negative, as t goes to infinity, c1 er1 t and c2 er2 t each go to zero. Therefore, the maximum temperature that can be attained in the attic will be
t lim x(t) = 90.4 F. 37. In this problem, we combine the idea exploded in interconnected tanks problems, rate of change = rate in  rate out, with the Newton's law of cooling dT = K(T  M). dt Let x(t) and y(t) denote temperatures in rooms A and B, respectively. Room A. It gets temperature only from the heater with a rate rate in = 80, 000 Btu/h 1/4 = 20 /h. 1000 Btu (5.25) (5.24) Temperature goes out of the room A into the room B and outside with different coefficients of proportionality in (5.25): K1 = 1/2 and K2 = 1/4, respectively. Therefore, rate out = rate into B + rate outside 1 3 1 1 (x  y) + (x  0) = x  y. = 2 4 4 2 Thus, (5.24) implies that x = 20  3 1 x y 4 2 = 20  3 1 x + y. 4 2 279 Chapter 5
Room B. Similarly, we obtain y = 1000 1 1 1 7 2 + (x  y)  (y  0) = 2 + x  y. 1000 2 5 2 10 Hence, the system governing the temperature exchange is x = 20  (3/4)x + (1/2)y, y = 2 + (1/2)x  (7/10)y. We find the critical points of this system by solving 20  (3/4)x + (1/2)y = 0, 2 + (1/2)x  (7/10)y = 0 3x  2y = 80, 5x + 7y = 20 x = 600/11 , y = 460/11 . Therefore, (600/11, 460/11) is the only critical point of the system. Analyzing the direction field, we conclude that (600/11, 460/11) is an asymptotically stable node. Hence, lim y(t) = 460 41.8 F. 11 t (One can also find an explicit solution y(t) = 460/11 + c1 er1 t + c2 er2 t , where r1 < 0, r2 < 0, to conclude that y(t) 460/11 as t .) 39. Let y be an arbitrary function differentiable as many times as necessary. Note that, for a differential operator, say, A, A[y] is a function, and so we can use commutative, associative, and distributive laws operating such functions. (a) It is straightforward that (A + B)[y] := A[y] + B[y] = B[y] + A[y] =: (B + A)[y]. To prove commutativity of the multiplication, we will use the linearity of the differential operator D, that is, D[x+y] = D[x]+D[y] and the fact that D i D j = D i+j = D j D i . For the latter, D iD j [y] := D i D j [y] = y (j) 280
(i) = y (i+j) = y (i) (j) = D j D i [y] =: D j D i [y]. Exercises 5.2
Thus we have
2 2 (AB)[y] := A B[y] =
j=0 2 2 j aj D j
i=0 i bi D i [y]
2 2 j i=0 :=
j=0 2 aj D
2 bi D [y] :=
i=0 2 j=0 2 aj D bi D i [y] =
j=0 i=0 2 aj D j bi D i [y] =
i=0 j=0 2 2 i j=0 2 2 j bi D i aj D j [y]
2 i j=0 =
i=0 bi D aj D [y] aj D j
j=0 =:
i=0 bi D aj D j [y] =:
i=0 bi D i [y] = B A[y] =: (BA)[y]. (b) We have {(A + B) + C} [y] := (A + B)[y] + C[y] := (A[y] + B[y]) + C[y] = A[y] + (B[y] + C[y]) =: A[y] + (B + C)[y] =: {A + (B + C)} [y] and {(AB)C} [y] := (AB) C[y] := A B C[y] =: A (BC)[y] =: {A(BC)} [y]. (c) Using the linearity of differential operators, we obtain {A(B + C)} [y] := A (B + C)[y] := A B[y] + C[y] = A B[y] +A C[y] =: (AB)[y] + (AC)[y] =: {(AB) + (AC)} [y]. 41. As it was noticed in Example 2, we can treat a "polynomial" in D, that is, an expression of the form p(D) =
n i=0 ai D i , as a regular polynomial, i.e., p(r) = n i=0 ai r i , while per forming arithmetic operations. Hence, the factorization problem for p(D) is equivalent to the factorization problem for p(r), which is the same as finding its roots. (a) r = 3 32  4(4) 3 5 = = 4, 1 2 2 D 2 + 3D  4 = (D + 4)(D  1). 281 Chapter 5
(b) r = (c) r = 1 9 12  4(6) 1 5 = = 3, 2 D 2 + D  6 = (D + 3)(D  2). 2 2 92  4(5)2 9 11 = = 5, 1/2 2D 2 + 9D  5 = (D + 5)(2D  1). 4 4 2 D  2 = (D + 2)(D  2). (d) r = 2 EXERCISES 5.3: Solving Systems and HigherOrder Equations Numerically, page 261 1. We isolate y (t) first and obtain an equivalent equation y (t) = 3y(t)  ty (t) + t2 . Denoting x1 := y, x2 := y we conclude that x1 = y = x2 , x2 = (y ) = y = 3y  ty + t2 = 3x1  tx2 + t2 , with initial conditions x1 (0) = y(0) = 3, x2 (0) = y (0) = 6. Therefore, given initial value problem is equivalent to x1 = x2 , x2 = 3x1  tx2 + t2 , x1 (0) = 3, 3. Isolating y (4) (t), we get y (4) (t) = y (3) (t)  7y(t) + cos t . In this problem, we need four new variables for y(t), y (t), y (t), and y (3) (t). Thus we denote x1 = y, x2 = y , x3 = y , and x4 = y (3) . x2 (0) = 6. The initial conditions then become x1 (0) = y(0) = 1, x2 (0) = y (0) = 1, x3 (0) = y (0) = 0, x4 (0) = y (3) (0) = 2. We have x1 = y = x2 , 282 Exercises 5.3
x2 = (y ) = y = x3 , x3 = (y ) = y (3) = x4 , x4 = y (3) = y (4) = y (3)  7y + cos t = x4  7x1 + cos t . Hence, the required initial value problem for a system in normal form is x1 = x2 , x2 = x3 , x3 = x4 , x4 = x4  7x1 + cos t, x1 (0) = x2 (0) = 1, x3 (0) = 0, x4 (0) = 2 . 5. First we express the given system as x = x  y + 2t, y = x  y  1. Setting x1 = x, x2 = x , x3 = y, x4 = y we obtain x1 = x = x2 , x2 = x = x2  x3 + 2t , x3 = y = x4 , x4 = y = x1  x3  1 x1 = x2 , x2 = x2  x3 + 2t , x3 = x4 , x4 = x1  x3  1 with initial conditions x1 (3) = 5, x2 (3) = 2, x3 (3) = 1, and x4 (3) = 1. 7. In an equivalent form, we have a system x = y + t, 2y  2x + 1 . y = 5 Setting x1 = x, x2 = x , x3 = x , x4 = y, x5 = y , 283 Chapter 5
we obtain a system in normal form x1 = x2 , x2 = x3 , x3 = x4 + t , x4 = x5 , 1 x5 = (2x4  2x3 + 1) 5 with initial conditions x1 (0) = x2 (0) = x3 (0) = 4, x4 (0) = x5 (0) = 1. 9. To see how the improved Euler's method can be extended let's recall, from Section 3.6, the improved Euler's method (pages 127128 of the text). For the initial value problem x = f (t, x), x(t0 ) = x0 , the recursive formulas for the improved Euler's method are tn+1 = tn + h, h xn+1 = xn + [f (tn , xn ) + f (tn + h, xn + hf (tn , xn ))] , 2 where h is the step size. Now suppose we want to approximate the solution x1 (t), x2 (t) to the system x1 = f1 (t, x1 , x2 ) that satisfies the initial conditions x1 (t0 ) = a1 , x2 (t0 ) = a2 . and x2 = f2 (t, x1 , x2 ), Let x1;n and x2;n denote approximations to x1 (tn ) and x2 (tn ), respectively, where tn = t0 + nh for n = 0, 1, 2, . . .. The recursive formulas for the improved Euler's method are obtained by forming the vector analogue of the scalar formula. We obtain tn+1 = tn + h, 284 Exercises 5.3
x1;n+1 = x1;n + h [f1 (tn , x1;n , x2;n ) 2 +f1 (tn + h, x1;n + hf1 (tn , x1;n , x2;n ), x2;n + hf2 (tn , x1;n , x2;n ))], h = x2;n + [f2 (tn , x1;n , x2;n ) 2 +f2 (tn + h, x1;n + hf1 (tn , x1;n , x2;n ), x2;n + hf2 (tn , x1;n , x2;n ))]. x2;n+1 The approach can be used more generally for systems of m equations in normal form. Suppose we want to approximate the solution x1 (t), x2 (t), . . ., xm (t) to the system x1 = f1 (t, x1 , x2 , . . . , xm ) , x2 = f2 (t, x1 , x2 , . . . , xm ) , . . . xm = fm (t, x1 , x2 , . . . , xm ) , with the initial conditions x1 (t0 ) = a1 , x2 (t0 ) = a2 , ... , xm (t0 ) = am . We adapt the recursive formulas above to obtain tn+1 = tn + h, x1;n+1 n = 0, 1, 2, . . . ; h = x1;n + [f1 (tn , x1;n , x2;n , . . . , xm;n ) + f1 (tn + h, x1;n + hf1 (tn , x1;n , x2;n , . . . , xm;n ), 2 x2;n + hf2 (tn , x1;n , x2;n , . . . , xm;n ), . . . , xm;n + hfm (tn , x1;n , x2;n , . . . , xm;n ))] , h = x2;n + [f2 (tn , x1;n , x2;n , . . . , xm;n ) + f2 (tn + h, x1;n + hf1 (tn , x1;n , x2;n , . . . , xm;n ), 2 x2;n + hf2 (tn , x1;n , x2;n , . . . , xm;n ), . . . , xm;n + hfm (tn , x1;n , x2;n , . . . , xm;n ))] , h [fm (tn , x1;n , x2;n , . . . , xm;n ) + fm (tn + h, x1;n + hf1 (tn , x1;n , x2;n , . . . , xm;n ), 2 x2;n + hf2 (tn , x1;n , x2;n , . . . , xm;n ), . . . , xm;n + hfm (tn , x1;n , x2;n , . . . , xm;n ))] . x2;n+1 . . . xm;n+1 = xm;n + 11. See the answer in the text. 285 Chapter 5
13. See the answer in the text. 15. See the answer in the text. 17. Let x1 := u and x2 := v, and denote the independent variable by t (in order to be consistent with formulas in Section 5.3). In new notation, we have an initial value problem x1 = 3x1  4x2 , x2 = 2x1  3x2 , x1 (0) = x2 (0) = 1 for a system in normal form. Here f1 (t, x1 , x2 ) = 3x1  4x2 , f2 (t, x1 , x2 ) = 2x1  3x2 . Thus formulas for ki,j 's in vectorized RungeKutta algorithm become k1,1 = h(3x1;n  4x2;n ), k2,1 = h(2x1;n  3x2;n ), k1,1 k2,1 k1,2 = h 3 x1;n +  4 x2;n + 2 2 k1,1 k2,1  3 x2;n + k2,2 = h 2 x1;n + 2 2 k1,2 k2,2  4 x2;n + k1,3 = h 3 x1;n + 2 2 k1,2 k2,2  3 x2;n + k2,3 = h 2 x1;n + 2 2 k1,4 = h [3 (x1;n + k1,3 )  4 (x2;n + k2,3 )] , k2,4 = h [2 (x1;n + k1,3 )  3 (x2;n + k2,3 )] . With the inputs t0 = 0, x1;0 = x2;0 = 1, and step size h = 1 we compute k1,1 = h(3x1;0  4x2;0 ) = 3(1)  4(1) = 1, k2,1 = h(2x1;0  3x2;0 ) = 2(1)  3(1) = 1, k1,1 k2,1 1 k1,2 = h 3 x1;0 +  4 x2;0 + =3 1+ 2 2 2 286 1 2 1 = , 2 , , , , 4 1+ Exercises 5.3
k2,2 = h 2 x1;0 + k1,3 k2,3 k1,4 k2,4 1 1 1 3 1+ = , 2 2 2 1/2 1/2 3 =3 1+ 4 1+ = , 2 2 4 1/2 1/2 3 =2 1+ 3 1+ = , 2 2 4 3 3 1 = h [3 (x1;0 + k1,3 )  4 (x2;0 + k2,3 )] = 3 1 + 4 1+ = , 4 4 4 3 3 1 = h [2 (x1;0 + k1,3 )  3 (x2;0 + k2,3 )] = 2 1 + 3 1+ = . 4 4 4 k1,1 2 k1,2 = h 3 x1;0 + 2 k1,2 = h 2 x1;0 + 2 k2,1 2 k2,2  4 x2;0 + 2 k2,2  3 x2;0 + 2  3 x2;0 + =2 1+ Using the recursive formulas, we find t1 = t0 + h = 0 + 1 = 1 and x1;1 = x1;0 + x2;1 3 1 (1) + 2(1/2) + 2(3/4) + (1/4) (k1,1 + 2k1,2 + 2k1,3 + k1,4 ) = 1 + = , 6 6 8 1 (1) + 2(1/2) + 2(3/4) + (1/4) 3 = x2;0 + (k2,1 + 2k2,2 + 2k2,3 + k2,4 ) = 1 + = 6 6 8 as approximations to x1 (1) and x2 (1) with step h = 1. We repeat the algorithm with h = 2m , m = 1, 2, . . . . The results of these computations are listed in Table 5A. Table 5A: Approximations of the solution to Problem 17.
m 0 1 2 h = 2m 1.0 0.5 0.25 x1 (1; h) 0.375 0.36817 0.36789 x2 (1; h) 0.375 0.36817 0.36789 We stopped at m = 2, since x1 (1; 21 )  x1 (1; 22 ) = x2 (1; 21 )  x2 (1; 22 ) = 0.36817  0.36789 = 0.00028 < 0.001 . Hence u(1) = v(1) 0.36789 . 287 Chapter 5
18. For starting values we take t0 = 0, x0,1 = 10, and x0,2 = 15, which are determined by the initial conditions. Here h = 0.1, and f1 (t, x1 , x2 ) = (0.1)x1 x2 , f2 (t, x1 , x2 ) = x1 . Now, using the definitions of tn , xi;n , ki,1 , ki,2 , ki,3 , and ki,4 on page 258 of the text, we have k1,1 = hf1 (tn , x1;n , x2;n ) = h(0.1)x1;n x2;n , k2,1 = hf2 (tn , x1;n , x2;n ) = hx1;n , h k1,1 k2,1 k1,1 , x2;n + k1,2 = hf1 tn + , x1;n + = h(0.1) x1;n + x2;n + 2 2 2 2 h k1,1 k2,1 k1,1 , x2;n + k2,2 = hf2 tn + , x1;n + = h x1;n + , 2 2 2 2 h k1,2 k2,2 k1,2 , x2;n + = h(0.1) x1;n + x2;n + k1,3 = hf1 tn + , x1;n + 2 2 2 2 h k1,2 k2,2 k1,2 , x2;n + = h x1;n + , k2,3 = hf2 tn + , x1;n + 2 2 2 2 k1,4 = hf1 (tn + h, x1;n + k1,3 , x2;n + k2,3 ) = h(0.1) (x1;n + k1,3 ) (x2;n + k2,3 ) , k2,4 = hf2 (tn + h, x1;n + k1,3 , x2;n + k2,3 ) = h (x1;n + k1,3 ) . Using these values, we find tn+1 = tn + h = tn + 0.1 , 1 x1;n+1 = x1;n + (k1,1 + 2k1,2 + 2k1,3 + k1,4 ) , 6 1 x2;n+1 = x2;n + (k2,1 + 2k2,2 + 2k2,3 + k2,4 ) . 6 In Table 5B we give approximate values for tn , x1;n , and x2;n . From Table 5B we see that the strength of the guerrilla troops, x1 , approaches zero, therefore with the combat effectiveness coefficients of 0.1 for guerrilla troops and 1 for conventional troops the conventional troops win. 19. See the answer in the text. 288 k2,1 2 , k2,2 2 , Exercises 5.3 Table 5B: Approximations of the solutions to Problem 18.
tn 0 0.1 0.2 0.3 0.4 0.5 x1;n 10 3.124 1.381 0.707 0.389 0.223 x2;n 15 9.353 7.254 6.256 5.726 5.428 21. First, we convert given initial value problem to an initial value problem for a normal system. Let x1 (t) = H(t), x2 (t) = H (t). Then H (t) = x2 (t), x1 (0) = H(0) = 0, x2 (0) = H (0) = 0, and we get x1 = x2 , 60  x1 = (77.7)x2 + x1 (0) = x2 (0) = 0 (19.42)x2 , 2 x1 = x2 , x2 = [60  x1  (19.42)x2] /77.7 , 2 x1 (0) = x2 (0) = 0. Thus f1 (t, x1 , x2 ) = x2 , f2 (t, x1 , x2 ) = [60  x1  (19.42)x2 ] /77.7, t0 = 0, x1;0 = 0, and 2 x2;0 = 0. With h = 0.5, we need (5  0)/0.5 = 10 steps to approximate the solution over the interval [0, 5]. Taking n = 0 in the vectorized RungeKutta algorithm, we approximate the solution at t = 0.5. k1,1 = hx2;0 = 0.5(0) = 0, k2,1 = h 60  x1;0  (19.42)x2 /77.7 = 0.5 60  (0)  (19.42)(0)2 /77.7 = 0.38610 , 2;0 k2,1 0.38610 k1,2 = h x2;0 + = 0.5 (0) + = 0.09653 , 2 2 k2,2 k1,1 = h 60  x1;0 + 2 k2,2 2 k2,1  (19.42) x2;0 + 2 0.38144 2
2 /77.7 = 0.38144 , k1,3 = h x2;0 + = 0.5 (0) + = 0.09536 , 289 Chapter 5
k2,3 k1,2 = h 60  x1;0 + 2 k2,2  (19.42) x2;0 + 2
2 /77.7 = 0.38124 , k1,4 = h (x2;0 + k2,3 ) = 0.5 ((0) + 0.38124) = 0.19062 , k2,4 = h 60  (x1;0 + k1,3 )  (19.42) (x2;0 + k2,3 )2 /77.7 = 0.36732 . Using the recursive formulas, we find t1 = t0 + h = 0 + 0.5 = 0.5 1 x1 (0.5) x1;1 = x1;0 + (k1,1 + 2k1,2 + 2k1,3 + k1,4 ) = 0.09573 , 6 1 x2 (0.5) x2;1 = x2;0 + (k2,1 + 2k2,2 + 2k2,3 + k2,4 ) = 0.37980 . 6 Next, we repeat the procedure with n = 1, 2, . . . , 9. The results of these computations (the values of x1;n only) are presented in Table 5C. Table 5C: Approximations of the solution to Problem 21.
n 0 1 2 3 4 5 tn 0 0.5 1.0 1.5 2.0 2.5 x1;n H(tn) 0 0.09573 0.37389 0.81045 1.37361 2.03111 n 6 7 8 9 10 tn 3.0 3.5 4.0 4.5 5.0 x1;n H(tn) 2.75497 3.52322 4.31970 5.13307 5.95554 23. Let x1 = y and x2 = y to give the initial value problem x1 = f1 (t, x1 , x2 ) = x2 , x2 = f2 (t, x1 , x2 ) = x1 (1 + rx2 ) , 1 x1 (0) = a, x2 (0) = 0. Now, using the definitions of tn , xi;n , ki,1 , ki,2 , ki,3 , and ki,4 on page 258 of the text, we have k1,1 = hf1 (tn , x1;n , x2;n ) = hx2;n , 290 Exercises 5.3
k2,1 = hf2 (tn , x1;n , x2;n ) = hx1;n 1 + rx2 , 1;n h k1,1 k2,1 k2,1 , x2;n + k1,2 = hf1 tn + , x1;n + = h x2;n + , 2 2 2 2 k2,2 = hf2 h k1,1 k2,1 , x2;n + tn + , x1;n + 2 2 2 k1,1 = h x1;n + 2 = h x2;n + k2,2 , 2 k1,2 2 1 + r x1;n + k1,2 2
2 k1,1 1 + r x1;n + 2 2 , h k1,2 k2,2 , x2;n + k1,3 = hf1 tn + , x1;n + 2 2 2 h k1,2 k2,2 , x2;n + k2,3 = hf2 tn + , x1;n + 2 2 2 = h x1;n + , k1,4 = hf1 (tn + h, x1;n + k1,3 , x2;n + k2,3 ) = h (x2;n + k2,3 ), k2,4 = hf2 (tn + h, x1;n + k1,3 , x2;n + k2,3 ) = h (x1;n + k1,3 ) 1 + r (x1;n + k1,3 )2 . Using these values, we find tn+1 = tn + h = tn + 0.1 , 1 x1;n+1 = x1;n + (k1,1 + 2k1,2 + 2k1,3 + k1,4 ) , 6 1 x2;n+1 = x2;n + (k2,1 + 2k2,2 + 2k2,3 + k2,4 ) . 6 In Table 5D we give the approximate period for r = 1 and 2 with a = 1, 2 and 3, from this we see that the period varies as r is varied or as a is varied. Table 5D: Approximate period of the solution to Problem 23.
r 1 2 a=1 4.8 4.0 a=2 3.3 2.4 a=3 2.3 1.7 25. With x1 = y, x2 = y , and x3 = y , the initial value problem can be expressed as the system x1 = x2 , x2 = x3 , x3 = t  x3  x2 1 , x1 (0) = 1, x2 (0) = 1, x3 (0) = 1. 291 Chapter 5
Here f1 (t, x1 , x2 , x3 ) = x2 , f2 (t, x1 , x2 , x3 ) = x3 , f3 (t, x1 , x2 , x3 ) = t  x3  x2 . 1 Since we are computing the approximations for c = 1, the initial value for h in Step 1 of the algorithm in Appendix E of the text is h = (1  0)20 = 1. The equations in Step 3 are k1,1 = hf1 (t, x1 , x2 , x3 ) = hx2 , k2,1 = hf2 (t, x1 , x2 , x3 ) = hx3 , k3,1 = hf3 (t, x1 , x2 , x3 ) = h t  x3  x2 , 1 k1,1 k2,1 h k1,2 = hf1 t + , x1 + , x2 + , x3 + 2 2 2 k1,1 k2,1 h , x2 + , x3 + k2,2 = hf2 t + , x1 + 2 2 2 k3,1 2 k3,1 2 k2,1 , 2 k3,1 = h x3 + , 2 = h x2 + =h t+ k3,1 k1,1 h  x3   x1 + 2 2 2
2 k1,1 k2,1 k3,1 h , x2 + , x3 + k3,2 = hf3 t + , x1 + 2 2 2 2 h k1,3 = hf1 t + , x1 + 2 h k2,3 = hf2 t + , x1 + 2 k3,3 = hf3 k1,2 , x2 + 2 k1,2 , x2 + 2 k2,2 , x3 + 2 k2,2 , x3 + 2 k3,2 2 k3,2 2 , k2,2 , 2 k3,2 = h x3 + , 2 = h x2 + k3,2 k1,2 h  x1 + = h t +  x3  2 2 2
2 k1,2 k2,2 k3,2 h , x2 + , x3 + t + , x1 + 2 2 2 2 , k1,4 = hf1 (t + h, x1 + k1,3 , x2 + k2,3 , x3 + k3,3 ) = h (x2 + k2,3 ) , k2,4 = hf2 (t + h, x1 + k1,3 , x2 + k2,3 , x3 + k3,3 ) = h (x3 + k3,3 ) , k3,4 = hf3 (t + h, x1 + k1,3 , x2 + k2,3 , x3 + k3,3 ) = h t + h  x3  k3,3  (x1 + k1,3 )2 . Using the starting values t0 = 0, a1 = 1, a2 = 0, and a3 = 1, we obtain the first approximations x1 (1; 1) = 1.29167 , x2 (1; 1) = 0.28125 , 292 Exercises 5.4
x3 (1; 1) = 0.03125 . Repeating the algorithm with h = 21 , 22 , 23 we obtain the approximations in Table 5E. Table 5E: Approximations of the Solution to Problem 25.
n 0 1 2 3 h 1.0 0.5 0.25 0.125 y(1) x1 (1; 2n ) 1.29167 1.26039 1.25960 1.25958 x2(1; 2n ) 0.28125 0.34509 0.34696 0.34704 x3(1; 2n ) 0.03125 0.06642 0.06957 0.06971 We stopped at n = 3 since x1 (1; 23 )  x1 (1; 22 ) 1.25958  1.25960 = = 0.00002 < 0.01 , 3 ) x1 (1; 2 1.25958 x2 (1; 23 )  x2 (1; 22 ) 0.34704  0.34696 = = 0.00023 < 0.01 , x2 (1; 23 ) 0.34704 0.06971 + 0.06957 x3 (1; 23 )  x3 (1; 22 ) = = 0.00201 < 0.01 . 3 ) x3 (1; 2 0.06971 Hence y(1) x1 1; 23 = 1.25958 , with tolerance 0.01 . 27. See the answer in the text. 29. See the answer in the text. EXERCISES 5.4: Introduction to the Phase Plane, page 274 and 1. Substitution of x(t) = e3t , y(t) = et into the system yields d 3t dx = e = 3e3t = 3 et dt dt
3 = 3y 3 , 293 Chapter 5
d t dy = e = et = y. dt dt Thus, given pair of functions is a solution. To sketch the trajectory of this solution, we express x as a function of y. x = e3t = et
3 = y3 for y = et > 0. Since y = et is an increasing function, the flow arrows are directed away from the origin. See Figure B.29 in the answers of the text. 3. In this problem, f (x, y) = x  y, g(x, y) = x2 + y 2  1. To find the critical point set, we solve the system x  y = 0, x +y 1 = 0 Eliminating y yields 1 x = . 2 Substituting x into the first equation, we find the corresponding value for y. Thus the critical points of the given system are (1/ 2, 1/ 2) and (1/ 2, 1/ 2). 2x2 = 1 5. In this problem, f (x, y) = x2  2xy, g(x, y) = 3xy  y 2,
2 2 x = y, x + y 2 = 1.
2 and so we find critical points by solving the system x2  2xy = 0, 3xy  y
2 = 0 x(x  2y) = 0, y(3x  y) = 0. From the first equation we conclude that either x = 0 or x = 2y. Substituting these values into the second equation, we get x=0 x = 2y y[3(0)  y] = 0 y[3(2y)  y] = 0 y 2 = 0 5y 2 = 0 y = 0; y = 0, x = 2(0) = 0. Therefore, (0, 0) is the only critical point. 294 Exercises 5.4
6. We see by Definition 1 on page 266 of the text that we must solve the system of equations given by y 2  3y + 2 = 0, (x  1)(y  2) = 0. By factoring the first equation above, we find that this system becomes (y  1)(y  2) = 0, (x  1)(y  2) = 0. Thus, we observe that if y = 2 and x is any constant, then the system of differential equations given in this problem will be satisfied. Therefore, one family of critical points is given by the line y = 2. If y = 2, then the system of equations above simplifies to y  1 = 0, and x  1 = 0. Hence, another critical point is the point (1, 1). 7. Here f (x, y) = y  1, g(x, y) = ex+y . Thus the phase plane equation becomes dy ex+y ex ey = = . dx y1 y1 Separating variables yields (y  1)ey dy = ex dx yey + C = ex or (y  1)ey dy = ex + yey = C. ex dx 9. The phase plane equation for this system is g(x, y) ex + y dy = = . dx f (x, y) 2y  x We rewrite this equation in symmetric form, (ex + y) dx + (2y  x) dy = 0, and check it for exactness. M = [(ex + y)] = 1, y y 295 Chapter 5 N = (2y  x) = 1. x x Therefore, the equation is exact. We have F (x, y) = M(x, y) = N(x, y) dy = (2y  x) dy = y 2  xy + g(x); F (x, y) = y 2  xy + g(x) = y + g (x) =  (ex + y) x x g(x) = (ex ) dx = ex . g (x) = ex Hence, a general solution to the phase plane equation is given implicitly by F (x, y) = y 2  xy  ex = C where c is an arbitrary constant. 11. In this problem, f (x, y) = 2y and g(x, y) = 2x. Therefore, the phase plane equation for given system is dy 2x x = = . dx 2y y Separation variables and integration yield y dy = x dx y dy = x dx y 2  x2 = c. or ex + xy  y 2 = C = c, 1 2 1 2 y = x +C 2 2 Thus, the trajectories are hyperbolas if c = 0 and, for c = 0, the lines y = x. In the upper halfplane, y > 0, we have x = 2y > 0 and, therefore, x(t) increases. In the lower halfplane, x < 0 and so x(t) decreases. This implies that solutions flow from the left to the right in the upper halfplane and from the right to the left in the lower halfplane. See Figure B.30 in the text. 13. First, we will find the critical points of this system. Therefore, we solve the system (y  x)(y  1) = 0, (x  y)(x  1) = 0. 296 Exercises 5.4
Notice that both of these equations will be satisfied if y = x. Thus, x = C and y = C, for any fixed constant C, will be a solution to the given system of differential equations and one family of critical points is the line y = x. We also see that we have a critical point at the point (1, 1). (This critical point is, of course, also on the line y = x.) Next we will find the integral curves. Therefore, we must solve the first order differential equation given by dy dy/dt (x  y)(x  1) = = dx dx/dt (y  x)(y  1) we have (y  1)dy = By completing the square, we obtain (x  1)2 + (y  1)2 = c, where c = 2C +2. Therefore, the integral curves are concentric circles with centers at the point (1, 1), including the critical point for the system of differential equations. The trajectories associated with the constants c = 1, 4, and 9, are sketched in Figure B.31 in the answers of the text. Finally we will determine the flow along the trajectories. Notice that the variable t imparts a flow to the trajectories of a solution to a system of differential equations in the same manner as the parameter t imparts a direction to a curve written in parametric form. We will find this flow by determining the regions in the xyplane where x(t) is increasing (moving from left to right on each trajectory) and the regions where x(t) is decreasing (moving from right to left on each trajectory). Therefore, we will use four cases to study the equation dx/dt = (yx)(y1), the first equation in our system. 297 (1  x)dx dy 1x = . dx y1 We can solve this last differential equation by the method of separation of variables. Thus, y2 x2 y = x +C 2 2 x2  2x + y 2  2y = 2C. Chapter 5
Case 1 : y > x and y < 1. (This region is above the line y = x but below the line y = 1.) In this case, y  x > 0 but y  1 < 0. Thus, dx/dt = (y  x)(y  1) < 0. Hence, x(t) will be decreasing here. Therefore, the flow along the trajectories will be from right to left and so the movement is clockwise. Case 2 : y > x and y > 1. (This region is above the lines y = x and y = 1.) In this case, we see that y  x > 0 and y  1 > 0. Hence, dx/dt = (y  x)(y  1) > 0. Thus, x(t) will be increasing and the flow along the trajectories in this region will still be clockwise. Case 3 : y < x and y < 1. (This region is below the lines y = x and y = 1.) In this case, y  x < 0 and y  1 < 0. Thus, dx/dt > 0 and so x(t) is increasing. Thus, the movement is from left to right and so the flow along the trajectories will be counterclockwise. Case 4 : y < x and y > 1. (This region is below the line y = x but above the line y = 1.) In this case, y  x < 0 and y  1 > 0. Thus, dx/dt < 0 and so x(t) will be decreasing here. Therefore, the flow is from right to left and, thus, counterclockwise here also. Therefore, above the line y = x the flow is clockwise and below that line the flow is counterclockwise. See Figure B.31 in the answers of the text. 15. From Definition 1 on page 266 of the text, we must solve the system of equations given by 2x + y + 3 = 0, 3x  2y  4 = 0. By eliminating y in the first equation we obtain x+2 =0 and by eliminating x in the first equation we obtain y + 1 = 0. Thus, we observe that x = 2 and y = 1 will satisfy both equations. Therefore (2, 1) is a critical point. From Figure B.32 in the answers of the text we see that all solutions passing near the point (2, 1) do not stay close to it therefore the critical point (2, 1) is unstable. 298 Exercises 5.4
17. For critical points, we solve the system f (x, y) = 0, g(x, y) = 0 2x + 13y = 0, x  2y = 0 2(2y) + 13y = 0, x = 2y y = 0, x = 0. Therefore, the system has just one critical point, (0, 0). The direction field is shown in Figure B.33 in the text. From this picture we conclude that (0, 0) is a center (stable). 19. We set v = y . Then y = (y ) = v and so given equation is equivalent to the system y = v, v y = 0 y = v, v = y. In this system, f (y, v) = v and g(y, v) = y. For critical points we solve f (y, v) = v = 0, g(y, v) = y = 0 y = 0, v=0 and conclude that, in yvplane, the system has only one critical point, (0, 0). In the upper halfplane, y = v > 0 and, therefore, y increases and solutions flow to the right; similarly, solutions flow to the left in the lower halfplane. See Figure B.34 in the answers of the text. The phase plane equation for the system is dv/dx y dv = = dy dy/dx v v dv = y dy v 2  y 2 = c. Thus, the integral curves are hyperbolas for c = 0 and lines v = y for c = 0. On the line v = y, the solutions flow into the critical point (0, 0), whereas solutions flow away from (0, 0) on v = y. So, (0, 0) is a saddle point (unstable). 21. First we convert the given equation into a system of first order equations involving the functions y(t) and v(t) by using the substitution v(t) = y (t) v (t) = y (t). Therefore, this equation becomes the system y = v, v = y  y 5 = y (1 + y 4) . 299 Chapter 5
To find the critical points, we solve the system of equations given by v = 0 and y (1 + y 4 ) = 0. This system is satisfied only when v = 0 and y = 0. Thus, the only critical point is the point (0, 0). To find the integral curves, we solve the first order equation given by dv dv/dt y  y 5 = = . dy dy/dt v This is a separable equation and can be written as v dv = y  y 5 dy v2 y2 y6 =  +C 2 2 6 (c = 6C), 3v 2 + 3y 2 + y 6 = c where we have integrated to obtain the second equation above. Therefore, the integral curves for this system are given by the equations 3v 2 + 3y 2 + y 6 = c for each positive constant c. To determine the flow along the trajectories, we will examine the equation dy/dt = v. Thus, we see that dy dy > 0 when v > 0, and < 0 when v < 0. dt dt Therefore, y will be increasing when v > 0 and decreasing when v < 0. Hence, above the yaxis the flow will be from left to right and below the xaxis the flow will be from right to left. Thus, the flow on these trajectories will be clockwise (Figure B.35 in the answers of the text). Thus (0, 0) is a center (stable). 23. With v = y , v = y , the equation transforms to the system y = v, v +yy = 0
4 y = v, v = y 4  y. (5.26) Therefore, f (y, v) = v and g(y, v) = y 4  y = y(y 3  1). We find critical points by solving v = 0, y(y 3  1) = 0 v = 0, y = 0 or y = 1. Hence, system (5.26) has two critical points, (0, 0) and (1, 0). In the upper half plane, y = v > 0 and so solutions flow to the right; similarly, solutions flow to the left in the lower halfplane. See Figure B.36 in the text for the direction field. This 300 Exercises 5.4
figure indicates that (0, 0) is a stable critical point (center) whereas (1, 0) is a saddle point (unstable). 25. This system has two critical points, (0, 0) and (1, 0), which are solutions to the system y = 0, x + x3 = 0. The direction field for this system is depicted in Figure B.37. From this figure we conclude that (a) the solution passing through the point (0.25, 0.25) flows around (0, 0) and thus is periodic; (b) for the solution (x(t), y(t)) passing through the point (2, 2), y(t) as t , and so this solution is not periodic; (c) the solution passing through the critical point (1, 0) is a constant (equilibrium) solution and so is periodic. 27. The direction field for given system is shown in Figure B.38 in the answers of the text. From the starting point, (1, 1), following the direction arrows the solution flows down and to the left, crosses the xaxis, has a turning point in the fourth quadrant, and then does to the left and up toward the critical point (0, 0). Thus we predict that, as t , the solution (x(t), y(t)) approaches (0, 0). 29. (a) The phase plane equation for this system is 3y dy = . dx x It is separable. Separating variables and integrating, we get 3dx dy = y x ln y = 3 ln x + C y = cx3 . So, integral curves are cubic curves. Since in the right halfplane x = x > 0, in the left halfplane x < 0, the solutions flow to the right in the right halfplane and to the left 301 Chapter 5
in the left halfplane. Solutions starting on the yaxis stay on it (x = 0); they flow up if the initial point is in the upper halfplane (because y = y > 0) and flow down if the initial point in the lower halfplane. This matches the figure for unstable node. (b) Solving the phase plane equation for this system, we get 4x dy = dx y y dy = 4x dx y 2 + 4x2 = C. Thus the integral curves are ellipses. (Also, notice that the solutions flow along these ellipses in clockwise direction because x increases in the upper halfplane and decreases in the lower halfplane.) Therefore, here we have a center (stable). (c) Solving 5x + 2y > 0 and x  4y > 0 we find that x increases in the halfplane y > 5x and decreases in the halfplane y < 5x, and y increases in the halfplane y < x/4 and decreases in the halfplane y > x/4. This leads to the scheme for the solution's flows. Thus all solutions approach the critical point (0, 0), as t , which corresponds to a stable node. (d) An analysis, similar to that in (c), shows that all the solutions flow away from (0, 0). Among pictures shown in Figure 5.7, only the unstable node and the unstable spiral have this feature. Since the unstable node is the answer to (a), we have the unstable spiral in this case. (e) The phase plane equation 4x  3y dy = , dx 5x  3y has two linear solutions, y = 2x and y = 2x/3. (One can find them by substituting y = ax into the above phase plane equation and solving for a.) Solutions starting from a point on y = 2x in the first quadrant, have x = 5x  3(2x) = x < 0 and so flow toward (0, 0); similarly, solutions, starting from a point on this line in the third quadrant, have x = x > 0 and, again, flow to (0, 0). On the other line, y = 2x/3, the picture is opposite: in the first quadrant, x = 5x  3(2x/3) = 3x > 0, and x < 0 in the third quadrant. Therefore, there are two lines, passing through the critical point (0, 0), such 302 Exercises 5.4
that solutions to the system flow into (0, 0) on one of them and flow away from (0, 0) on the other. This is the case of a saddle (unstable) point. (f) The only remaining picture is the asymptotically stable spiral. (One can also get a diagram for solution's flows with just one matching picture in Figure 5.7.) 31. (a) Setting y = v and so y = v , we transform given equation to a first order system dy = v, dx dv = f (y). dx (b) By the chain rule, dv dv dx dv = = dy dx dy dx dy f (y) = dx v dv f (y) = . dy v This equation is separable. Separation variables and integration yield v dv = f (y) dy v dv = f (y) dy 1 2 v = F (y) + K, 2 where F (y) is an antiderivative of f (y). Substituting back v = y gives the required. 33. Since S(t) and I(t) represent population and we cannot have a negative population, we are only interested in the first quadrant of the SIplane. (a) In order to find the trajectory corresponding to the initial conditions I(0) = 1 and S(0) = 700, we must solve the first order equation dI/dt aSI  bI aS  b dI = = = dS dS/dt aSI aS b 1 dI = 1 + . dS aS curves given by I(S) = S + b ln S + C. a 303 (5.27) By integrating both sides of equation (5.27) with respect to S, we obtain the integral Chapter 5
A sketch of this curve for a = 0.003 and b = 0.5 is shown in Figure B.39 in the answers of the text. (b) From the sketch in Figure B.39 in the answers of the text we see that the peak number of infected people is 295. (c) The peak number of infected people occurs when dI/dS = 0. From equation (5.27) we have b 1 dI = 0 = 1 + . dS aS Solving for S we obtain S= 0.5 b = 167 people. a 0.003 35. (a) We denote v(t) = x (t) to transform the equation d2 x 1 = x + dt2 x to an equivalent system of two first order differential equations, that is dx = v, dt 1 dv = x + . dt x (b) The phase plane equation in xvplane for the system in (a) is x + 1/(  x) dv = . dx v This equation is separable. Separating variables and integrating, we obtain v dv = 1 dx x 1 1 2 v =  x2  ln   x + C1 2 2 v = C  x2  2 ln(  x) . x + v dv = x + 1 x dx v 2 = C  x2  2 ln   x (The absolute value sign is not necessary because x < .) 304 Exercises 5.4
(c) To find critical points for the system in (a), we solve v = 0, x + 1 =0 x v = 0, x2  x + 1 = 0 v = 0, 2  4 . x= 2 For 0 < < 2, 2  4 < 0 and so both roots are complex numbers. However, for > 2 there are two distinct real solutions,  2  4 x1 = 2 and the critical points are  2  4 ,0 2 2  4 , 2 and x2 = + and + 2  4 ,0 . 2 (d) The phase plane diagrams for = 1 and = 3 are shown in Figures B.40 and B.41 in the answers section of the text. (e) From Figures B.40 we conclude that, for = 1, all solution curves approach the vertical line x = 1(= ). This means that the bar is attracted to the magnet. The case = 3 is more complicated. The behavior of the bar depends on the initial displacement x(0) and the initial velocity v(0) = x (0). From Figure B.41 we see that (with v(0) = 0) if x(0) is small enough, then the bar will oscillate about the position x = x1 ; if x(0) is close enough to , then the bar will be attracted to the magnet. It is also possible that, with an appropriate combination of x(0) and v(0), the bar will come to rest at the saddle point (x2 , 0). 37. (a) Denoting y = v, we have y = v , and (with m = = k = 1) (16) can be written as a system y = v, v = y + y, 0, = if y < 1, v = 0, if y < 1, v = 0, sign(y), if y 1, v = 0, sign(v), if v = 0 y + sign(y), if y 1, v = 0, y  sign(v), if v = 0. 305 Chapter 5
(b) The condition v = 0 corresponds to the third case in (5.28), i.e., the system has the form y = v, v = y  sign(v). The phase plane equation for this system is dv dv/dt y  sign(v) = = . dy dy/dt v We consider two cases. 1) v > 0. In this case sign(v) = 1 and we have y  1 dv = dy v where c = 2C. 2) v < 0. In this case sign(v) = 1 and we have dv y + 1 = dy v v dv = (y  1)dy (y  1)dy v 2 + (y  1)2 = c. v dv = (y + 1)dy (y + 1)dy v 2 + (y + 1)2 = c, v dv =  1 2 1 v =  (y + 1)2 + C 2 2 v dv =  1 2 1 v =  (y  1)2 + C 2 2 (c) The equation v 2 + (y + 1)2 = c defines a circle in the yvplane centered at (1, 0) and of the radius c if c > 0, and it is the empty set if c < 0. The condition v > 0 means that we have to take only the half of these circles lying in the upper half plane. Moreover, the first equation, y = v, implies that trajectories flow from left to right. Similarly, in the lower half plane, v < 0, we have concentric semicircles v 2 + (y  1)2 = c, c 0, centered at (1, 0) and flowing from right to left. 306 Exercises 5.5
(d) For the system found in (a), f (y, v) = v, 0, if y < 1, v = 0, g(y, v) = y + sign(y), if y 1, v = 0, y  sign(v), if v = 0. Since f (y, v) = 0 v = 0 and g(y, 0) = 0, if y < 1, y + sign(y), if y 1, we consider two cases. If y < 1, then g(y, 0) 0. This means that any point of the interval 1 < y < 1 is a critical point. If y 1, then g(y, 0) = y + sign(y) which is 0 if y = 1. Thus the critical point set is the segment v = 0, 1 y 1. (e) According to (c), the mass released at (7.5, 0) goes in the lower half plane from right to left along a semicircle centered at (1, 0). The radius of this semicircle is 7.5  1 = 6.5, and its other end is (1  6.5, 0) = (5.5, 0). From this point, the mass goes from left to right in the upper half plane along the semicircle centered at (1, 0) and of the radius 1  (5.5) = 4.5, and comes to the point (1 + 4.5, 0) = (3.5, 0). Then the mass again goes from right to left in the lower half plane along the semicircle centered at (1, 0) and of the radius 3.5  1 = 2.5, and comes to the point (1  2.5, 0) = (1.5, 0). From this point, the mass goes in the upper half plane from left to right along the semicircle centered at (1, 0) and of the radius 1  (1.5) = 0.5, and comes to the point (1 + 0.5, 0) = (0.5, 0). Here it comes to rest because   0.5 < 1, and there is not a lower semicircle starting at this point. See the colored curve in Figure B.42 of the text. EXERCISES 5.5: Coupled MassSpring Systems, page 284 1. For the mass m1 there is only one force acting on it; that is the force due to the spring with constant k1 . This equals k1 (x  y). Hence, we get m1 x = k1 (x  y). 307 Chapter 5
For the mass m2 there are two forces acting on it: the force due to the spring with constant k2 is k2 y; and the force due to the spring with constant k1 is k1 (y  x). So we get m2 y = k1 (x  y)  k2 y. So the system is m1 x = k1 (y  x), m2 y = k1 (y  x)  k2 y, or, in operator form, m1 D 2 + k1 [x]  k1 y = 0, k1 x + m2 D 2 + (k1 + k2 ) [y] = 0. With m1 = 1, m2 = 2, k1 = 4, and k2 = 10/3, we get (D 2 + 4) [x]  4y = 0, 4x + (2D 2 + 22/3) [y] = 0, with initial conditions: x(0) = 1, x (0) = 0, y(0) = 0, y (0) = 0. (5.28) Multiplying the second equation of the system given in (5.28) by 4, applying (2D 2 + 22/3) to the first equation of this system, and adding the results, we get D2 + 4 The characteristic equation is 3r 4 + 23r 2 + 20 = 0, 308 22 [x]  16x = 0 3 46 2 40 D + [x] = 0 2D 4 + 3 3 3D 4 + 23D 2 + 20 [x] = 0. 2D 2 + Exercises 5.5
which is a quadratic in r 2 . So r =
2 23 23 17 529  240 = . 6 6 Since 20/3 and 1 are negative, the roots of the characteristic equation are i1 and i2 , where 1 = Hence x(t) = c1 cos 1 t + c2 sin 1 t + c3 cos 2 t + c4 sin 2 t. Solving the first equation of the system given in (5.28) for y, we get y(t) = 1 1 D 2 + 4 [x] = 4 4
2 2 1 + 4 c1 cos 1 t + 1 + 4 c2 sin 1 t 2 2 + 2 + 4 c3 cos 2 t + 2 + 4 c4 sin 2 t . 20 , 3 2 = 1. Next we substitute into the initial conditions. Setting x(0) = 1, x (0) = 0 yields 1 = c1 + c3 , 0 = c2 1 + c4 2 . From the initial conditions y(0) = 0, y (0) = 0, we get 1 2 2 1 + 4 c1 + 2 + 4 c3 , 4 1 2 2 0= 1 1 + 4 c2 + 2 2 + 4 c4 . 4 0= The solution to the above system is c2 = c4 = 0, which yields the solutions x(t) =  y(t) = 9 cos 17 8 20 t cos t , 3 17 20 6 t cos t . 3 17 309 c1 =  9 , 17 c3 =  8 , 17 6 cos 17 Chapter 5
3. We define the displacements of masses from equilibrium, x, y, and z, as in Example 2. For each mass, there are two forces acting on it due to Hook's law. For the mass on the left, F11 = kx for the mass in the middle, F21 = k(y  x) finally, for the mass on the right, F31 = k(z  y) and F32 = kz. and F22 = k(z  y); and F12 = k(y  x); Applying Newton's second law for each mass, we obtain the following system mx = kx + k(y  x), my = k(y  x) + k(z  y), mz = k(z  y)  kz, or, in operator form, mD 2 + 2k [x]  ky = 0, kx + mD 2 + 2k [y]  kz = 0, ky + mD 2 + 2k [z] = 0. From the first equation, we express y= 1 mD 2 + 2k [x] k (5.29) and substitute this expression into the other two equations to get 1 mD 2 + 2k [x]  kz = 0, k  mD 2 + 2k [x] + mD 2 + 2k [z] = 0. kx + mD 2 + 2k 310 Exercises 5.5
The first equation yields z = x + and so  mD 2 + 2k [x] + mD 2 + 2k 1 mD 2 + 2k 2 k
2 1 mD 2 + 2k k 2 [x] = 1 mD 2 + 2k 2 k 2  1 [x], (5.30)  1 [x] 1 mD 2 + 2k k2
2 = mD 2 + 2k  2 [x] = 0. The characteristic equation for this homogeneous linear equation with constant coefficients is mr 2 + 2k which splits onto two equations, mr 2 + 2k = 0 and 1 2 mr 2 + 2k  2 = 0 2 k mr 2 + 2k  2k r = i (2  2)k , m mr 2 + 2k  2k 2 = 0 mr 2 + 2k + 2k = 0 r = i (2 + 2)k . m (5.32)
2 1 mr 2 + 2k k2 2 2 = 0, r = i 2k m (5.31) Solutions (5.31) and (5.32) give normal frequences 1 1 = 2 2k , m 1 2 = 2 (2  2)k , m 1 3 = 2 (2 + 2)k . m Thus, a general solution x(t) has the form x(t) = x1 (t) + x2 (t) + x3 (t), where functions xj (t) = c1j cos(2j t) + c2j sin(2j t). Note that xj 's satisfy the following differential equations: mD 2 + 2k [x1 ] = 0, 311 Chapter 5
mD 2 + 2k  mD 2 + 2k + 2k [x2 ] = 0, (5.33) 2k [x3 ] = 0. For normal modes, we find solutions yj (t) and zj (t), corresponding to xj , j = 1, 2, and 3 by using (5.29), (5.30), and identities (5.33). 1 : 1 mD 2 + 2k [x1 ] 0, k 1 2 z1 = mD 2 + 2k  1 [x1 ] = x1 ; k y1 = 2 : 1 1 mD 2 + 2k [x2 ] = mD 2 + 2k  2k + 2 [x2 ] = 2x2 , k k 1 1 2 2 mD 2 + 2k  1 [x2 ] = mD 2 + 2k  2 + 1 [x2 ] = x2 ; z2 = k k y2 = 3 : 1 1 mD 2 + 2k [x3 ] = mD 2 + 2k + 2k  2 [x3 ] =  2x3 , k k 1 1 2 2 mD 2 + 2k  1 [x3 ] = mD 2 + 2k  2 + 1 [x3 ] = x3 ; z3 = k k y3 = 5. This spring system is similar to the system in Example 2 on page 282 of the text, except the middle spring has been replaced by a dashpot. We proceed as in Example 1. Let x and y represent the displacement of masses m1 and m2 to the right of their respective equilibrium positions. The mass m1 has a force F1 acting on its left side due to the left spring and a force F2 acting on its right side due to the dashpot. Applying Hooke's law, we see that F1 = k1 x. Assuming as we did in Section 4.1 that the damping force due to the dashpot is proportional to the magnitude of the velocity, but opposite in direction, we have F2 = b (y  x ) , 312 Exercises 5.5
where b is the damping constant. Notice that velocity of the arm of the dashpot is the difference between the velocities of mass m2 and mass m1 . The mass m2 has a force F3 acting on its left side due to the dashpot and a force F4 acting on its right side due to the right spring. Using similar arguments, we find F3 = b (y  x ) and F4 = k2 y. Applying Newton's second law to each mass gives m1 x (t) = F1 + F2 = k1 x(t) + b [y (t)  x (t)] , m2 y (t) = F3 + F4 = b [y (t)  x (t)]  k2 y. Plugging in the constants m1 = m2 = 1, k1 = k2 = 1, and b = 1, and simplifying yields x (t) + x (t) + x(t)  y (t) = 0, x (t) + y (t) + y (t) + y(t) = 0. (5.34) The initial conditions for the system will be y(0) = 0 (m2 is held in its equilibrium position), x(0) = 2 (m1 is pushed to the left 2 ft), and x (0) = y (0) = 0 (the masses are simply released at time t = 0 with no additional velocity). In operator notation this system becomes (D 2 + D + 1) [x]  D[y] = 0, D[x] + y (t) + (D 2 + D + 1) [y] = 0. By multiplying the first equation above by D and the second by (D 2 + D + 1) and adding the resulting equations, we can eliminate the function y(t). Thus, we have D2 + D + 1 2  D 2 [x] = 0 D2 + D + 1 + D [x] = 0 D2 + D + 1  D D2 + 1 (D + 1)2 [x] = 0. This last equation is a fourth order linear differential equation with constant coefficients whose associated auxiliary equation has roots r = 1, 1, i, and i. Therefore, the solution to this differential equation is x(t) = c1 et + c2 tet + c3 cos t + c4 sin t 313 Chapter 5 x (t) = (c1 + c2 )et  c2 tet  c3 sin t + c4 cos t x (t) = (c1  2c2 )et + c2 tet  c3 cos t  c4 sin t. To find y(t), note that by the first equation of the system given in (5.34), we have y (t) = x (t) + x (t) + x(t). Substituting x(t), x (t), and x (t) into this equation yields y (t) = (c1  2c2 )et + c2 tet  c3 cos t  c4 sin t +(c1 + c2 )et  c2 tet  c3 sin t + c4 cos t + c1 et + c2 tet + c3 cos t + c4 sin t y (t) = (c1  c2 )et + c2 tet  c3 sin t + c4 cos t. By integrating both sides of this equation with respect to t, we obtain y(t) = (c1  c2 )et  c2 tet  c2 et + c3 cos t + c4 sin t + c5 , where we have integrated c2 tet by parts. Simplifying yields y(t) = c1 et  c2 tet + c3 cos t + c4 sin t + c5 . To determine the five constants, we will use the four initial conditions and the second equation in system (5.34). (We used the first equation to determine y). Substituting into the second equation in (5.34) gives  (c1 + c2 )et  c2 tet  c3 sin t + c4 cos t + (c1 + 2c2 )et  c2 tet  c3 cos t  c4 sin t + (c1  c2 )et + c2 tet  c3 sin t + c4 cos t + c1 et  c2 tet + c3 cos t + c4 sin t + c5 = 0, which reduces to c5 = 0. Using the initial conditions and the fact that c5 = 0, we see that x(0) = c1 + c3 = 2, y(0) = c1 + c3 = 0, 314 x (0) = (c1 + c2 ) + c4 = 0, y (0) = (c1  c2 ) + c4 = 0 . Exercises 5.5
By solving these equations simultaneously, we find c1 = 1, c2 = 1, c3 = 1, and c4 = 0. Therefore, the solution to this springmassdashpot system is x(t) = et  tet  cos t, 7. In operator notations, D 2 + 5 [x]  2y = 0, 2x + D 2 + 2 [y] = 3 sin 2t. Multiplying the first equation by (D 2 + 2) and the second equation by 2, and adding the results, we obtain D2 + 2 D 2 + 5  4 [x] = 6 sin 2t D 2 + 6 [x] = 6 sin 2t . y(t) = et + tet  cos t. D 4 + 7D 2 + 6 [x] = 6 sin 2t D2 + 1 (5.35) Since the characteristic equation, (r 2 + 1)(r 2 + 6) = 0, has the roots r = i and r = i 6, a general solution to the corresponding homogeneous equation is given by xh (t) = c1 cos t + c2 sin t + c3 cos 6t + c4 sin 6t . Due to the righthand side in (5.35), a particular solution has the form xp (t) = A cos 2t + B sin 2t . In order to simplify computations, we note that both functions, cos 2t and sin 2t, and so xp (t), satisfy the differential equation (D 2 + 4)[x] = 0. Thus, D2 + 1 D 2 + 6 [xh ] = (D 2 + 4)  3 (D 2 + 4) + 2 [xh ] = 2 (D 2 + 4)  3 [xh ] = 6xh = 6A cos 2t  6B sin 2t = 6 sin 2t 315 Chapter 5 and A = 0, B = 1 xh (t) =  sin 2t x(t) = xh (t) + xp (t) = c1 cos t + c2 sin t + c3 cos 6t + c4 sin 6t  sin 2t . From the first equation in the original system, we have y(t) = 1 (x + 5x) 2 = 2c1 cos t + 2c2 sin t  1 1 1 c3 cos 6t  c4 sin 6t  sin 2t . 2 2 2 We determine constants c1 and c3 using the initial conditions x(0) = 0 and y(0) = 1. 0 = x(0) = c1 + c3 , 1 = y(0) = 2c1  c3 /2 c3 = c1 , 2c1  (c1 ) /2 = 1 c3 = 2/5, c1 = 2/5. To find c2 and c4 , compute x (t) and y (t), evaluate these functions at t = 0, and use the other two initial conditions, x (0) = y (0) = 0. This yields 0 = x (0) = c2 + 6c4  2, 0 = y (0) = 2c2  6c4 /2  1 Therefore, the required solution is 4 2 2 6 sin 6t  sin 2t , x(t) = cos t + sin t  cos 6t + 5 5 5 5 8 1 6 4 1 sin 6t  sin 2t . y(t) = cos t + sin t + cos 6t  5 5 5 10 2 9. Writing the equations of this system in operator form we obtain mg + k [x1 ]  kx2 = 0, l mg + k [x2 ] = 0. kx1 + mD 2 + l mD 2 + Applying {mD 2 + (mg/l + k)} to the first equation, multiplying the second equation by k, and then adding, results in mD 2 + 316 mg +k l
2 c4 = 6/5, c2 = 4/5.  k 2 [x1 ] = 0. Exercises 5.6
This equation has the auxiliary equation mr 2 + with roots i and mg +k l
2  k 2 = mr 2 + mg l mr 2 + mg + 2k = 0 l g/l g/l and i (g/l) + (2k/m). As discussed on page 211 of the text (g/l) + (2k/m) are the normal angular frequencies. To find the normal frequencies we divide each one by 2 and obtain 1 2 EXERCISES 5.6: g l and 1 2 g 2k + . l m Electrical Circuits, page 291 1. In this problem, R = 100 , L = 4 H, C = 0.01 F, and E(t) = 20 V. Therefore, the equation (4) on page 287 of the text becomes 4 d(20) dI d2 I + 100I = =0 + 100 2 dt dt dt d2 I dI + 25I = 0. + 25 2 dt dt The roots of the characteristic equation, r 2 + 25r + 25 = 0, are r= and so a general solution is I(t) = c1 e(255 21)t/2 25 (25)2  4(25)(1) 25 5 21 = , 2 2 + c2 e(25+5 21)t/2 . To determine constants c1 and c2 , first we find the initial value I (0) using given I(0) = 0 and q(0) = 4. Substituting t = 0 into equation (3) on page 287 of the text (with dq/dt replaced by I(t)), we obtain L d[I(t)] 1 + RI(t) + q(t) = E(t) dt C 1 (4) = 20 4I (0) + 100(0) + 0.01 I (0) = 95. 317 Chapter 5
Thus, I(t) satisfies I(0) = 0, I (0) = 95. Next, we compute c1 (25  5 21) (25521)t/2 c2 (25 + 5 21) (25+521)t/2 I (t) = e e + , 2 2 substitute t = 0 into formulas for I(t) and I (t), and obtain the system 0 = I(0) = c1 + c2 , 95 = I (0) = c1 (25  5 21)/2 + c2 (25 + 5 21)/2 c1 = 19/ 21 , c2 = 19/ 21. So, the solution is 19 e(255 21)t/2  e(25+5 21)t/2 . I(t) = 21 3. In this problem L = 4, R = 120, C = (2200)1 , and E(t) = 10 cos 20t. Therefore, we see that 1/C = 2200 and E (t) = 200 sin 20t. By substituting these values into equation (4) on page 287 of the text, we obtain the equation 4 By simplifying, we have d2 I dI + 2200I = 200 sin 20t. + 120 dt2 dt dI d2 I + 550I = 50 sin 20t. + 30 2 dt dt (5.36) The auxiliary equation associated with the homogeneous equation corresponding to (5.36) above is r 2 + 30r + 550 = 0. This equation has roots r = 15 5 13i. Therefore, the transient current, that is Ih (t), is given by Ih (t) = e15t C1 cos 5 13t + C2 sin 5 13t . By the method of undetermined coefficients, a particular solution, Ip (t), of equation (5.36) will be of the form Ip (t) = ts [A cos 20t + B sin 20t]. Since neither y(t) = cos 20t nor y(t) = sin 20t is a solution to the homogeneous equation (that is the system is not at resonance), we can let s = 0 in Ip (t). Thus, we see that Ip (t), the steadystate current, has the form Ip (t) = A cos 20t + B sin 20t. 318 Exercises 5.6
To find the steadystate current, we must, therefore, find A and B. To accomplish this, we observe that Ip (t) = 20A sin 20t + 20B cos 20t, Ip (t) = 400A cos 20t  400B sin 20t. Plugging these expressions into equation (5.36) yields Ip (t) + 30Ip (t) + 550I(t) = 400A cos 20t  400B sin 20t  600A sin 20t + 600B cos 20t +550A cos 20t + 550B sin 20t = 50 sin 20t (150A + 600B) cos 20t + (150B  600A) sin 20t = 50 sin 20t. By equating coefficients we obtain the system of equations 15A + 60B = 0, 60A + 15B = 5. By solving these equations simultaneously for A and B, we obtain A = 4/51 and B = 1/51. Thus, we have the steadystate current given by 1 4 cos 20t  sin 20t. Ip (t) = 51 51 As was observed on page 290 of the text, there is a correlation between the RLC series circuits and mechanical vibration. Therefore, we can discuss the resonance frequency of the RLC series circuit. To do so we associate the variable L with m, R with b, and 1/C with k. Thus, we see that the resonance frequency for an RLC series circuit is given by r /(2), where r = provided that R2 < 2L/C. For this problem R2 = 14, 400 < 2L/C = 17, 600 . Therefore, we can find the resonance frequency of this circuit. To do so we first find R2 1 2200 14400  2 =  = 10. CL 2L 4 32 Hence the resonance frequency of this circuit is 10/(2) = 5/. r = 319 R2 1  2, CL 2L Chapter 5
5. In this problem, C = 0.01 F, L = 4 H, and R = 10 . Hence, the equation governing the RLC d2 I 1 d dI E0 + I= (E0 cos t) =  sin t . + 10 dt2 dt 0.01 dt 4 The frequency response curve M() for an RLC curcuit is determined by 4 M() = 1 [(1/C)  L 2 ]2 + R2 2 , circuit is which comes from the comparison Table 5.3 on page 290 of the text and equation (13) in Section 4.9. Therefore M() = 1 = [(1/0.01)  4 2 ]2 + (10)2 2 1 . (100  4 2 )2 + 100 2 The graph of this function is shown in Figure B.43 in the answers of the text. M() has its maximal value at the point 0 = x0 , where x0 is the point where the quadratic function (100  4x)2 + 100x attains its minimum (the first coordinate of the vertex). We find that 0 = 175 8 and M(0 ) = 2 0.02 . 25 15 7. This spring system satisfies the differential equation 7 d2 x dx + 3x = 10 cos 10t. +2 dt2 dt Since we want to find an RLC series circuit analog for the spring system with R = 10 ohms, we must find L, 1/C, and E(t) so that the differential equation L d2 q dq 1 + 10 + q = E(t) 2 dt dt C corresponds to the one above. Thus, we want E(t) = 50 cos 10t volts, L = 35 henrys, and C = 1/15 farads. 11. For this electric network, there are three loops. Loop 1 is through a 10V battery, a 10 resistor, and a 20H inductor. Loop 2 is through a 10V battery, a 10 resistor, a 5 resistor, and a (1/30)F capacitor. Loop 3 is through a 5 resistor, a (1/30)F capacitor, and a 20H inductor. 320 Exercises 5.6
Therefore, applying Kirchhoff's second law to this network yields the three equations given by Loop 1 : Loop 2 : Loop 3 : dI2 = 10, dt 10I1 + 5I3 + 30q3 = 10, dI2 5I3 + 30q3  20 = 0. dt 10I1 + 20 Since the equation for Loop 2 minus the equation for Loop 1 yields the remaining equation, we will use the first and second equations above for our calculations. By examining a junction point, we see that we also have the equation I1 = I2 +I3 . Thus, we have I1 = I2 +I3 . We begin by dividing the equation for Loop 1 by 10 and the equation for Loop 2 by 5. Differentiating the equation for Loop 2 yields the system I1 + 2 dI2 = 1, dt dI1 dI3 2 + + 6I3 = 0, dt dt where I3 = q3 . Since I1 = I2 + I3 and I1 = I2 + I3 , we can rewrite the system using operator notation in the form (2D + 1)[I2 ] + I3 = 1, (2D)[I2 ] + (3D + 6)[I3 ] = 0. If we multiply the first equation above by (3D + 6) and then subtract the second equation, we obtain {(3D + 6)(2D + 1)  2D} [I2 ] = 6 6D 2 + 13D + 6 [I2 ] = 6. This last differential equation is a linear equation with constant coefficients whose associated equation, 6r 2 +13r+6 = 0, has roots 3/2, 2/3. Therefore, the solution to the homogeneous equation corresponding to the equation above is given by I2h (t) = c1 e3t/2 + c2 e2t/3 . By the method of undetermined coefficients, the form of a particular solution to the differential equation above will be I2p (t) = A. By substituting this function into the differential equation, 321 Chapter 5
we see that a particular solution is given by I2p (t) = 1. Thus, the current, I2 , will satisfy the equation I2 (t) = c1 e3t/2 + c2 e2t/3 + 1. As we noticed above, I3 can now be found from the first equation 3 2 I3 (t) = (2D + 1)[I2 ] + 1 = 2  c1 e3t/2  c2 e2t/3  c1 e3t/2 + c2 e2t/3 + 1 + 1 2 3 1 I3 (t) = 2c1 e3t/2 + c2 e2t/3 . 3 To find I1 , we will use the equation I1 = I2 + I3 . Therefore, we have I1 (t) = c1 e3t/2 + c2 e2t/3 + 1 + 2c1 e3t/2 + I1 (t) = 3c1 e3t/2 + 4 c2 e2t/3 + 1. 3 1 c2 e2t/3 3 We will use the initial condition I2 (0) = I3 (0) = 0 to find the constants c1 and c2 . Thus, we have I2 (0) = c1 + c2 + 1 = 0 and I3 (0) = 2c1 + 1 c2 = 0. 3 Solving these two equations simultaneously yields c1 = 1/5 and c2 = 6/5. Therefore, the equations for the currents for this electric network are given by 3 3t/2 e  5 1 I2 (t) = e3t/2  5 2 3t/2 I3 (t) = e  5 I1 (t) = 8 2t/3 e + 1, 5 6 2t/3 e + 1, 5 2 2t/3 e . 5 13. In this problem, there are three loops. Loop 1 is through a 0.5 H inductor and a 1 resistor. Loop 2 is through is through a 0.5 H inductor, a 0.5 F capacitor, and a voltage source supplying the voltage cos 3t V at time t. Loop 3 is through a 1 resistor, a 0.5 F capacitor, and the 322 Exercises 5.6
voltage source. We apply Kirchhoff's voltage law, EL + ER + EC = E(t), to Loop 1 and Loop 2 to get two equations connecting currents in the network. (Similarly to Example 2 and Problem 11, there is no need to apply Kirchhoff's voltage law to Loop 3 because the resulting equation is just a linear combination of those for other two loops.) Loop 1: EL + ER = 0 Loop 2: EL + EC = cos 3t 0.5 q3 dI1 + = cos 3t dt 0.5 dI1 + 4q3 = 2 cos 3t. dt (5.38) 0.5 dI1 + 1 I2 = 0 dt dI1 + 2I2 = 0. dt (5.37) Additionally, at joint points, by Kirchhoff's current law, I1 + I2 + I3 = 0 I1 + I2 + dq3 = 0. dt (5.39) Putting (5.37)(5.39) together yields the following system: dI1 + 2I2 = 0, dt dI1 + 4q3 = 2 cos 3t, dt dq3 I1 + I2 + =0 dt or, in operator form, D[I1 ] + 2I2 = 0, D[I1 ] + 4q3 = 2 cos 3t, I1 + I2 + D[q3 ] = 0 with the initial condition I1 (0) = I2 (0) = I3 (0) = 0 (I3 = dq3 /dt). From the first equation, I2 = (1/2)D[I1 ], which (when substituted into the third equation) leads to the system D[I1 ] + 4q3 = 2 cos 3t, (D + 2)[I1 ] + 2D[q3 ] = 0. 323 Chapter 5
Multiplying the first equation by D, the second equation by 2, and subtracting the results, we eliminate q3 : D 2 + 2D + 4 [I1 ] = 6 sin 3t. The roots of the characteristic equation, r 2 + 2r + 4 = 0, are r = 1 3i, and so a general solution to the corresponding homogeneous equation is I1h = C1 et cos 3t + C2 et sin 3t. A particular solution has the form I1p = A cos 3t + B sin 3t. Substitution into the equation yields (5A + 6B) cos 3t + (6A  5B) sin 3t = 6 sin 3t Therefore, 30 36 cos 3t + sin 3t. = C1 et cos 3t + C2 et sin 3t + 61 61 Substituting this solution into (5.37)we find that 1 dI1 I2 =  2 dt 54 C1 3 + C2 t 45 C1  C2 3 t e cos 3t + e sin 3t  cos 3t + sin 3t. = 2 2 61 61 The initial condition, I1 (0) = I2 (0) = 0 yields C1 + 36/61 = 0, (C1  C2 3)/2  45/61 = 0 Thus C1 = 36/61, C2 = 42 3/61. I1 = I1h + I1p 5A + 6B = 0, 6A  5B = 6 A = 36/61, B = 30/61. D 2 + 2(D + 2) [I1 ] = 6 sin 3t 30 36 t 42 3 t 36 e sin 3t + cos 3t + sin 3t, I1 =  e cos 3t  61 61 61 61 54 45 t 39 3 t 45 I2 = e cos 3t  e sin 3t  cos 3t + sin 3t, 61 61 61 61 24 81 3 3 t 81 I3 = I1  I2 =  et cos 3t  e sin 3t + cos 3t  sin 3t. 61 61 61 61 324 Exercises 5.7
EXERCISES 5.7: Dynamical Systems, Poincar` Maps, and Chaos, page 301 e 1. Let = 3/2. Using system (3) on page 294 of the text with A = F = 1, = 0, and = 3/2, we define the Poincar map e xn = sin(3n) + 4 4 1 = sin(3n) + = , (9/4)  (4/4) 5 5 3 3 vn = cos(3n) = (1)n , 2 2 for n = 0, 1, 2, . . . . Calculating the first few values of (xn , vn ), we find that they alternate between (4/5, 3/2) and (4/5, 3/2). Consequently, we can deduce that there is a subharmonic solution of period 4. Let = 3/5. Using system (3) on page 294 of the text with A = F = 1, = 0, and = 3/5, we define the Poincar map e xn = sin 6n 6n 1 = sin + 5 (9/25)  1 5 6n 3 6n = (0.6) cos , vn = cos 5 5 5  1.5625 , e for n = 0, 1, 2, . . . . Calculating the first few values of (xn , vn ), we find that the Poincar map cycles through the points (1.5625, 0.6), (2.1503, 0.4854), (0.6114, 0.1854), (2.5136, 0.1854), (0.9747, 0.4854), n = 0, 5, 10, . . . , n = 1, 6, 11, . . . , n = 2, 7, 12, . . . , n = 3, 8, 13, . . . , n = 4, 9, 14, . . . . Consequently, we can deduce that there is a subharmonic solution of period 10. 3. With A = F = 1, = 0, = 1, b = 0.1, and = 0 (because tan = ( 2  1)/b = 0) the solution (5) to equation (4) becomes x(t) = e0.05t sin Thus v(t) = x (t) = e
0.05t 3.99 t + 10 sin t. 2 3.99 cos 2 3.99 t 2 0.05 sin 3.99 t+ 2 + 10 cos t 325 Chapter 5
v
500 400 300 200 100 x
80 60 40 20 0 Figure 5A: Poincar section for Problem 3. e and, therefore, xn = x(2n) e0.1n sin(1.997498n), vn = v(2n) e0.1n (0.05 sin(1.997498n) + 0.998749 cos(1.997498n)) + 10. The values of xn and vn for n = 0, 1, . . . , 20 are listed in Table 5F, and points (xn , vn ) are shown in Figure 5A. When n , the points (xn , vn ) become unbounded because of e0.1n term. 5. We want to construct the Poincar map using t = 2n for x(t) given in equation (5) on e page 295 of the text with A = F = 1, = 0, = 1/3, and b = 0.22. Since 2  1 tan = = 4.040404 , b we take = tan1 (4.040404) = 1.328172 and get xn = x(2n) = e0.22n sin(0.629321n)  (1.092050) sin(1.328172), vn = x (2n) = 0.11e0.22n sin(0.629321n) + (1.258642)e0.22n cos(0.629321n) +(1.092050) cos(1.328172). 326 Exercises 5.7 Table 5F: Poincar map for Problem 3. e
n 0 1 2 3 4 5 6 7 8 9 10 xn 0 0.010761 0.029466 0.060511 0.110453 0.189009 0.310494 0.495883 0.775786 1.194692 1.817047 vn 10.998749 11.366815 11.870407 12.559384 13.501933 14.791299 16.554984 18.967326 22.266682 26.778923 32.949532 n 11 12 13 14 15 16 17 18 19 20 xn 2.735915 4.085318 6.057783 8.929255 13.09442 19.11674 27.79923 40.28442 58.19561 83.83579 vn 41.387469 52.925111 68.700143 90.267442 119.75193 160.05736 215.15152 290.45581 393.37721 534.03491 In Table 5G we have listed the first 21 values of the Poincar map. e As n gets large, we see that xn (1.092050) sin(1.328172) 1.060065 , vn (1.092050) cos(1.328172) 0.262366 . Hence, as n , the Poincar map approaches the point (1.060065, 0.262366). e 7. Let A, and A , denote the values of constants A, in solution formula (2), corresponding to initial values (x0 , v0 ) and (x , v0 ), respectively. 0 (i) From recursive formulas (3) we conclude that xn  F/( 2  1) = A sin(2n + ), vn / = A cos(2n + ), and so (A, 2n+) are polar coordinates of the point (vn /, xn F/( 2 1)) in vxplane. Similarly, (A , 2n+ ) represent polar coordinates of the point (vn /, x F/( 2 1)). n 327 Chapter 5 Table 5G: Poincar map for Problem 5. e
n 0 1 2 3 4 5 6 7 8 9 10 xn 1.060065 0.599847 1.242301 1.103418 0.997156 1.074094 1.070300 1.052491 1.060495 1.061795 1.059271 vn 1.521008 0.037456 0.065170 0.415707 0.251142 0.228322 0.278664 0.264458 0.257447 0.263789 0.263037 n 11 12 13 14 15 16 17 18 19 20 xn 1.059944 1.060312 1.059997 1.060030 1.060096 1.060061 1.060058 1.060068 1.060065 1.060064 vn 0.261743 0.262444 0.262491 0.262297 0.262362 0.262385 0.262360 0.262364 0.262369 0.262366 Therefore, (vn /, x  F/( 2  1)) (vn /, xn  F/( 2  1)) n as A A and if A = 0 or as A 0 (regardless of ) if A = 0. Note that the convergence is uniform with respect to n. (One can easily see this from the distance formula in polar coordinates.) This is equivalent to x  F/( 2  1) xn  F/( 2  1), n vn / vn / x xn , n vn vn uniformly with respect to n. (ii) On the other hand, A and satisfy A sin + F/( 2  1) = x , 0 A cos = v0 A = (x  F/( 2  1))2 + (v0 /)2 , 0 cos = v0 / (A ) . Therefore, A is a continuous function of (x , v0 ) and so A A as (x , v0 ) (x0 , v0 ). 0 0 If (x0 , v0 ) is such that A = 0, then , as a function of (x , v0 ), is also continuous at 0 (x0 , v0 ) and, therefore, as (x , v0 ) (x0 , v0 ). 0 328 Exercises 5.7
Combining (i) and (ii) we conclude that (x , vn ) (xn , vn ) n as (x , v0 ) (x0 , v0 ) 0 uniformly with respect to n. Thus, if (x , v0 ) is close to (x0 , v0 ), (x , vn ) is close to (xn , vn ) 0 n for all n. 9. (a) When x0 = 1/7, the doubling modulo 1 map gives x1 = 2 2 (mod 1) = , 7 7 8 1 (mod 1) = , 7 7 4 4 (mod 1) = , 7 7 x2 = 4 4 (mod 1) = , 7 7 2 2 (mod 1) = , 7 7 1 8 (mod 1) = , 7 7 x3 = x5 = x4 = x6 = x7 = This is the sequence 2 2 etc. (mod 1) = , 7 7 1 2 4 1 k , , , , . . . . For x0 = , k = 2, . . . , 6, we obtain 7 7 7 7 7 2 4 1 2 , , , ,... , 7 7 7 7 4 1 2 4 , , , ,... , 7 7 7 7 6 5 3 6 , , , ,... . 7 7 7 7 3 6 5 3 , , , ,... , 7 7 7 7 5 3 6 5 , , , ,... , 7 7 7 7 These sequences fall into two classes. The first has the repeating sequence the second has the repeating sequence (c) To see what happens, when x0 = Then, x1 = 2 1 3 3 (mod 1) = (mod 1) = , 4 2 2 3 6 5 , , . 7 7 7 1 2 4 , , and 7 7 7 k 3 3 , let's consider the special case when x0 = 2 = . j 2 2 4 329 Chapter 5
x2 = 2 x3 = 0, x4 = 0, etc. Observe that x2 = 22 In general, xj = 2j Consequently, xn = 0 for n j. 11. (a) A general solution to equation (6) is given by x(t) = xh (t) + xp (t), where xh (t) = Ae0.11t sin 9879t + k 2j (mod 1) = k (mod 1) = 0. 3 22 (mod 1) = 3 (mod 1) = 0. 1 (mod 1) = 1 (mod 1) = 0, 2 is the transient term (a general solution to the corresponding homogeneous equation) and xp (t) = 1 sin t + 0.22 1 sin 2t + , 1 + 2(0.22)2 tan =  1 , 0.22 2 is the steadystate term (a particular solution to (6)). (xp (t) can be found, say, by applying formula (7), Section 4.12, and using Superposition Principle of Section 4.7.) Differentiating x(t) we get 1 v(t) = xh (t) + xp (t) = xh (t) + cos t + 0.22 2 cos 2t + . 1 + 2(0.22)2 The steadystate solution does not depend on initial values x0 and v0 ; these values affect only constants A and in the transient part. But, as t , xh (t) and xh (t) tend to zero and so the values of x(t) and v(t) approach the values of xp (t) and xp (t), respectively. Thus the limit set of points (x(t), v(t)) is the same as that of (xp (t), xp (t)) which is independent of initial values. 330 Review Problems
(b) Substitution t = 2n into xp (t) and xp (t) yields xn = x(2n) = xh (2n) + vn = v(2n) = xh (2n) + 1 1 + 0.22 1 + 2(0.22)2 sin 22n + , 2 cos 22n + . 1 + 2(0.22)2 As n , xh (2n) 0 and xh (2n) 0. Therefore, for n large, xn 1 1 + 2(0.22)2 2 1 vn cos 22n + = c + 2a cos 2 2n + . + 0.22 1 + 2(0.22)2 sin 22n + = a sin 2 2n + , (c) From part (b) we conclude that, for n large x2 a2 sin2 2 2n + n and (vn  c)2 2a2 cos2 2 2n + . Dividing the latter by 2 and summing yields x2 + n (vn  c)2 a2 sin2 2 2n + + cos2 2 2n + 2 = a2 , and the error (coming from the transient part) tends to zero as n . Thus any limiting point of the sequence (xn , vn ) satisfies the equation x2 + (v  c)2 = a2 , 2 which is an ellipse centered at (0, c) with semiaxes a and a 2. REVIEW PROBLEMS: page 304 1. Expressing the system in the operator notation gives D[x] + D 2 + 1 [y] = 0, D 2 [x] + D[y] = 0. 331 Chapter 5
Eliminating x by applying D to the first equation and subtracting the second equation from it yields D D 2 + 1  D [y] = 0 Thus on integrating 3 times we get y(t) = C3 + C2 t + C1 t2 . We substitute this solution into the first equation of given system to get x =  (y + y) =  (2C1 ) + (C3 + C2 t + C1 t2 ) =  (C3 + 2C1 ) + C2 t + C1 t2 . Integrating we obtain x(t) =  (C3 + 2C1 ) + C2 t + C1 t2 dt = C4  (C3 + 2C1 )t  1 1 C2 t2  C1 t3 . 2 3 D 3 [y] = 0. Thus the general solution of the given system is x(t) = C4  (C3 + 2C1 )t  y(t) = C3 + C2 t + C1 t2 . 3. Writing the system in operator form yields (2D  3)[x]  (D + 1)[y] = et , (4D + 15)[x] + (3D  1)[y] = et . (5.40) 1 1 C2 t2  C1 t3 , 2 3 We eliminate y by multiplying the first equation by (3D  1), the second by (D + 1), and summing the results. {(2D  3)(3D  1) + (4D + 15)(D + 1)} [x] = (3D  1)[et ] + (D + 1)[et ] (D 2 + 9)[x] = et . Since the characterictic equation, r 2 + 9 = 0, has roots r = 3i, a general solution to the corresponding homogeneous equation is xh (t) = c1 cos 3t + c2 sin 3t. 332 Review Problems
We look for a particular solution of the form xp (t) = Aet . Substituting this function into the equation, we obtain Aet + 9Aet = et and so A= 1 10 xp (t) = et , 10 et . 10 To find y, we multiply the first equation in (5.40) by 3 and add to the second equation. This x(t) = xh (t) + xp (t) = c1 cos 3t + c2 sin 3t + yields 2(D + 3)[x]  4y = 3et + et . Thus y = 3 1 1 (D + 3)[x]  et  et 2 4 4 3(c1  c2 ) 11 t 1 t 3(c1 + c2 ) cos 3t  sin 3t  e  e . = 2 2 20 4 5. Differentiating the second equation, we obtain y = z . We eliminate z from the first and the third equations by substituting y for z and y for z into them: x = y  y, y =y x or, in operator notation, D[x]  (D  1)[y] = 0, x + (D 2  D)[y] = 0. We eliminate y by applying D to the first equation and adding the result to the second equation: D 2 [x]  D(D  1)[y] + x + (D 2  D)[y] = 0 D 2 + 1 [x] = 0. x  y + y = 0, y y +x=0 (5.41) This equation is the simple harmonic equation, and its general solution is given by x(t) = C1 cos t + C2 sin t. 333 Chapter 5
Substituting x(t) into the first equation of the system (5.41) yields y  y = C1 sin t + C2 cos t. The general solution to the corresponding homogeneous equation, y  y = 0, is yh (t) = C3 et . We look for a particular solution to (5.42) of the form yp (t) = C4 cos t+C5 sin t. Differentiating, we obtain yp (t) = C4 sin t + C5 cos t. Thus the equation (5.42) becomes C1 sin t + C2 cos t = yp  y = (C4 sin t + C5 cos t)  (C4 cos t + C5 sin t) = (C5  C4 ) cos t  (C5 + C4 ) sin t. Equating the coefficients yields C5  C4 = C2 , C5 + C4 = C1 (by adding the equations) 2C5 = C1 + C2 C5 = (5.43) C1 + C2 . 2 (5.42) From the second equation in (5.43), we find C4 = C1  C5 = C1  C2 . 2 Therefore, the general solution to the equation (5.42) is y(t) = yh (t) + yp (t) = C3 et + C1  C2 C1 + C2 cos t + sin t. 2 2 Finally, we find z(t) from the second equation: z(t) = y (t) = C1 + C2 C1  C2 cos t + sin t 2 2 C1 + C2 C1  C2 sin t + cos t. = C3 et  2 2 C3 et + Hence, the general solution to the given system is x(t) = C1 cos t + C2 sin t, 334 Review Problems
C1  C2 C1 + C2 cos t + sin t, 2 2 C1 + C2 C1  C2 sin t + cos t. z(t) = C3 et  2 2 y(t) = C3 et + To find constants C1 , C2 , and C3 , we use the initial conditions. So we get 0 = x(0) = C1 cos 0 + C2 sin 0 = C1 , C1  C2 C1  C2 C1 + C2 0 = y(0) = C3 e0 + cos 0 + sin 0 = C3 + , 2 2 2 C1 + C2 C1  C2 C1 + C2 sin 0 + cos 0 = C3 + , 2 = z(0) = C3 e0  2 2 2 which simplifies to C1 = 0, C1  C2 + 2C3 = 0, C1 + C2 + 2C3 = 4. Solving we obtain C1 = 0, C2 = 2, C3 = 1 and so x(t) = 2 sin t, y(t) = et  cos t + sin t, z(t) = et + cos t + sin t. 7. Let x(t) and y(t) denote the mass of salt in tanks A and B, respectively. The only difference between this problem and the problem in Section 5.1 is that a brine solution flows in tank A instead of pure water. This change affects the input rate for tank A only, adding 6 L/min 0.2 kg/L = 1.2 kg/min to the original (y/12) kg/min. Thus the system (1) on page 242 becomes 1 1 y + 1.2 , x = x+ 3 12 1 1 y = x  y. 3 3 Following the solution in Section 5.1, we express x = 3y + y from the second equation and substitute it into the first equation. (3y + y) =  1 1 (3y + y) + y + 1.2 3 12 3y + 2y + 1 y = 1.2 . 4 335 Chapter 5
A general solution to the corresponding homogeneous equation is given in (3) on page 243 of the text: yh (t) = c1 et/2 + c2 et/6 . A particular solution has the form yp (t) C, which results 3(C) + 2(C) + 1 C = 1.2 4 C = 4.8 . Therefore, yp (t) 4.8, and a general solution to the system is y(t) = yh (t) + yp (t) = c1 et/2 + c2 et/6 + 4.8 , c1 c2 x(t) = 3y (t) + y(t) =  et/2 + et/6 + 4.8 . 2 2 We find constants c1 and c2 from the initial conditions, x(0) = 0.1 and y(0) = 0.3 . Substitution yields the system  c1 c2 + + 4.8 = 0.1 , 2 2 c1 + c2 + 4.8 = 0.3 . Solving, we obtain c1 = 49/20, c2 = 139/20, and so 49 t/2 139 t/6  + 4.8 , e e 40 40 49 t/2 139 t/6 e e y(t) =  + 4.8 . 20 20 x(t) =  9. We first rewrite the given differential equation in an equivalent form as y = 1 5 + et y  2y . 3 Denoting x1 (t) = y(t), x2 (t) = y (t), and x3 (t) = y (t), we conclude that x1 = y = x2 , x2 = (y ) = y = x3 , 1 5 + et x1  2x2 , x3 = (y ) = y = 3 336 Review Problems
that is, x1 = x2 , x2 = x3 , 1 5 + et x1  2x2 . x3 = 3 11. This system is equivalent to x =ty y , y =x x . Next, we introduce, as additional unknowns, derivatives of x(t) and y(t): x1 (t) := x(t), x4 (t) := y(t), With new variables, the system becomes x = (x ) =: x3 = t  y  y =: t  x5  x6 , y = (y ) =: x6 = x  x =: x2  x3 . Also, we have four new equations connecting xj 's: x1 = x =: x2 , x2 = (x ) = x =: x3 , x4 = y =: x5 , x5 = (y ) = y =: x6 . Therefore, the answer is x1 = x2 , x2 = x3 , x3 = t  x5  x6 , 337 x2 (t) := x (t), x5 (t) := y (t), x3 (t) := x (t), x6 (t) := y (t). Chapter 5
x4 = x5 , x5 = x6 , x6 = x2  x3 . 13. With the notation used in (1) on page 264 of the text, f (x, y) = 4  4y, g(x, y) = 4x, and the phase plane equation (see equation (2) on page 265 of the text) can be written as g(x, y) 4x x dy = = = . dx f (x, y) 4  4y y1 This equation is separable. Separating variables yields (y  1) dy = x dx (y  1) dy = x dx (y  1)2 + C = x2 or x2  (y  1)2 = C, where C is an arbitrary constant. We find the critical points by solving the system f (x, y) = 4  4y = 0, g(x, y) = 4x = 0 So, (0, 1) is the unique critical point. For y > 1, dx = 4(1  y) < 0, dt which implies that trajectories flow to the left. Similarly, for y < 1, trajectories flow to the right. Comparing the phase plane diagram with those given on Figure 5.12 on page 270 of the text, we conclude that the critical point (0, 1) is a saddle (unstable) point. 15. Some integral curves and the direction field for the given system are shown in Figure 5B. Comparing this picture with Figure 5.12 on page 270 of the text, we conclude that the origin is an asymptotically stable spiral point. 338 y = 1, x = 0. Review Problems 1 y 0.5 1 0.5 0 0.5 x 1 0.5 1 Figure 5B: Integral curves and the direction field for Problem 15. 17. A trajectory is a path traced by an actual solution pair (x(t), y(t)) as t increases; thus it is a directed (oriented) curve. An integral curve is the graph of a solution to the phase plane equation; it has no direction. All trajectories lie along (parts of) integral curves. A given integral curve can be the underlying point set for several different trajectories. 19. We apply Kirchhoff's voltage law to Loops 1 and 2. Loop 1 contains a capacitor C and a resistor R2 ; note that the direction of the loop is opposite to that of I2 . Thus we have q  R2 I2 = 0 C q = R2 I2 , C where q denotes the charge of the capacitor. Loop 2 consists of an inductor L and two resistors R1 and R2 ; note that the loop direction is opposite to the direction of I3 . Therefore, R2 I2  R1 I3  LI3 = 0 R2 I2 = R1 I3 + LI3 . 339 Chapter 5
For the top juncture, all the currents flow out, and the Kirchhoff's current law gives I1  I2  I3 = 0 I1 + I2 + I3 = 0. Therefore, the system, describing the current in RLC, is q = R2 I2 , C R2 I2 = R1 I3 + LI3 , I1 + I2 + I3 = 0. With given data, R1 = R2 = 1 , L = 1 H, and C = 1 F, and the relation I1 = dq/dt, this system becomes q = I2 , I2 = I3 + I3 , q + I2 + I3 = 0. Replacing in the last two equations I2 by q, we get I3 + I3  q = 0, q + q + I3 = 0. We eliminate q by substituting q = I3 + I3 into the second equation and obtain I3 + 2I3 + 2I3 = 0. The characteristic equation, r 2 + 2r + 2 = 0, has roots r = 1 i and so, a general solution to this homogeneous equation is I3 = et (A cos t + B sin t). Thus I2 = q = I3 + I3 = et (A cos t + B sin t) + et (A sin t + B cos t) + et (A cos t + B sin t) = et (B cos t  A sin t) and I1 = dq = et (B cos t  A sin t) + et (B sin t  A cos t) dt = et [(A  B) sin t  (A + B) cos t]. 340 CHAPTER 6: Theory of Higher Order Linear Differential Equations
EXERCISES 6.1: Basic Theory of Linear Differential Equations, page 324 1. Putting the equation in standard form, y  we find that p1 (x) 0, 3 p2 (x) =  , x p3 (x) = ex , x and q(x) = x2  1 . x 3 ex x2  1 y + y= , x x x Functions p2 (x), p3 (x), and q(x) have only one point of discontinuity, x = 0, while p1 (x) is continuous everywhere. Therefore, all these functions are continuous on (, 0) and (0, ). Since the initial point, x0 = 2, belongs to (, 0), Theorem 1 guarantees the existence of a unique solution to the given initial value problem on (, 0). 3. For this problem, p1 (x) = 1, p2 (x) = x  1, and g(x) = tan x. Note that p1 (x) is contin uous everywhere, p2 (x) is continuous for x 1, and g(x) is continuous everywhere except at odd multiples of /2. Therefore, these three functions are continuous simultaneously on the intervals 1, , 2 3 , 2 2 , 3 5 , 2 2 ,... . Because 5, the initial point, is in the interval (3/2, 5/2), Theorem 1 guarantees that we have a unique solution to the initial value problem on this interval. 5. Dividing the equation by x x + 1, we obtain 1 1 y + y = 0. y  x x+1 x+1 341 Chapter 6 Thus p1 (x) 0, p2 (x) = 1/(x x + 1), p3 (x) = 1/ x + 1, and g(x) 0. Functions p1 (x) and q(x) are continuous on whole real line; p3 (x) is defined and continuous for x > 1; p2 (x) is defined and continuous for x > 1 and x = 0. Therefore, all these function is continuous on (1, 0) and (0, ). The initial point lies on (0, ), and so, by Theorem 1, the given initial value problem has a unique solution on (0, ). 7. Assume that c1 , c2 , and c3 are constants for which c1 e3x + c2 e5x + c3 ex 0 on (, ). (6.1) If we show that this is possible only if c1 = c2 = c3 = 0, then linear independence will follow. Evaluating the linear combination in (6.1) at x = 0, x = ln 2, and x =  ln 2, we find that constants c1 , c2 , and c3 satisfy c1 + c2 + c3 = 0, 1 8c1 + 32c2 + c3 = 0, 2 1 1 c1 + c2 + 2c3 = 0. 8 32 This system is a homogeneous system of linear equations whose determinant 1 8 1 1 = 2 32 1/2 32 1/2 1/32 2  8 1/2 1/8 2 + 8 32 1/8 1/32 1/8 1/32 = 2827 = 0. 64 Hence it has the unique trivial solution, that is, c1 = c2 = c3 = 0. 9. Let y1 = sin2 x, y2 = cos2 x, and y3 = 1. We want to find c1 , c2 , and c3 , not all zero, such that c1 y1 + c2 y2 + c3 y3 = c1 sin2 x + c2 cos2 x + c3 1 = 0, for all x in the interval (, ). Since sin2 x + cos2 x = 1 for all real numbers x, we can choose c1 = 1, c2 = 1, and c3 = 1. Thus, these functions are linearly dependent. 342 Exercises 6.1
11. Let y1 = x1 , y2 = x1/2 , and y3 = x. We want to find constants c1 , c2 , and c3 such that c1 y1 + c2 y2 + c3 y3 = c1 x1 + c2 x1/2 + c3 x = 0, for all x on the interval (0, ). This equation must hold if x = 1, 4, or 9 (or any other values for x in the interval (0, )). By plugging these values for x into the equation above, we see that c1 , c2 , and c3 must satisfy the three equations c1 + c2 + c3 = 0, c1 + 2c2 + 4c3 = 0, 4 c1 + 3c2 + 9c3 = 0. 9 Solving these three equations simultaneously yields c1 = c2 = c3 = 0. Thus, the only way for c1 x1 + c2 x1/2 + c3 x = 0 for all x on the interval (0, ), is for c1 = c2 = c3 = 0. Therefore, these three functions are linearly independent on (0, ). 13. A linear combination, c1 x + c2 x2 + c3 x3 + c4 x4 , is a polynomial of degree at most four, and so, by the fundamental theorem of algebra, it cannot have more than four zeros unless it is the zero polynomial (that is, it has all zero coefficients). Thus, if this linear combination vanishes on an interval, then c1 = c2 = c3 = c4 = 0. Therefore, the functions x, x2 , x3 , and x4 are linearly independent on any interval, in particular, on (, ). 15. Since, by inspection, r = 3, r = 1, and r = 4 are the roots of the characteristic equation, r 3 + 2r 2  11r  12 = 0, the functions e3x , ex , and e4x form a solution set. Next, we check that these functions are linearly independent by showing that their Wronskian is never zero. e3x W e ,e ,e
3x x 4x ex e
x e4x 4e
4x 1 =e e e
3x x 4x 1 1 1 = 84e2x , 16 (x) = 3e 3x 3 1 4 9 9e3x ex 16e4x which does not vanish. Therefore, {e3x , ex , e4x } is a fundamental solution set and, by Theorem 4, a general solution to the given differential equation is y = C1 e3x + C2 ex + C3 e4x . 343 Chapter 6
17. Writing the given differential equation, x3 y  3x2 y + 6xy  6y = 0, in standard form (17), we see that its coefficients, 3/x, 6/x2 , and 6/x3 are continuous on the specified interval, which is x > 0. Next, substituting x, x2 , and x3 into the differential equation, we verify that these functions are indeed solutions. x3 (x)  3x2 (x) + 6x(x)  6(x) = 0  0 + 6x  6x = 0, x3 (x2 )  3x2 (x2 ) + 6x(x2 )  6(x2 ) = 0  6x2 + 12x2  6x2 = 0, x3 (x3 )  3x2 (x3 ) + 6x(x3 )  6(x3 ) = 6x3  18x3 + 18x3  6x3 = 0. Evaluating the Wronskian yields x x2 W x, x2 , x3 (x) = 0 = x 2 x3 6x  x2 x3 2 6x = x 6x2  4x3 = 2x3 . 1 2x 3x2 2x 3x2 2 6x Thus W [x, x2 , x3 ] (x) = 0 on (0, ) and so {x, x2 , x3 } is a fundamental solution set for the given differential equation. We involve Theorem 2 to conclude that y = C1 x + C2 x2 + C3 x3 is a general solution. 19. (a) Since {ex , ex cos 2x, ex sin 2x, } is a fundamental solution set for the associated homogeneous differential equation and since yp = x2 is a solution to the nonhomogeneous equation, by the superposition principle, we have a general solution given by y(x) = C1 ex + C2 ex cos 2x + C3 ex sin 2x + x2 . 344 Exercises 6.1
(b) To find the solution that satisfies the initial conditions, we must differentiate the general solution y(x) twice with respect to x. Thus, we have y (x) = C1 ex  C2 ex cos 2x  2C2 ex sin 2x  C3 ex sin 2x + 2C3 ex cos 2x + 2x = C1 ex + (C2 + 2C3 ) ex cos 2x + (2C2  C3 ) ex sin 2x + 2x , y (x) = C1 ex + (C2  2C3 ) ex cos 2x  2 (C2 + 2C3 ) ex sin 2x  (2C2  C3 ) ex sin 2x + 2 (2C2  C3 ) ex cos 2x + 2 = C1 ex + (3C2  4C3 ) ex cos 2x + (4C2  3C3 ) ex sin 2x + 2 . Plugging the initial conditions into these formulas, yields the equations y(0) = C1 + C2 = 1, y (0) = C1  C2 + 2C3 = 1, y (0) = C1  3C2  4C3 + 2 = 3. By solving these equations simultaneously, we obtain C1 = 1, C2 = 0, and C3 = 1. Therefore, the solution to the initial value problem is given by y(x) = ex + ex sin 2x + x2 . 21. In the standard form, given equation becomes y + 1 1 3  ln x y  3y= . 2 x x x3 Since its coefficients are continuous on (0, ), we can apply Theorems 2 and 4 to conclude that a general solution to the corresponding homogeneous equation is yh (x) = C1 x + C2 x ln x + C3 x(ln x)2 and a general solution to the given nonhomogeneous equation is y(x) = yp (x) + yh (x) = ln x + C1 x + C2 x ln x + C3 x(ln x)2 . 345 Chapter 6
To satisfy the initial conditions, first we find 1 + C1 + C2 (ln x + 1) + C3 (ln x)2 + 2 ln x , x 2 ln x 2 1 C2 y (x) =  2 + + C3 + . x x x x y (x) = Substituting the initial conditions, y(1) = 3, y (1) = 3, and y (1) = 0, we get the system 3 = y(1) = C1 , 3 = y (1) = 1 + C1 + C2 , 0 = y (1) = 1 + C2 + 2C3 Thus, y(x) = ln x + 3x  x ln x + x(ln x)2 is the desired solution. 23. Substituting y1 (x) = sin x and y2 (x) = x into the given differential operator yields L[sin x] = (sin x) + (sinx) + x(sin x) =  cos x + cos x + x sin x = x sin x, L[x] = (x) + (x) + x(x) = 0 + 1 + x2 = x2 + 1. Note that L[y] is a linear operator of the form (7). So, we can use the superposition principle. (a) Since 2x sin x  x2  1 = 2(x sin x)  (x2 + 1), by the superposition principle, y(x) = 2y1(x)  y2 (x) = 2 sin x  x is a solution to L[y] = 2x sin x  x2  1. (b) We can express 4x2 + 4  6x sin x = 4(x2 + 1)  6(x sin x). Hence, y(x) = 4y2 (x)  6y1 (x) = 4x  6 sin x is a solution to L[y] = 4x2 + 4  6x sin x. 346 C1 = 3, C1 + C2 = 2, C2 + 2C3 = 1 C1 = 3, C2 = 1, C3 = 1. Exercises 6.1
25. Clearly, it is sufficient to prove (9) just for two functions, y1 and y2 . Using the linear property of differentiation, we have L [y1 + y2 ] = [y1 + y2 ](n) + p1 [y1 + y2 ](n1) + + pn [y1 + y2 ] = y1 + y2
(n) (n) (n) + p1 y 1 (n1) + y2 (n1) + + pn [y1 + y2 ]
(n) (n1) = y 1 + p1 y 1 Next, we verify (10). (n1) + + pn y 1 + y 2 + p1 y 2 + + pn y2 = L [y1 ] + L [y1 ] . L [cy] = [cy](n) + p1 [cy](n1) + + pn [cy] = cy (n) + p1 cy (n1) + + pn cy = c y (n) + p1 y (n1) + + pn y = cL [y] . 27. A linear combination c0 + c1 x + c2 x2 + + cn xn of the functions from the given set is a polynomial of degree at most n and so, by the fundamental theorem of algebra, it cannot have more than n zeros unless it is the zero polynomial, i.e., it has all zero coefficients. Thus, if this linear combination vanishes on a whole interval (a, b), then it follows that c0 = c1 = c2 = . . . = cn = 0. Therefore, the set of functions {1, x, x2 , . . . , xn } is linearly independent on any interval (a, b). 29. (a) Assuming that functions f1 , f2 , . . . , fm are linearly dependent on (, ), we can find their nontrivial linear combination vanishing identically on (, ), i.e., c1 f1 + c2 f2 + + cm fm 0 on (, ), where not all cj 's are zeros. In particular, this linear combination vanishes on (1, 1), which contradicts the assumption that f1 , f2 , . . . , fm are linearly independent on (1, 1). (b) Let f1 (x) := x  1 , f2 (x) := x  1 . 347 Chapter 6
On (1, 1) (even on (, 1) ) we have f1 (x) f2 (x) or, equivalently, f1 (x) + f2 (x) 0 and so these functions are linearly dependent on (1, 1). However, their linear combination c1 f1 (x) + c2 f2 (x) = (c2  c1 ) (x  1), x 1; (c1 + c2 ) (x  1), x > 1 cannot vanish identically on (, ) unless c1  c2 = 0 and c1 + c2 = 0, which implies c1 = c2 = 0. 31. (a) Linearity of differentiation and the product rule yield y (x) = (v(x)ex ) = v (x)ex + v(x) (ex ) = [v (x) + v(x)] ex , y (x) = [v (x) + v(x)] ex + [v (x) + v(x)] (ex ) = [v (x) + 2v (x) + v(x)] ex , y (x) = [v (x) + 2v (x) + v(x)] ex + [v (x) + 2v (x) + v(x)] (ex ) = [v (x) + 3v (x) + 3v (x) + v(x)] ex . (b) Substituting y, y , y , and y into the differential equation (32), we obtain [v + 3v + 3v + v] ex  2 [v + 2v + v] ex  5 [v + v] ex + 6vex = 0 [(v + 3v + 3v + v)  2 (v + 2v + v)  5 (v + v) + 6v] ex = 0 v + v  6v = 0, where we have used the fact that the function ex is never zero. Let v =: w. Then v = w , v = w , and so the above equation becomes w + w  6w = 0. (6.2) (c) The auxiliary equation for (6.2), r 2 + r  6 = 0, has the roots r = 3 and r = 2. Therefore, a general solution to this differential equation is w(x) = C1 e3x + C2 e2x , where C1 and C2 are arbitrary constants. Choosing, say, C1 = 3, C2 = 0 and C1 = 0, C2 = 2, we find two linearly independent solutions, w1 (x) = 3e3x 348 and w2 (x) = 2e2x . Exercises 6.1
Integration yields v1 (x) = v2 (x) = w1 (x) dx = w2 (x) dx = 3e3x dx = e3x , 2e2x dx = e2x , where we have chosen zero integration constants. (d) With functions v1 (x) and v2 (x) obtained in (c), we have y1 (x) = v1 (x)ex = e3x ex = e2x , y2 (x) = v2 (x)ex = e2x ex = e3x . To show that the functions ex , e2x , and e3x are linearly independent on (, ), we can use the approach similar to that in Problem 7. Alternatively, since these functions are solutions to the differential equation (32), one can apply Theorem 3, as we did in Problem 15. To this end, ex W e ,e
x 2x e2x 2e
2x e3x 3e
3x 1 =e e
x 2x 3x 1 1 = 30e2x = 0 4 9 ,e 3x (x) = e x e 1 2 3 1 ex 4e2x 9e3x on (, ) and so the functions ex , e2x , and e3x are linearly independent on (, ). 33. Let y(x) = v(x)e2x . Differentiating y(x), we obtain y (x) = [v (x) + 2v(x)] e2x , y (x) = [v (x) + 4v (x) + 4v(x)] e2x , y (x) = [v (x) + 6v (x) + 12v (x) + 8v(x)] e2x . Substituting these expressions into the given differential equation yields [(v + 6v + 12v + 8v)  2 (v + 4v + 4v) + (v + 2v)  (2v)] e2x = 0 [v + 4v + 5v ] e2x = 0 v + 4v + 5v = 0. With w(x) := v (x), the above equation becomes w (x) + 4w (x) + 5w(x) = 0. 349 Chapter 6
The roots of the auxiliary equation, r 2 + 4r + 5 = 0, for this second order equation are r = 2 i. Therefore, {w1 (x), w2 (x)} = e2x cos x, e2x sin x form a fundamental solution set. Integrating, we get v1 (x) = v2 (x) = w1 (x) = w2 (x) = e2x (sin x  2 cos x) , 5 e2x (2 sin x + cos x) e2x sin x dx =  , 5 e2x cos x dx = where we have chosen integration constants to be zero. Thus, functions f (x) = e2x , e2x (sin x  2 cos x) 2x sin x  2 cos x e = , 5 5 e2x (2 sin x + cos x) 2x 2 sin x + cos x e = y2 (x) = v2 (x)f (x) = 5 5 y1 (x) = v1 (x)f (x) = are three linearly independent solutions to the given differential equation. 35. First, let us evaluate the Wronskian of the system {x, sin x, cos x} to make sure that the result of Problem 34 can be applied. x W [x, sin x, cos x] = 1 sin x cos x cos x  sin x cos x  sin x  sin x  cos x
2 2 0  sin x  cos x = x  sin x cos x  sin x  cos x = x  cos x  sin x  ( sin x cos x + sin x cos x) = x. Thus, W [x, sin x, cos x] = 0 on (, 0) and (0, ). Therefore, on either of these two intervals, {x, sin x, cos x} is a fundamental solution set for the third order linear differential equation 350 Exercises 6.2
given in Problem 34. Expanding the determinant over its last column yields x 1 sin x cos x y y y = y W [x, sin x, cos x]  y cos x  sin x x 1 sin x cos x sin x 1 cos x  sin x sin x 0  sin x  cos x 0  cos x cos x  sin x sin x y x +y sin x 0  cos x cos x sin x  0  sin x  cos x  y 0  sin x  cos x 0  cos x 0  cos x sin x cos x  cos x sin x  sin x  cos x  cos x sin x cos x  sin x  cos x  cos x sin x  sin x  cos x sin x = xy  y x +y x y = xy + y  xy + y = 0. EXERCISES 6.2: Homogeneous Linear Equations with Constant Coefficients, page 331 1. The auxiliary equation r 3 + 2r 2  8r = 0 r r 2 + 2r  8 = r(r  2)(r + 4) = 0 has the roots r = 0, 2, and 4. Thus a general solutions to the differential equation has the form y = c1 + c2 e2x + c3 e4x . 3. The auxiliary equation for this problem is 6r 3 + 7r 2  r  2 = 0. By inspection we see that r = 1 is a root to this equation and so we can factor it as follows 6r 3 + 7r 2  r  2 = (r + 1)(6r 2 + r  2) = (r + 1)(3r + 2)(2r  1) = 0. Thus, we see that the roots to the auxiliary equation are r = 1, 2/3, and 1/2. These roots are real and nonrepeating. Therefore, a general solution to this problem is given by z(x) = c1 ex + c2 e2x/3 + c3 ex/2 . 351 Chapter 6
5. We can factor the auxiliary equation, r 3 + 3r 2 + 28r + 26 = 0, as follows: r 3 + 3r 2 + 28r + 26 = (r 3 + r 2 ) + (2r 2 + 2r) + (26r + 26) = r 2 (r + 1) + 2r(r + 1) + 26(r + 1) = (r + 1)(r 2 + 2r + 26) = 0. Thus either r + 1 = 0 r = 1 or r 2 + 2r + 26 = 0 r = 1 5i. Therefore, a general solution is given by y(x) = c1 ex + c2 ex cos 5x + c3 ex sin 5x . 7. Factoring the characteristic polynomial yields 2r 3  r 2  10r  7 = (2r 3 + 2r 2 ) + (3r 2  3r) + (7r  7) = 2r 2 (r + 1)  3r(r + 1)  7(r + 1) = (r + 1)(2r 2  3r  7). Thus the roots of the characteristic equation, 2r 3  r 2  10r  7 = 0, are r+1=0 r = 1 , r= 3 32  4(2)(7) 3 65 = , 4 4 2r 2  3r  7 = 0 and a general solution is y(x) = c1 ex + c2 e(3+ 65)x/4 + c3 e(3 65)x/4 . 9. In the characteristic equation, r 3  9r 2 + 27r  27 = 0, we recognize a complete cube, namely, (r  3)3 = 0. Thus, it has just one root, r = 3, of multiplicity three. Therefore, a general solution to the given differential equation is given by u(x) = c1 e3x + c2 xe3x + c3 x2 e3x . 11. Since r 4 + 4r 3 + 6r 2 + 4r + 1 = (r + 1)4 , the characteristic equation becomes (r + 1)4 = 0, and it has the root r = 1 of multiplicity four. Therefore, the functions ex , xex , x2 ex , and x3 ex form a fundamental solution set and a general solution to the given differential equation is y(x) = c1 ex + c2 xex + c3 x2 ex + c4 x3 ex = c1 + c2 x + c3 x2 + c4 x3 ex . 352 Exercises 6.2
13. The auxiliary equation in this problem is r 4 +4r 2 +4 = 0. This can be factored as (r 2 + 2) = 0. Therefore, this equation has roots r = 2i,  2i, 2i,  2i, which we see are repeated and complex. Therefore, a general solution to this problem is given by y(x) = c1 cos 2x + c2 x cos 2x + c3 sin 2x + c4 x sin 2x .
2 15. The roots to this auxiliary equation, (r  1)2 (r + 3)(r 2 + 2r + 5)2 = 0, are r = 1, 1, 3, 1 2i, 1 2i , where we note that 1 and 1 2i are repeated roots. Therefore, a general solution to the differential equation with the given auxiliary equation is y(x) = c1 ex + c2 xex + c3 e3x + (c4 + c5 x)ex cos 2x + (c6 + c7 x)ex sin 2x . 17. From the differential operator, replacing D by r, we obtain the characteristic equation (r + 4)(r  3)(r + 2)3 (r 2 + 4r + 5)2 r 5 = 0 , whose roots r+4=0 r3=0 (r + 2) = 0 r5 = 0
3 r = 4, r = 3, r = 2 of multiplicity 3, r = 2 i of multiplicity 2, r = 0 of multiplicity 5. (r 2 + 4r + 5)2 = 0 Therefore, a general solution is given by y(x) = c1 e4x + c2 e3x + c3 + c4 x + c5 x2 e2x + (c6 + c7 x) e2x cos x + (c8 + c9 x) e2x sin x +c10 + c11 x + c12 x2 + c13 x3 + c14 x4 . 19. First, we find a general solution to the given equation. Solving the auxiliary equation, r 3  r 2  4r + 4 = (r 3  r 2 )  (4r  4) = (r  1)(r 2  4) = (r  1)(r + 2)(r  2) = 0, 353 Chapter 6
yields the roots r = 1, 2, and 2. Thus a general solution has the form y(x) = c1 ex + c2 e2x + c3 e2x . Next, we find constants c1 , c2 , and c3 such that the solution satisfies the initial conditions. Differentiating y(x) and substituting the initial conditions, we obtain the system y(0) = c1 ex + c2 e2x + c3 e2x
x=0 = c1 + c2 + c3 = 4, = c1  2c2 + 2c3 = 1, = c1 + 4c2 + 4c3 = 19. y (0) = c1 ex  2c2 e2x + 2c3 e2x y (0) = c1 ex + 4c2 e2x + 4c3 e2x Solving yields c1 = 1, c2 = 2, x=0 x=0 c3 = 3. With these coefficients, the solution to the given initial problem is y(x) = ex  2e2x  3e2x . 21. By inspection, r = 2 is a root of the characteristic equation, r 3  4r 2 + 7r  6 = 0. Factoring yields r 3  4r 2 + 7r  6 = (r  2)(r 2  2r + 3) = 0. Therefore, the other two roots are the roots of r 2  2r + 3 = 0, which are r = 1 so a general solution to the given differential equation is given by y(x) = c1 e2x + c2 cos 2x + c3 sin 2x ex . Differentiating, we obtain y = 2c1 e2x + y = 4c1 e2x + c2 + c3 2 cos 2x + c3  c2 2 sin 2x ex , 2c3 2  c2 cos 2x  2c2 2 + c3 sin 2x ex . 2i, and Hence, the initial conditions yield y(0) = c1 + c2 = 1, c1 = 1, c2 = 0, c3 =  2 . y (0) = 2c1 + c2 + c3 2 = 0, y (0) = 4c1  c2 + 2c3 2 = 0 354 Exercises 6.2
Substituting these constants into the general solution, we get the answer y(x) = e2x  23. Rewriting the system in operator form yields (D 3  1) [x] + (D + 1)[y] = 0, (D  1)[x] + y = 0. Multiplying the second equation in this sytem by (D + 1) and subtracting the result from the first equation, we get D 3  1  (D + 1)(D  1) [x] = D 2 (D  1)[x] = 0. Since the roots of the characteristic equation, r 2 (r  1) = 0 are r = 0 of multiplicity two and r = 1, a general solution x(t) is given by x(t) = c1 + c2 t + c3 et . From the second equation in the original system, we obtain y(t) = x(t)  x (t) = c1 + c2 t + c3 et  c1 + c2 t + c3 et 25. A linear combination of the given functions c0 erx + c1 xerx + c2 x2 erx + + cm1 xm1 erx = c0 + c1 x + c2 x2 + + cm xm erx (6.3) = (c1  c2 ) + c2 t. 2ex sin 2x . vanishes on an interval if and only if its polynomial factor, c0 + c1 x + c2 x2 + + cm1 xm1 , vanishes on this interval (the exponential factor, erx , is never zero). But, as we have proved in Problem 27, Section 6.1, the system of monomials {1, x, . . . , xn } is linearly independent on any interval. Thus, the linear combination (6.3) vanishes on an inteval if and only if it has all zero coefficients, i.e., c0 = c1 = . . . = cm1 = 0. Therefore, the system {erx , xerx , . . . , xm1 erx } is linearly independent on any interval, in particular, on (, ). 355 Chapter 6
27. Solving the auxiliary equation, r 4 + 2r 3  3r 2  r + (1/2) = 0, using computer software yields the roots r1 = 1.119967680, r2 = 0.2963247800, r3 = 0.5202201098, r4 = 2.896072350 . Thus, all the roots are real and distinct. A general solution to the given equation is, therefore, y(x) = c1 er1 x + c2 er2 x + c3 er3 x + c4 er4 x c1 e1.120x + c2 e0.296x + c3 e0.520x + c4 e2.896x . 29. The auxiliary equation in this problem is r 4 + 2r 3 + 4r 2 + 3r + 2 = 0. Let g(r) = r 4 + 2r 3 + 4r 2 + 3r + 2 g (r) = 4r 3 + 6r 2 + 8r + 3. Then the Newton's recursion formula (2) in Appendix A of the text becomes rn+1 = rn 
4 3 2 rn + 2rn + 4rn + 3rn + 2 . 3 2 4rn + 6rn + 8rn + 3 With initial guess r0 = 1 + i, this formula yields (1 + i)4 + 2(1 + i)3 + 4(1 + i)2 + 3(1 + i) + 2 0.481715 + 0.837327i , 4(1 + i)3 + 6(1 + i)2 + 8(1 + i) + 3 r 4 + 2r 3 + 4r 2 + 3r1 + 2 r2 = r1  1 3 1 2 1 0.052833 + 0.763496i , 4r1 + 6r1 + 8r1 + 3 r 4 + 2r 3 + 4r 2 + 3r2 + 2 r3 = r2  2 3 2 2 2 0.284333 + 0.789859i , 4r2 + 6r2 + 8r2 + 3 . . . r 4 + 2r 3 + 4r 2 + 3r6 + 2 0.500000 + 0.866025i , r7 = r6  6 3 6 2 6 4r6 + 6r6 + 8r6 + 3 r 4 + 2r 3 + 4r 2 + 3r7 + 2 r8 = r7  7 3 7 2 7 0.500000 + 0.866025i . 4r7 + 6r7 + 8r7 + 3 r1 = (1 + i)  Therefore, first two roots of the auxiliary equation are r 0.5 + 0.866i 356 and r = 0.5 + 0.866i = 0.5  0.866i . Exercises 6.2
Similarly, we find other two roots. With the initial guess r0 = 1  2i, we find that r1 = (1  2i)  . . . r6 0.499994  1.322875i , r7 0.500000  1.322876i , r8 0.500000  1.322876i . Therefore, the other two roots are r 0.5  1.323i and r = 0.5  1.323i = 0.5 + 1.323i . (1  2i)4 + 2(1  2i)3 + 4(1  2i)2 + 3(1  2i) + 2 4(1  2i)3 + 6(1  2i)2 + 8(1  2i) + 3 0.830703  1.652798i , Thus, the auxiliary equation has four complex roots, and a general solution to the given differential equation is given by y(x) c1 e0.5x cos(0.866x) + c2 e0.5x sin(0.866x) + c3 e0.5x cos(1.323x) + c4 e0.5x sin(1.323x) . 31. (a) If we let y(x) = xr , then we see that y = rxr1 , y = r(r  1)xr2 = (r 2  r)xr2 , y = r(r  1)(r  2)xr3 = (r 3  3r 2 + 2r)xr3 . Thus, if y = xr is a solution to this third order CauchyEuler equation, then we must have x3 (r 3  3r 2 + 2r)xr3 + x2 (r 2  r)xr2  2xrxr1 + 2xr = 0 (r 3  3r 2 + 2r)xr + (r 2  r)xr  2rxr + 2xr = 0 (r 3  2r 2  r + 2)xr = 0. (6.5) (6.4) Therefore, in order for y = xr to be a solution to the equation with x > 0, we must have r 3  2r 2  r + 2 = 0. Factoring this equation yields r 3  2r 2  r + 2 = (r 3  2r 2)  (r  2) = (r  2)(r 2  1) = (r  2)(r + 1)(r  1) = 0. 357 Chapter 6
Equation (6.5) will equal zero and, therefore, the differential equation will be satisfied for r = 1 and r = 2. Thus, three solutions to the differential equation are y = x, y = x1 , and y = x2 . Since these functions are linearly independent, they form a fundamental solution set. (b) Let y(x) = xr . In addition to (6.4), we need the fourth derivative of y(x). y (4) = (y ) = r(r  1)(r  2)(r  3)xr4 = (r 4  6r 3 + 11r 2  6r)xr4 . Thus, if y = xr is a solution to this fourth order CauchyEuler equation, then we must have x4 (r 4  6r 3 + 11r 2  6r)xr4 + 6x3 (r 3  3r 2 + 2r)xr3 +2x2 (r 2  r)xr2  4xrxr1 + 4xr = 0 (r 4  6r 3 + 11r 2  6r)xr + 6(r 3  3r 2 + 2r)xr + 2(r 2  r)xr  4rxr + 4xr = 0 (r 4  5r 2 + 4)xr = 0. (6.6) Therefore, in order for y = xr to be a solution to the equation with x > 0, we must have r 4  5r 2 + 4 = 0. Factoring this equation yields r 4  5r 2 + 4 = (r 2  4)(r 2  1) = (r  2)(r + 2)(r  1)(r + 1) = 0. Equation (6.6) will be satisfied if r = 1, 2. Thus, four solutions to the differential equation are y = x, y = x1 , y = x2 , and y = x2 . These functions are linearly independent, and so form a fundamental solution set. (c) Substituting y = xr into this differential equation yields (r 3  3r 2 + 2r)xr  2(r 2  r)xr + 13rxr  13xr = 0 (r 3  5r 2 + 17r  13)xr = 0. Thus, in order for y = xr to be a solution to this differential equation with x > 0, we must have r 3  5r 2 + 17r  13 = 0. By inspection we find that r = 1 is a root to this equation. Therefore, we can factor this equation as follows (r  1)(r 2  4r + 13) = 0. 358 Exercises 6.2
We find the remaining roots by using the quadratic formula. Thus, we obtain the roots r = 1, 2 3i. From the root r = 1, we obtain the solution y = x. From the roots r = 2 3i, by applying the hint given in the problem, we see that a solution is given by y(x) = x2+3i = x2 {cos(3 ln x) + i sin(3 ln x)} . Therefore, by Lemma 2 on page 172 of the text, we find that two realvalued solutions to this differential equation are y(x) = x2 cos(3 ln x) and y(x) = x2 sin(3 ln x). Since these functions and the function y(x) = x are linearly independent, we obtain the fundamental solution set x, x2 cos(3 ln x), x2 sin(3 ln x) . 33. With suggested values of parameters m1 = m2 = 1, k1 = 3, and k2 = 2, the system (34)(35) becomes x + 5x  2y = 0, y  2x + 2y = 0. (6.7) (a) Expressing y = (x + 5x) /2 from the first equation and substituting this expression into the second equation, we obtain 1 (x + 5x)  2x + (x + 5x) = 0 2 x(4) + 5x  4x + 2 (x + 5x) = 0 as it is stated in (36). (b) The characteristic equation corresponding to (6.8) is r 4 + 7r 2 + 6 = 0. This equation is of quadratic type. Substitution s = r 2 yields s2 + 7s + 6 = 0 Thus r = 1 = i s = 1, 6. r = 6 = i 6 , x(4) + 7x + 6x = 0, (6.8) and and a general solution to (6.8) is given by x(t) = c1 cos t + c2 sin t + c3 cos 6t + c4 sin 6t . 359 Chapter 6
(c) As we have mentioned in (a), the first equation in (6.7) implies that y = (x + 5x) /2. Substituting the solution x(t) yields y(t) = 1 2 c1 cos t + c2 sin t + c3 cos 6t + c4 sin 6t +5 c1 cos t + c2 sin t + c3 cos 6t + c4 sin 6t 1 c1 cos t  c2 sin t  6c3 cos 6t  6c4 sin 6t = 2 +5 c1 cos t + c2 sin t + c3 cos 6t + c4 sin 6t c3 c4 cos 6t  sin 6t . = 2c1 cos t + 2c2 sin t  2 2 (d) Initial conditions x(0) = y(0) = 1 and x (0) = y (0) = 0 imply the system of linear equations for c1 , c2 , c3 , and c4 . Namely, x(0) = c1 + c3 = 1, y(0) = 2c1  (c3 /2) = 1, x (0) = c2 + c4 6 = 0, y (0) = 2c2  (c4 6/2) = 0 Thus, the solution to this initial value problem is x(t) = 2 3 cos t + cos 6t , 5 5 y(t) = 1 6 cos t  cos 6t . 5 5 c1 = 3/5, c3 = 2/5, c2 = 0, c4 = 0. 35. Solving the characteristic equation yields EIr 4  k = 0 r2 = or or r4 = k EI k EI k k 4 = i .  EI EI k EI k 4 r= EI r2 =  r=
4 The first two roots are real numbers, the other two are pure imaginary numbers. Therefore, a general solution to the vibrating beam equation is y(x) = C1 e k/(EI)x + C2 e k/(EI)x + C3 sin 360
4 k x EI + C4 cos 4 k x . EI Exercises 6.3
Using the identities eax = cosh ax + sinh ax, eax = cosh ax  sinh ax, we can express the solution in terms of hyperbolic and trigonometric functions as follows. y(x) = C1 e k/(EI)x + C2 e k/(EI)x + C3 sin = C1 cosh
4 4 k x EI + C4 cos
4 4 k x EI  sinh + C4 cos
4 4 k x EI + sinh 4 k x EI + C2 cosh +C3 sin k x EI k x EI k x EI
4 4 k x EI = c1 cosh 4 k x EI + c2 sinh 4 k x EI + c3 sin 4 k x EI + c4 cos k x , EI where c1 := C1 + C2 , c2 := C1  C2 , c3 := C3 , and c4 := C4 are arbitrary constants. EXERCISES 6.3: Undetermined Coefficients and the Annihilator Method, page 337 1. The corresponding homogeneous equation for this problem is y  2y  5y + 6y = 0 which has the associated auxiliary equation given by r 3  2r 2  5r + 6 = 0. By inspection we see that r = 1 is a root to this equation. Therefore, this equation can be factored as follows r 3  2r 2  5r + 6 = (r  1)(r 2  r  6) = (r  1)(r  3)(r + 2) = 0. Thus, the roots to the auxiliary equation are given by r = 1, 3, and 2, and a general solution to the homogeneous equation is yh (x) = c1 ex + c2 e3x + c3 e2x . The nonhomogeneous term, g(x) = ex +x2 , is the sum of an exponential term and a polynomial term. Therefore, according to Section 4.5, this equation has a particular solution of the form yp (x) = xs1 C1 ex + xs2 C2 + C3 x + C4 x2 . 361 Chapter 6
Since ex is a solution to the associated homogeneous equation and xex is not, we set s1 = 1. Since none of the terms x2 , x, or 1 is a solution to the associated homogeneous equation, we set s2 = 0. Thus, the form of a particular solution is yp (x) = C1 xex + C2 + C3 x + C4 x2 . 3. The associated homogeneous equation for this equation is y + 3y  4y = 0. This equation has the corresponding auxiliary equation y 3 + 3r 2  4 = 0, which, by inspection, has r = 1 as one of its roots. Thus, the auxiliary equation can be factored as follows (r  1)(r 2 + 4r + 4) = (r  1)(r + 2)2 = 0. From this we see that the roots to the auxiliary equation are r = 1, 2, 2. Therefore, a general solution to the homogeneous equation is yh (x) = c1 ex + c2 e2x + c3 xe2x . The nonhomogeneous term is g(x) = e2x . Therefore, a particular solution to the original differential equation has the form yp (x) = xs c1 e2x . Since both e2x and xe2x are solutions to the associated homogeneous equation, we set s = 2. (Note that this means that r = 2 will be a root of multiplicity three of the auxiliary equation associated with the operator equation A[L[y]](x) = 0, where A is an annihilator of the nonhomogeneous term g(x) = e2x and L is the linear operator L := D 3 + 3D 2  4.) Thus, the form of a particular solution to this equation is yp (x) = C1 x2 e2x . 5. In the solution to Problem 1, we determined that a general solution to the homogeneous differential equation associated with this problem is yh (x) = c1 ex + c2 e3x + c3 e2x , and that a particular solution has the form yp (x) = C1 xex + C2 + C3 x + C4 x2 . 362 Exercises 6.3
By differentiating yp (x), we find yp (x) = C1 xex + C1 ex + C3 + 2C4 x yp (x) = C1 xex + 2C1 ex + 2C4 yp (x) = C1 xex + 3C1 ex . Substituting these expressions into the original differential equation, we obtain yp (x)  2yp (x)  5yp (x) + 6yp (x) = C1 xex + 3C1 ex  2C1 xex  4C1 ex  4C4 5C1 xex  5C1 ex  5C3  10C4 x + 6C1 xex + 6C2 + 6C3 x + 6C4 x2 = ex + x2 6C1 ex + (4C4  5C3 + 6C2 ) + (10C4 + 6C3 )x + 6C4 x2 = ex + x2 . Equating coefficients yields 6C1 = 1 C1 = C4 = C3 = C2 = 1 , 6 1 , 6 10 5 10C4 = = , 6 36 18 4(1/6) + 5(5/18) 37 4C4 + 5C3 = = . 6 6 108 6C4 = 1 10C4 + 6C3 = 0 4C4  5C3 + 6C2 = 0 Thus, a general solution to the nonhomogeneous equation is given by y(x) = yh (x) + yp (x) = c1 ex + c2 e3x + c3 e2x  37 1 x 1 2 5 xe + x + x+ . 6 6 18 108 7. In Problem 3, a general solution to the associated homogeneous equation was found to be yh (x) = c1 ex + c2 e2x + c3 xe2x , and the form of a particular solution to the nonhomogeneous equation was yp (x) = C1 x2 e2x . 363 Chapter 6
Differentiating yp (x) yields yp (x) = 2C1 xe2x  2C1 x2 e2x = 2C1 (x  x2 )e2x yp (x) = 4C1 (x  x2 )e2x + 2C1 (1  2x)e2x = 2C1 (2x2  4x + 1)e2x yp (x) = 4C1 (2x2  4x + 1)e2x + 2C1 (4x  4)e2x = 4C1 (2x2 + 6x  3)e2x . By substituting these expressions into the nonhomogeneous equation, we obtain yp (x) + 3yp (x)  4yp (x) = 4C1 (2x2 + 6x  3)e2x +6C1 (2x2  4x + 1)e2x  4C1 x2 e2x = e2x 6C1 e2x = e2x . By equating coefficients, we see that C1 = 1/6. Thus, a general solution to the nonhomogeneous differential equation is given by y(x) = yh (x) + yp (x) = c1 ex + c2 e2x + c3 xe2x  1 2 2x . xe 6 9. Solving the auxiliary equation, r 3  3r 2 + 3r  1 = (r  1)3 = 0, we find that r = 1 is its root of multiplicity three. Therefore, a general solution to the associated homogeneous equation is given by yh (x) = c1 ex + c2 xex + c3 x2 ex . The nonhomogeneous term, ex , suggests a particular solution of the form yp (x) = Axs ex , where we have to choose s = 3 since the root r = 1 of the auxiliary equation is of multiplicity three. Thus yp (x) = Ax3 ex . Differentiating yp (x) yields yp (x) = A x3 + 3x2 ex , yp (x) = A x3 + 6x2 + 6x ex , yp (x) = A x3 + 9x2 + 18x + 6 ex . 364 Exercises 6.3
By substituting these expressions into the original equation, we obtain yp  3yp + 3yp  y = ex A x3 + 9x2 + 18x + 6 ex  3 A x3 + 6x2 + 6x ex +3 A x3 + 3x2 ex  Ax3 ex = ex 6Aex = ex A= 1 , 6 1 3 x xe . 6 and so yp (x) = x3 ex /6. A general solution to the given equation then has the form y(x) = yh (x) + yp (x) = c1 ex + c2 xex + c3 x2 ex + 11. The operator D 5 , that is, the fifth derivative operator, annihilates any polynomial of degree at most four. In particular, D 5 annihilates the polynomial x4  x2 + 11. 13. According to (i) on page 334 of the text, the operator [D  (7)] = (D + 7) annihilates the exponential function e7x . 15. The operator (D2) annihilates the function f1 (x) := e2x and the operator (D1) annihilates the function f2 (x) := ex . Thus, the composition of these operators, namely, (D  2)(D  1), annihilates both of these functions and so, by linearity, it annihilates their algebraic sum. 17. This function has the same form as the functions given in (iv) on page 334 of the text. Here we see that = 1, = 2, and m  1 = 2. Thus, the operator (D  {1})2 + 22 annihilates this function. 19. Given function as a sum of two functions. The first term, xe2x , is of the type (ii) on the page 334 of the text with m = 2 and r = 2; so [D  (2)]2 = (D + 2)2 annihilates this function. The second term, xe5x sin 3x, is annihilated by (D  (5))2 + 32
2 2 2 3 = (D + 1)2 + 4 3 = (D + 5)2 + 9 according to (iv). Therefore, the composition [(D + 2)2 (D + 5)2 + 9] annihilates the function xe2x + xe5x sin 3x. 365 Chapter 6
21. In operator form, the given equation can be written as D 2  5D + 6 [u] = cos 2x + 1. The function g(x) = cos 2x + 1 is a sum of two functions: cos 2x is of the type (iii) on page 334 of the text with = 2, and so it is innihilated by (D 2 + 4); 1, as a constant, is annihilated by D. Therefore, the operator D(D 2 + 4) innihilates the righthand side, g(x). Applying this operator to both sides of the differential equation given in this problem yields D D2 + 4 D 2  5D + 6 [u] = D D 2 + 4 [cos 2x + 1] = 0 D D 2 + 4 (D  3)(D  2)[u] = 0. This last equation has the associated auxiliary equation r (r 2 + 4) (r  3)(r  2) = 0, which has roots r = 2, 3, 0, 2i. Thus, a general solution to the differential equation associated with this auxiliary equation is u(x) = c1 e2x + c2 e3x + c3 cos 2x + c4 sin 2x + c5 . The homogeneous equation, u  5u + 6u = 0, associated with the original problem, has as its corresponding auxiliary equation r 2  5r + 6 = (r  2)(r  3) = 0. Therefore, the solution to the homogeneous equation associated with the original problem is uh (x) = c1 e2x + c2 e3x . Since a general solution to this original problem is given by u(x) = uh (x) + up (x) = c1 e2x + c2 e3x + up (x) and since u(x) must be of the form u(x) = c1 e2x + c2 e3x + c3 cos 2x + c4 sin 2x + c5 , we see that up (x) = c3 cos 2x + c4 sin 2x + c5 . 23. The function g(x) = e3x  x2 is annihilated by the operator A := D 3 (D  3). Applying the operator A to both sides of the differential equation given in this problem yields A [y  5y + 6y] = A e3x  x2 = 0 366 Exercises 6.3 D 3 (D  3)(D 2  5D + 6)[y] = D 3 (D  3)2 (D  2)[y] = 0. This last equation has the associated auxiliary equation r 3 (r  3)2 (r  2) = 0, which has roots r = 0, 0, 0, 3, 3, 2. Thus, a general solution to the differential equation associated with this auxiliary equation is y(x) = c1 e2x + c2 e3x + c3 xe3x + c4 x2 + c5 x + c6 . The homogeneous equation, y  5y + 6y = 0, associated with the original problem, is the same as in Problem 21 (with u replaced by y). Therefore, the solution to the homogeneous equation associated with the original problem is yh (x) = c1 e2x +c2 e3x . Since a general solution to this original problem is given by y(x) = yh (x) + yp (x) = c1 e2x + c2 e3x + yp (x) and since y(x) must be of the form y(x) = c1 e2x + c2 e3x + c3 xe3x + c4 x2 + c5 x + c6 , we see that yp (x) = c3 xe3x + c4 x2 + c5 x + c6 . 25. First, we rewrite the equation in operator form, that is, D 2  6D + 9 [y] = sin 2x + x (D  3)2 [y] = sin 2x + x . In this problem, the righthand side is a sum of two functions. The first function, sin 2x, is annihilated by (D 2 + 4), and the operator D 2 annihilates the term x. Thus A := D 2 (D 2 + 4) annihilates the function sin 2x+x. Applying this operator to the original equation (in operator form) yields D 2 (D 2 + 4)(D  3)2 [y] = D 2 (D 2 + 4)[sin 2x + x] = 0. (6.9) 367 Chapter 6
This homogeneous equation has associated characteristic equation r 2 (r 2 + 4)(r  3)2 = 0 with roots 2i, and double roots r = 0 and r = 3. Therefore, a general solution to (6.9) is given by y(x) = c1 e3x + c2 xe3x + c3 + c4 x + c5 cos 2x + c6 sin 2x . (6.10) Since the homogeneous equation, (D  3)2 [y] = 0, which corresponds to the original equation, has a general solution yh (x) = c1 e3x + c2 xe3x , the "tail" in (6.10) gives the form of a particular solution to the given equation. 27. Since y + 2y + 2y = D 2 + 2D + 2 [y] = (D + 1)2 + 1 [y], the auxiliary equation in this problm is (r +1)2 +1 = 0, whose roots are r = 1i. Therefore, a general solution to the homogeneous equation, corresponding to the original equation, is yh (x) = (c1 cos x + c2 sin x) ex . Applying the operator D 3 {(D+1)2 +1} to the given equation, which annihilates its righthand side, yields D 3 (D + 1)2 + 1 (D + 1)2 + 1 [y] = D 3 (D + 1)2 + 1
2 ex cos x + x2 = 0 (6.11) D 3 (D + 1)2 + 1 [y] = 0. The corresponding auxiliary equation, r 3 [(r + 1)2 + 1]2 = 0 has a root r = 0 of multiplicity three and double roots r = 1 i. Therefore, a general solution to (6.11) is given by y(x) = (c1 cos x + c2 sin x) ex + (c3 cos x + c4 sin x) xex + c5 x2 + c6 x + c7 . Since y(x) = yh (x) + yp (x), we conclude that yp (x) = (c3 cos x + c4 sin x) xex + c5 x2 + c6 x + c7 . 368 Exercises 6.3
29. In operator form, the equation becomes D 3  2D 2 + D [z] = D(D  1)2 [z] = x  ex . (6.12) Solving the corresponding auxiliary equation, r(r  1)2 = 0, we find that r = 0, 1, and 1. Thus zh (x) = C1 + C2 ex + C3 xex is a general solution to the homogeneous equation associated with the original equation. To annihilate the righthand side in (6.12), we apply the operator D 2 (D  1) to this equation. Thus we obtain D 2 (D  1)D(D  1)2 [z] = D 2 (D  1) [x  ex ] D 3 (D  1)3 = 0. Solving the corresponding auxiliary equation, r 3 (r  1)3 = 0, we see that r = 0 and r = 1 are its roots of multiplicity three. Hence, a general solution is given by z(x) = c1 + c2 x + c3 x2 + c4 ex + c5 xex + c6 x2 ex . This general solution, when compared with zh (x), gives zp (x) = c2 x + c3 x2 + c6 x2 ex . 31. Writing this equation in operator form yields D 3 + 2D 2  9D  18 [y] = 18x2  18x + 22 . Since, D 3 + 2D 2  9D  18 = D 2 (D + 2)  9(D + 2) = (D + 2) D 2  9 = (D + 2)(D  3)(D + 3), (6.13) becomes (D + 2)(D  3)(D + 3)[y] = 18x2  18x + 22 . The auxiliary equation in this problem is (r + 2)(r  3)(r + 3) = 0 with roots r = 2, 3, and 3. Hence, a general solution to the corresponding homogeneous equation has the form yh (x) = c1 e2x + c2 e3x + c3 e3x . 369 (6.13) Chapter 6
Since the operator D 3 annihilates the nonhomogeneous term in the original equation and r = 0 is not a root of the auxiliary equation, we seek for a particular solution of the form yp (x) = C0 x2 + C1 x + C2 . Substituting yp into the given equation (for convenience, in operator form) yileds D 3 + 2D 2  9D  18 C0 x2 + C1 x + C2 = 18x2  18x + 22 0 + 2 (2C0 )  9 [2C0 x + C1 ]  18 C0 x2 + C1 x + C2 = 18x2  18x + 22 18C0 x2 + (18C1  18C0 )x + (18C2  9C1 + 4C0 ) = 18x2  18x + 22. Equating coefficients, we obtain the system 18C0 = 18, 18C1  18C0 = 18, 18C2  9C1 + 4C0 = 22 Thus, yp (x) = x2  1 and y(x) = yh (x) + yp (x) = c1 e2x + c2 e3x + c3 e3x + x2  1 is a general solution to the original nonhomogeneous equation. Next, we satisfy the initial conditions. Differentiation yields y (x) = 2c1 e2x + 3c2 e3x  3c3 e3x + 2x, y (x) = 4c1 e2x + 9c2 e3x + 9c3 e3x + 2. Therefore, 2 = y(0) = c1 + c2 + c3  1, 8 = y (0) = 2c1 + 3c2  3c3 , 12 = y (0) = 4c1 + 9c2 + 9c3 + 2 c1 + c2 + c3 = 1, 2c1 + 3c2  3c3 = 8, 4c1 + 9c2 + 9c3 = 14. C0 = 1, C1 = 0, C2 = 1. Solving this system, we find that c1 = 1, c2 = 2, and c3 = 0, and so y(x) = e2x  2e3x + x2  1 gives the solution to the given initial value problem. 370 Exercises 6.3
33. Let us write given equation in operator form. D 3  2D 2  3D + 10 [y] = (34x  16)e2x  10x2 + 6x + 34 . By inspection, r = 2 is a root of the characteristic equation, r 3  2r 2  3r + 10 = 0. Using, say, long division we find that r 3  2r 2  3r + 10 = (r + 2) r 2  4r + 5 = (r + 2) (r  2)2 + 1 and so the other two roots of the auxiliary equation are r = 2i. This gives a general solution to the corresponding homogeneous equation yh (x) = c1 e2x + (c2 cos x + c3 sin x) e2x . According to the nonhomogeneous term, we look for a particular solution to the original equation of the form yp (x) = x (C0 x + C1 ) e2x + C2 x2 + C3 x + C4 , where the factor x in the exponential term appears due to the fact that r = 2 is a root of the characteristic equation. Substituting yp (x) into the given equation and simplifying yield D 3  2D 2  3D + 10 [yp (x)] = (34x  16)e2x  10x2 + 6x + 34 (34C0 x + 17C1  16C0 ) e2x + 10C2 x2 + (10C3  6C2 )x +10C4  3C3  4C2 = (34x  16)e2x  10x2 + 6x + 34. Equating corresponding coefficients, we obtain the system 34C0 = 34, 17C1  16C0 = 16, 10C2 = 10, 10C3  6C2 = 6, 10C4  3C3  4C2 = 34 Thus, yp (x) = x2 e2x  x2 + 3 and y(x) = yh (x) + yp (x) = c1 e2x + (c2 cos x + c3 sin x) e2x + x2 e2x  x2 + 3 371 C0 = 1, C1 = 0, C2 = 1, C3 = 0, C4 = 3. Chapter 6
is a general solution to the given nonhomogeneous equation. Next, we find constants c1 , c2 , and c3 such that the initial conditions are satisfied. Differentiation yields y (x) = 2c1 e2x + [(2c2 + c3 ) cos x + (2c3  c2 ) sin x] e2x + (2x  2x2 )e2x  2x, y (x) = 4c1 e2x + [(3c2 + 4c3 ) cos x + (3c3  4c2 ) sin x] e2x + (2  8x + 4x2 )e2x  2. Therefore, 3 = y(0) = c1 + c2 + 3, 0 = y (0) = 2c1 + 2c2 + c3 , 0 = y (0) = 4c1 + 3c2 + 4c3 y(x) = x2 e2x  x2 + 3. 35. If a0 = 0, then equation (4) becomes an y (n) + an1 y (n1) + + a1 y = f (x) or, in operator form, an D n + an1 D n1 + + a1 D [y] = f (x) D an D n1 + an1 D n2 + + a1 [y] = f (x). (6.14) c1 + c2 = 0, 2c1 + 2c2 + c3 = 0, 4c1 + 3c2 + 4c3 = 0. The solution of this homogeneous linear system is c1 = c2 = c3 = 0. Hence, the answer is Since the operator D m+1 annihilates any polynomial f (x) = bm xm + + b0 , applying D m+1 to both sides in (6.14) yields D m+1 D an D n1 + an1 D n2 + + a1 [y] = D m+1 [f (x)] = 0 D m+2 an D n1 + an1 D n2 + + a1 [y] = 0. (6.15) The auxiliary equation, corresponding to this homogeneous equation is, r m+2 an r n1 + an1 r n2 + + a1 = 0. Since a1 = 0, an r n1 + an1 r n2 + + a1 372
r=0 (6.16) = a1 = 0, Exercises 6.3
which means that r = 0 is not a root of this polynomial. Thus, for the auxiliary equation (6.16), r = 0 is a root of exact multiplicity m + 2, and so a general solution to (6.15) is given by y(x) = c0 + c1 x + + cm+1 xm+1 + Y (x), (6.17) where Y (x), being associated with roots of an r n1 + an1 r n2 + + a1 = 0, is a general solution to (an D n1 + an1 D n2 + + a1 ) [y] = 0. (One can write down Y (x) explicitly but there is no need in doing this.) On the other hand, the auxiliary equation for the homogeneous equation, associated with (6.14), is r(an r n1 + an1 r n2 + + a1 ) = 0, and r = 0 is its simple root. Hence, a general solution yh (x) to the homogeneous equation is given by yh (x) = c0 + Y (x), (6.18) where Y (x) is the same as in (6.17). Since y(x) = yh (x) + yp (x), it follows from (6.17) and (6.18) that yp (x) = c1 x + + cm+1 xm+1 = x (c1 + + cm+1 xm ) , as stated. 37. Writing equation (4) in operator form yields an D n + an1 D n1 + + a0 [y] = f (x). The characteristic equation, corresponding to the associated homogeneous equation, is an r n + an1 r n1 + + a0 = 0. (6.20) (6.19) Suppose that r = i is a root of (6.20) of multiplicity s 0. (s = 0 means that r = i is not a root.) Then (6.20) can be factored as an r n + an1 r n1 + + a0 = r 2 + 2
s an r n2s + + a0 / 2s = 0 and so a general solution to the homogeneous equation is given by yh (x) = (c1 cos x + c2 sin x) + x(c3 cos x + c4 sin x) + + xs1 (c2s1 cos x + c2s sin x) + Y (x), (6.21) 373 Chapter 6
where Y (x) is the part of yh (x) corresponding to the roots of an r n2s + + a0 / 2s = 0. Since the operator (D 2 + 2 ) annihilates f (x) = a cos x + b sin x, applying this operator to both sides in (6.19), we obtain (D 2 + 2 ) an D n + an1 D n1 + + a0 [y] = (D 2 + 2 )[f (x)] = 0. The corresponding auxiliary equation, (r 2 + 2 ) an r n + an1 r n1 + + a0 = 0 r2 + 2
s+1 an r n2s + + a0 / 2s = 0 has r = i as its root of multiplicity s + 1. Therefore, a general solution to this equation is given by y(x) = (c1 cos x + c2 sin x) + x(c3 cos x + c4 sin x) + + xs1 (c2s1 cos x + c2s sin x) + xs (c2s+1 cos x + c2s+2 sin x) + Y (x). Since, y(x) = yh (x) + yp (x), comparing y(x) with yh (x) given in (6.21), we conclude that yp (x) = xs (c2s+1 cos x + c2s+2 sin x). All that remains is to note that, for any m < s, the functions xm cos x and xm sin x are presented in (6.21), meaning that they are solutions to the homogeneous equation corresponding to (6.19). Thus s is the smallest number m such that xm cos x and xm sin x are not solutions to the corresponding homogeneous equation. 39. Writing the system in operator form yields (D 2  1) [x] + y = 0, x + (D 2  1) [y] = e3t . Subtracting the first equation from the second equation multiplied by (D 2  1), we get D 2  1 [x] + D 2  1 374 D2  1
2 2 [y]  D 2  1 [x] + y = D 2  1 e3t  0 = 8e3t (6.22)  1 [y] = 8e3t D 2 D 2  2 [y] = 8e3t . Exercises 6.4 The auxiliary equation, r 2 (r 2  2) = 0, has roots r = 2 and a double root r = 0. Hence, yh (t) = c1 + c2 t + c3 e 2t + c4 e 2t is a general solution to the homogeneous equation coresponding to (6.22). A particular solution to (6.22) has the form yp (t) = Ae3t . Substitution yields D2 D2  2 and so Ae3x = D 4  2D 2 8e3x 3x yp (t) = Ae = , 63 y(t) = yp (t) + yh (t) = Ae3x = 81Ae3x  (2)9Ae3x = 63Ae3x = 8e3x 8e3x + c1 + c2 t + c3 e 2t + c4 e 2t 63 is a general solution to (6.22). We find x(t) from the second equation in the original system. x(t) = e3t + y(t)  y (t) 8e3x = e3t + + c1 + c2 t + c3 e 2t + c4 e 2t  63 3x e + c1 + c2 t  c3 e 2t  c4 e 2t . =  63 EXERCISES 6.4: 72e3x + 2c3 e 2t + 2c4 e 2t 63 Method of Variation of Parameters, page 341 1. To apply the method of variation of parameters, first we have to find a fundamental solution set for the corresponding homogeneous equation, which is y  3y + 4y = 0. Factoring the auxiliary polynomial, r 3  3r 2 + 4, yields r 3  3r 2 + 4 = r 3 + r 2  4r 2  4 = r 2 (r + 1)  4(r  1)(r + 1) = (r + 1)(r  2)2 . Therefore, r = 1, 2, and 2 are the roots of the auxiliary equation, and y1 = ex , y2 = e2x , and y3 = xe2x form a fundamental solution set. According to the variation of parameters method, we seek for a particular solution of the form yp (x) = v1 (x)y1 (x) + v2 (x)y2 (x) + v3 (x)y3 (x) = v1 (x)ex + v2 (x)e2x + v3 (x)xe2x . 375 Chapter 6
To find functions vj 's we need four determinants, the Wronskian W [y1 , y2 , y3](x) and W1 (x), W2 (x), and W3 (x) given in (10) on page 340 of the text. Thus we compute ex W e , e , xe
x 2x 2x e2x 2e
2x xe2x (1 + 2x)e e2x 2e
2x 2x 1 1 =e e e xe2x
x 2x 2x x = 9e3x , (x) = e x 1 2 1 + 2x 1 4 4 + 4x ex 4e2x (4 + 4x)e2x W1 (x) = (1)31 W e2x , xe2x (x) = W2 (x) = (1)
32 2x (1 + 2x)e ex
x = e4x , = (1 + 3x)ex , W e , xe x 2x (x) =  xe2x (1 + 2x)e
2x e ex W3 (x) = (1)33 W ex , e2x (x) = e2x ex 2e2x = 3ex . Substituting these expressions into the formula (11) for determining vj 's, we obtain v1 (x) = v2 (x) = v3 (x) = g(x)W1 (x) dx = W [ex , e2x , xe2x ] g(x)W2 (x) dx = W [ex , e2x , xe2x ] g(x)W3 (x) dx = W [ex , e2x , xe2x ] e2x e4x 1 3x e , dx = 3x 9e 27 e2x (1 + 3x)ex 1 dx =  3x 9e 9 2x x e 3e x dx = , 3x 9e 3 x x2 , (1 + 3x) dx =   9 6 where we have chosen zero integration constants. Then formula (12), page 340 of the text, gives a particular solution yp (x) = 1 3x x e e  27 x x2 + 9 6 e2x + x 2x 1 2x xe2x x2 e2x xe = e  + . 3 27 9 6 Note that the first two terms in yp (x) are solutions to the corresponding homogeneous equation. Thus, another (and simpler) answer is yp (x) = x2 e2x /6. 3. Let us find a fundamental solution set for the corresponding homogeneous equation, z + 3z  4z = 0. Factoring the auxiliary polynomial, r 3 + 3r 2  4, yields r 3 + 3r 2  4 = r 3  r 2 + 4r 2  4 = r 2 (r  1) + 4(r + 1)(r  1) = (r  1)(r + 2)2 . 376 Exercises 6.4
Therefore, r = 1, 2, and 2 are the roots of the auxiliary equation, and so the functions z1 = ex , z2 = e2x , and z3 = xe2x form a fundamental solution set. A particular solution then has the form zp (x) = v1 (x)z1 (x) + v2 (x)z2 (x) + v3 (x)z3 (x) = v1 (x)ex + v2 (x)e2x + v3 (x)xe2x . (6.23) To find functions vj 's we need four determinants, the Wronskian W [z1 , z2 , z3 ](x) and W1 (x), W2 (x), and W3 (x) given in (10) on page 340 of the text. Thus we compute ex W e ,e
x 2x e2x xe2x =e
3x 1 1 xe2x
2x 1 x = 9e3x , , xe 2x (x) = ex 2e2x (1  2x)e2x ex 4e2x (4x  4)e2x e2x 2e ex e
x 2x 1 2 1  2x 4 4x  4 W1 (x) = (1)31 W e2x , xe2x (x) = W2 (x) = (1)
32 (1  2x)e xe2x
2x = e4x , W e , xe x 2x (x) =  ex e
x (1  2x)e e2x
2x = (3x  1)ex , W3 (x) = (1)33 W ex , e2x (x) = 2e = 3ex . Substituting these expressions into the formula (11) on page 340 of the text, we obtain v1 (x) = v2 (x) = g(x)W1 (x) dx = W [ex , e2x , xe2x ] g(x)W2 (x) dx = W [ex , e2x , xe2x ] = v3 (x) = g(x)W3 (x) dx = W [ex , e2x , xe2x ] 1 9 e2x e4x 1 dx = ex , 3x 9e 9 2x x e (3x  1)e dx 9e3x x 7  e4x , 12 144 e2x (3ex ) 1 dx =  e4x . 3x 9e 12 (3x  1)e4x dx = Substituting these expressions into (6.23) yields zp (x) = 1 x x e e + 9 x 7 1 4x 2x 1 2x  e xe e . = e4x e2x  12 144 12 16 377 Chapter 6
5. Since the nonhomogeneous term, g(x) = tan x, is not a solution to a homogeneous linear differential equation with constant coefficients, we will find a particular solution by the method of variation of parameters. To do this, we must first find a fundamental solution set for the corresponding homogeneous equation, y + y = 0. Its auxiliary equation is r 3 + r = 0, which factors as r 3 + r = r(r 2 + 1). Thus, the roots to this auxiliary equation are r = 0, i. Therefore, a fundamental solution set to the homogeneous equation is {1, cos x, sin x} and yp (x) = v1 (x) + v2 (x) cos x + v3 (x) sin x. To accomplish this, we must find the four determinants W [1, cos x, sin x](x), W1 (x), W2 (x), W3 (x). That is, we calculate 1 W [1, cos x, sin x](x) = cos x sin x cos x = sin2 x + cos2 x = 1, 0  sin x 0  cos x  sin x W1 (x) = (1)31 W [cos x, sin x](x) = W2 (x) = (1)32 W [1, sin x](x) =  W3 (x) = (1)33 W [1, cos x](x) = 1 cos x sin x  sin x cos x 1 sin x 0 cos x cos x 0  sin x = cos2 x + sin2 x = 1, =  cos x, =  sin x. By using formula (11) on page 340 of the text, we can now find v1 (x), v2 (x), and v3 (x). Since g(x) = tan x, we have (assuming that all constants of integration are zero) v1 (x) = g(x)W1 (x) dx = W [1, cos x, sin x](x) g(x)W2 (x) dx = W [1, cos x, sin x](x) tan x dx = ln(sec x), v2 (x) = tan x( cos x) dx =  sin x dx = cos x, v3 (x) = sin2 x g(x)W3 (x) dx = tan x( sin x) dx =  dx W [1, cos x, sin x](x) cos x 1  cos2 x = dx = (cos x  sec x) dx = sin x  ln(sec x + tan x). cos x 378 Exercises 6.4
Therefore, we have yp (x) = v1 (x) + v2 (x) cos x + v3 (x) sin x = ln(sec x) + cos2 x + sin2 x  sin x ln(sec x + tan x) yp (x) = ln(sec x)  sin x ln(sec x + tan x) + 1. Since y 1 is a solution to the homogeneous equation, we may choose yp (x) = ln(sec x)  sin x ln(sec x + tan x). Note: We left the absolute value signs off ln(sec x) and ln(sec x + tan x) because of the stated domain: 0 < x < /2. 7. First, we divide the differential equation by x3 to obtain the standard form y  3x1 y + 6x2 y  6x3 y = x4 , x > 0, from which we see that g(x) = x4 . Given that {x, x2 , x3 } is a fundamental solution set for the corresponding homogeneous equation, we are looking for a particular solution of the form yp (x) = v1 (x)x + v3 (x)x2 + v3 (x)x3 . Evaluating determinants W [x, x2 , x3 ](x), W1 (x), W2 (x), and W3 (x) yileds x x2 W [x, x2 , x3 ](x) = 0 2 x3
2 (6.24) 1 2x 3x =x x2 2x 3x2 2 x3
2 6x 6x  x2 x3 2 6x = 2x3 , W1 (x) = (1)31 W [x2 , x3 ](x) = W2 (x) = (1)
32 2x 3x x = x4 , = 2x3 , W [x, x ](x) =  3 x3
2 1 3x W3 (x) = (1)33 W [x, x2 ](x) = x x2 1 2x = x2 . 379 Chapter 6
So, v1 (x) = v2 (x) = v3 (x) = g(x)W1 (x) dx = W [x, x2 , x3 ](x) g(x)W2 (x) dx = W [x, x2 , x3 ](x) g(x)W3 (x) dx = W [x, x2 , x3 ](x) x4 x4 1 dx =  2 + c1 , 3 2x 4x 4 3 1 x (2x ) dx = 3 + c2 , 3 2x 3x 4 2 1 x (x ) dx =  4 + c3 , 3 2x 8x where c1 , c2 , and c3 are constants of integration. Substitution back into (6.24) yields yp (x) =  1 + c1 x + 4x2 1 1 1 + c1 x + c2 x2 + c3 x3 . + c2 x2 +  4 + c3 x3 =  3 3x 8x 24x Since {x, x2 , x3 } is a fundamental solution set for the homogeneous equation, taking c1 , c2 , and c3 to be arbitrary constants, we obtain a general solution to the original nonhomogeneous equation. That is, y(x) =  1 + c1 x + c2 x2 + c3 x3 . 24x 9. To find a particular solution to the nonhomogeneous equation, we will use the method of variation of parameters. We must first calculate the four determinants W [ex , ex , e2x ](x), W1 (x), W2 (x), W3 (x). Thus, we have ex W [ex , ex , e2x ](x) = ex ex 0 W1 (x) = 1 e2x = (1)31 0 ex 2e2x ex 4e2x e2x = (1)32 ex e
x ex e2x = 4e2x + 2e2x + e2x + e2x  2e2x  4e2x = 6e2x , ex ex 2e2x ex 4e2x ex e2x ex 2e2x e2x 2e
2x = 2ex + ex = 3ex , ex 0 W2 (x) = ex 0 2e2x ex 1 4e2x ex ex 0 e
x x =  2e3x  e3x = e3x , W3 (x) = e e x x 0 1 = (1)33 ex ex e ex ex = 1  1 = 2. 380 Exercises 6.4
Therefore, according to formula (12) on page 340 of the text, a particular solution, yp (x), will be given by yp (x) = e
x 3ex g(x) dx + ex 2x 6e e3x g(x) dx + e2x 2x 6e 1 x e 6 2g(x) dx 6e2x 1 2x e 3 e2x g(x) dx. 1 yp (x) =  ex 2 ex g(x) dx + ex g(x) dx + 11. First, we find a fundamental solution set to the corresponding homogeneous equation, x3 y  3xy + 3y = 0. (6.25) Here we involve the procedure of solving CauchyEuler equations discussed in Problem 38, Section 4.3. Thus, let x = et . Then dx/dt = et = x and so the chain rule yields dy dy dx dy = =x , dt dx dt dx dy d dy dx dy d2 y d2 y d2 y d2 y d dy dy = x = + x 2 x == x + x2 2 = + x2 2 , = dt2 dt dt dx dx dt dx dx dx dx dt dx d2 y d2 y dy , x2 2 = 2  dx dt dt dy d2 y d2 y dx d3 y d3 y d d2 y d dy = + 3x 2 + x2 3 x = = x + x2 2 dt3 dt dt2 dx dx dx dt dx dx dx 2 3 2 3 d y dy dy dy dy d3 y dy d2 y dy dy +3  + x3 3 = 3 2  2 + x3 3 = x + 3x2 2 + x3 3 = dx dx dx dt dt2 dt dx dt dt dx 3 3 2 dy dy dy dy x3 3 = 3  3 2 + 2 . dx dt dt dt Substituting these expressions into (6.25), we obtain d2 y dy dy d3 y 3 2 +2 3 + 3y = 0 3 dt dt dt dt d2 y dy d3 y + 3y = 0. 3 2  dt3 dt dt The auxiliary equation corresponding to this linear homogeneous equation with constant coefficients is r 3  3r 2  r + 3 = 0 r 2 (r  3)  (r  3) = 0 (r  3)(r + 1)(r  1) = 0, 381 Chapter 6
whose roots are r = 1, 1, and 3. Therefore, the functions y1 (t) = et , y2 (t) = et , and y3 (t) = e3t form a fundamental solution set. Substituting back et = x we find that y1 (x) = et = x , y2 (x) = et = (et )
1 = x1 , y3 (x) = e3t = (et ) = x3 form a fundamental solution set for the homogeneous equation (6.25). Next, we apply the variation of parameters to find a particular solution to the original equation. A particular solution has the form yp (x) = v1 (x)x + v2 (x)x1 + v3 (x)x3 . (6.26) 3 To find functions v1 (x), v2 (x), and v3 (x) we use formula (11) on page 340 of the text. We compute x W [x, x1 , x3 ](x) = 0 x1 2x3 x3 =x x1 x3
2 1 x2 3x2 6x x2 3x2 2x3 x3 6x  x1 x3 2x3 6x = 16 , W1 (x) = (1)31 W [x1 , x3 ](x) = W2 (x) = (1)
32 x2 3x2 x 1 3x x = 4x , W [x, x ](x) =  3 = 2x3 , = 2x1 . W3 (x) = (1)33 W [x, x1 ](x) = x1 1 x2 Also, writing the given equation in standard form, y  3 3 y + 3 y = x cos x, 2 x x we see that the nonhomogeneous term is g(x) = x cos x. Thus, by (11), v1 (x) = 382 1 x cos x(4x) dx =  16 4 x2 cos x dx =  1 2 x sin x + 2x cos x  2 sin x + c1 , 4 Exercises 6.4
v2 (x) = = 1 x cos x(2x3 ) dx = 16 8 x4 cos x dx 1 4 x sin x + 4x3 cos x  12x2 sin x  24x cos x + 24 sin x + c2 , 8 x cos x(2x1 ) 1 1 dx = v3 (x) = cos x dx = sin x + c3 , 16 8 8 where c1 , c2 , c3 are constants of integration, and we have used integration by parts to evaluate v1 (x) and v2 (x). Substituting these functions into (6.26) and simplifying yields yp (x) =  (x2 sin x + 2x cos x  2 sin x) x + c1 x 4 x3 sin x (x4 sin x + 4x3 cos x  12x2 sin x  24x cos x + 24 sin x) x1 + c2 x1 + + c3 x3 + 8 8 = c1 x + c2 x1 + c3 x3  x sin x  3 cos x + 3x1 sin x . If we allow c1 , c2 , and c3 in the above formula to be arbitrary constants, we obtain a general solution to the original CauchyEuler equation. Thus, the answer is y(x) = c1 x + c2 x1 + c3 x3  x sin x  3 cos x + 3x1 sin x . 13. Since y1 Wk (x) = y1 . . .
(n2) y1 (n1) y1 . . . yk1 . . . yk1 . . . ... ...
(n2) yk1 (n1) yk1 0 yk+1 0 . . . 0 1 yk+1 . . .
(n2) yk+1 (n1) yk+1 . . . yn . . . yn . . . ... ...
(n2) yn (n1) yn , the kth column of this determinant consists of all zeros except the last entry, which is 1. Therefore, expanding Wk (x) by the cofactors in the kth column, we get Wk (x) = (0)C1,k + (0)C2,k + + (0)Cn1,n + (1)Cn,k y1 y1 = (1)(1)n+k . . . y1
(n2) . . . yk1 . . . yk1 . . . . . . yk1
(n2) yk+1 yk+1 . . . yk+1
(n2) . . . yn . . . yn . . . . . . yn
(n2) 383 Chapter 6
= (1)n+k W [y1 , . . . , yk1, yk+1, . . . , yn ] (x). Finally, (1)n+k = (1)(nk)+(2k) = (1)nk . REVIEW PROBLEMS: page 344 1. (a) In notation of Theorem 1, we have p1 (x) 0, p2 (x) =  ln x, p3 (x) = x, p4 (x) 2, and g(x) = cos 3x. All these functions, except p2 (x), are continuous on (, ), and p2 (x) is defined and continuous on (0, ). Thus, Theorem 1 guarantees the existence of a unique solution on (0, ). (b) By dividing both sides of the given differential equation by x2 1, we rewrite the equation in standard form, that is, x2 + 3 x+4 sin x ex y + 2 y + 2 y= 2 . y + 2 x 1 x 1 x 1 x 1 p2 (x) = x+4 , 2 1 x ex , x2  1 x2 + 3 . x2  1 Thus we see that sin x , p1 (x) = 2 x 1 p3 (x) = and g(x) = Functions p1 (x), p3 (x), and g(x) are defined and continuous on (, ) except x = 1; p2 (x) is defined and continuous on {x 4, x = 1}. Thus, the common domain for p1 (x), p2 (x), p3 (x), and g(x) is {x 4, x = 1}, and, in addition, these functions are continuous there. This set consists of three intervals, [4, 1), (1, 1), and (1, ). Theorem 1 guarantees the existence of a unique solution on each of these intervals. 3. A linear combination, c1 sin x + c2 x sin x + c3 x2 sin x + c4 x3 sin x = c1 + c2 x + c3 x2 + c4 x3 sin x (6.27) vanishes identically on (, ) if and only if the polynomial c1 + c2 x + c3 x2 + c4 x3 vanishes identically on (, ). Since the number of real zeros of a polynomial does not exceed 384 Review Problems
its degree, unless it's the zero polynomial, we conclude that the linear combination (6.27) vanishes identically on (, ) if and only if c1 = c2 = c3 = c4 = 0. This means that the given functions are linearly independent on (, ). 5. (a) Solving the auxiliary equation yields (r + 5)2 = 0 or (r + 5)2 (r  2)3 (r 2 + 1)2 = 0 (r  2)3 = 0 or (r 2 + 1)2 = 0. Thus, the roots of the auxiliary equation are r = 5 of multiplicity 2, r=2 r = i of multiplicity 3, of multiplicity 2. According to (22) on page 329 and (28) on page 330 of the text, the set of functions (assuming that x is the independent variable) e5x , xe5x , e2x , xe2x , x2 e2x , cos x, x cos x, sin x, x sin x forms an independent solution set. Thus, a general solution is given by c1 e5x + c2 xe5x + c3 e2x + c4 xe2x + c5 x2 e2x + c6 cos x + c7 x cos x + c8 sin x + c9 x sin x . (b) Solving the auxiliary equation yields r 4 = 0 or r 4 (r  1)2 (r 2 + 2r + 4)2 = 0 (r  1)2 = 0 or (r 2 + 2r + 4)2 = 0. Thus, the roots of the auxiliary equation are r=0 r=1 r = 1 of multiplicity 4, of multiplicity 2, 3i of multiplicity 2. 385 Chapter 6
Using (22) on page 329 and (28) on page 330 of the text, we conclude that the set of functions (with x as the independent variable) 1 , x , x2 , x3 , ex , xex , ex cos 3x, xex cos 3x, sin 3x, xex sin 3x forms an independent solution set. A general solution is given then by c1 + c2 x + c3 x2 + c4 x3 + c5 ex + c6 xex + c7 ex cos 3x + c8 xex cos 3x +c9 sin 3x + c10 xex sin 3x = c1 + c2 x + c3 x2 + c4 x3 + (c5 + c6 x)ex + (c7 + c8 x)ex cos 3x +(c9 + c10 x)ex sin 3x . 7. (a) D 3 , since the third derivative of a quadratic polynomial is identically zero. (b) The function e3x + x  1 is the sum of e3x and x  1. The function x  1 is annihilated by D 2 , the second derivative operator, and, according to (i) on page 334 of the text, (D  3) annihilates e3x . Therefore, the composite operator D 2 (D  3) = (D  3)D 2 annihilates both functions and, hence, there sum. (c) The function x sin 2x is of the form given in (iv) on page 334 of the text with m = 2, = 0, and = 2. Thus, the operator (D  0)2 + 22 annihilates this function. (d) We again use (iv) on page 334 of the text, this time with m = 3, = 2, and = 3, to conclude that the given function is annihilated by [D  (2)]2 + 32 386
3 2 = D2 + 4 2 = (D + 2)2 + 9 3 . Review Problems
(e) Representing the given function as a linear combination, x2  2x + xex + (sin 2x)  (cos 3x), we find an annihilator for each term. Thus, we have: x2  2x is annihilated by D 3 , xex sin 2x cos 3x is annihilated by [D  (1)]2 = (D + 1)2 (by (ii), page 334) , is annihilated by D 2 + 22 = D 2 + 4 is annihilated by D 2 + 32 = D 2 + 9 (by (iii), page 334) , (by (iii), page 334) . Therefore, the product D 3 (D + 1)2 (D 2 + 4)(D 2 + 9) annihilates the given function. 9. A general solution to the corresponding homogeneous equation, x3 y  2x2 y  5xy + 5y = 0, is given by yh (x) = c1 x + c2 x5 + c3 x1 . We now apply the variation of parameters method described in Section 6.4, and seek for a particular solution to the original nonhomogeneous equation in the form yp (x) = v1 (x)x + v2 (x)x5 + v3 (x)x1 . Since (x) = 1, (x5 ) = 5x4 , (x) = 0 , (x5 ) = 20x3 , (x1 ) = x2 , (x1 ) = 2x3 , the Wronskian W [x, x5 , x1 ](x) and determinants Wk (x) given in (10) on page 340 of the text become x W [x, x5 , x1 ](x) = 1 x5
3 x1 = (x) 2x
3 5x4 x2 5x4 x2 20x3 2x3  (1) x5 x1 0 20x x5 5x
4 20x3 2x3 = (x)(30x)  (18x2 ) = 48x2 , W1 (x) = (1)31 x1 x
2 = 6x3 , 387 Chapter 6
W2 (x) = (1)32 W3 (x) = (1)33 x x1 x5
4 1 x2 x 1 5x = 2x1 , = 4x5 . Now we divide both sides of the given equation by x3 to obtain an equation in standard form, that is, y  2x1 y  5x2 y + 5x3 y = x5 . Hence, the righthand side, g(x), in formula (1) on page 339 of the text equals to x5 . Applying formula (11), page 340 of the text, yields v1 (x) = v2 (x) = v3 (x) = Therefore, yp (x) = = 1 3 1 1 7 x x x+  x5 +  x1 x1 24 168 12 1 1 1 1   x2 =  x2 , 24 168 12 21 x5 (6x3 ) 1 1 3 x4 dx = x , dx =  2 48x 8 24 x5 (2x1 ) 1 1 7 x , dx = x8 dx =  2 48x 24 168 x5 (4x5 ) 1 1 x2 dx =  x1 . dx = 2 48x 12 12 and a general solution to the given equation is given by y(x) = yh (x) + yp (x) = c1 x + c2 x5 + c3 x1  1 2 x . 21 388 CHAPTER 7: Laplace Transforms
EXERCISES 7.2: Definition of the Laplace Transform, page 359 1. For s > 0, using Definition 1 on page 351 and integration by parts, we compute N N L {t} (s) =
0 e st t dt = lim
st N 0 N e
0 st t dt = lim N td 
0 est s
N 0 N = = te lim  N s
N + 1 s est dt = lim
0 N  test s  est s2 N 0 lim  1 NesN 1 esN +0 2 + 2 = 2 s s s s because, for s > 0, esN 0 and NesN = N/esN 0 as N . 3. For s > 6, we have N L {t} (s) =
0 est e6t dt =
0 e(6s)t dt = lim
N 0 N e(6s)t dt
0 = 5. For s > 0, N lim e 6s (6s)t = lim N e 1 1 1  =0 = . 6s 6s 6s s6 (6s)N N L {cos 2t} (s) =
0 est cos 2t dt = lim lim e
st N est cos 2t dt
0 = = N N lim (s cos 2t + 2 sin 2t) N 0 s2 + 4 sN s e (s cos 2N + 2 sin 2N) s  2 , = 2 s2 + 4 s +4 s +4 where we have used integration by parts to find an antiderivative of est cos 2t. 389 Chapter 7
7. For s > 2, L e2t cos 3t (s) =
0 est e2t cos 3t dt =
0 e(2s)t cos 3t dt e(2s)t ((2  s) cos 3t + 3 sin 3t) N = lim N 0 (2  s)2 + 9 s2 e(2s)N [(2  s) cos 3N + 3 sin 3N]  (2  s) = . = lim 2+9 N (2  s) (s  2)2 + 9 9. As in Example 4 on page 353 in the text, we first break the integral into separate parts. Thus, 2 L {f (t)} (s) =
0 est f (t) dt =
0 est 0 dt +
2 test dt =
2 test dt . An antiderivative of test was, in fact, obtained in Problem 1 using integration by parts. Thus, we have test dt =
2 N lim  test est  2 s s N 2 = lim N  esN 2e2s e2s NesN  2 + + 2 s s s s 2s + 1 s2 . = 2e2s e2s + 2 = e2s s s 2 1 + 2 s s = e2s 11. In this problem, f (t) is also a piecewise defined function. So, we split the integral and obtain L {f (t)} (s) =
0 est f (t) dt =
0 st est sin t dt + 0 est 0 dt =
0 s est sin t dt = which is valid for all s. e (s sin t  cos t) s2 + 1 = e s s2 e  (1) +1 = 2 , +1 s +1 13. By the linearity of the Laplace transform, L 6e3t  t2 + 2t  8 (s) = 6L e3t (s)  L t2 (s) + 2L {t} (s)  8L {1} (s). From Table 7.1 on page 358 in the text, we see that L e3t (s) = 390 1 1 = , s  (3) s+3 s > 3; Exercises 7.2
L t2 (s) = Thus the formula L 6e3t  t2 + 2t  8 (s) = 6 1 1 1 2 8 2 6 2  3 +2 2 8 =  3+ 2 , s+3 s s s s+3 s s s 2! s2+1 = 2 , s3 L {t} (s) = 1! s1+1 = 1 , s2 L {1} (s) = 1 , s s > 0. is valid for s in the intersection of the sets s > 3 and s > 0, which is s > 0. 15. Using the linearity of Laplace transform and Table 7.1 on page 358 in the text, we get L t3  tet + e4t cos t (s) = L t3 (s)  L tet (s) + L e4t cos t (s) 1! s4 3! + = 3+1  s (s  1)1+1 (s  4)2 + 12 6 1 s4 , = 4 + 2 s (s  1) (s  4)2 + 1 which is valid for s > 4. 17. Using the linearity of Laplace transform and Table 7.1 on page 358 in the text, we get L e3t sin 6t  t3 + et (s) = L e3t sin 6t (s)  L t3 (s) + L et (s) 3! 1 1 6 6 6  3+1 + = =  4+ , 2 + 62 2 + 36 (s  3) s s1 (s  3) s s1 valid for s > 3. 19. For s > 5, we have L t4 e5t  et cos 7t (s) = L t4 e5t (s)  L et cos 7t (s) = 4! s1 24 s1 .  =  4+1 5 (s  5) (s  5) (s  1)2 + 7 (s  1)2 + ( 7)2 21. Since the function g1 (t) 1 is continuous on (, ) and f (t) = g1 (t) for t in [0, 1], we conclude that f (t) is continuous on [0, 1) and continuous from the left at t = 1. The function g2 (t) (t  2)2 is also continuous on (, ), and so f (t) (which is the same as g2 (t) on (1, 10]) is continuous on (1, 10]. Moreover,
t1 lim f (t) = lim g2 (t) = g2 (1) = (1  2)2 = 1 = f (1), + +
t1 391 Chapter 7
which implies that f (t) is continuous from the right at t = 1. Thus f (t) is continuous at t = 1 and, therefore, is continuous at any t in [0, 10]. 23. All the functions involved in the definition of f (t), that is, g1 (t) 1, g2 (t) = t  1, and g3 (t) = t2  4, are continuous on (, ). So, f (t), being a restriction of these functions, on [0, 1), (1, 3), and (3, 10], respectively, is continuous on these three intervals. At points t = 1 and 3, f (t) is not defined and so is not continuous. But onesided limits lim f (t) = lim g1 (t) = g1 (1) = 1, 
t1 t1 t1 t1+ t3 lim f (t) = lim g2 (t) = g2 (1) = 0, + lim f (t) = lim g2 (t) = g2 (3) = 2,  
t3 t3 t3+ lim f (t) = lim g3 (t) = g3 (3) = 5, + exist and pairwise different. Therefore, f (t) has jump discontinuities at t = 1 and t = 3, and hence piecewise continuous on [0, 10]. 25. Given function is a rational function and, therefore, continuous on its domain, which is all reals except zeros of the denominator. Solving t2 + 7t + 10 = 0, we conclude that the points of discontinuity of f (t) are t = 2 and t = 5. These points are not in [0, 10]. So, f (t) is continuous on [0, 10]. 27. Since
t0 lim f (t) = lim + +
t0 1 = , t f (t) has infinite discontinuity at t = 0, and so neither continuous nor piecewise continuous [0, 10]. 29. (a) First observe that t3 sin t t3  for all t. Next, three applications of L'Hospital's rule show that t3 3t2 6t 6 = lim = lim 2 t = lim 3 t = 0 t t t e t e t e t e lim 392 Exercises 7.2
for all > 0. Thus, fixed > 0, for some T = T () > 0, we have t3  < et for all t > T , and so t3 sin t t3 < et , t > T. Therefore, t3 sin t is of exponential order , for any > 0. (b) Clearly, for any t, f (t) = 100e49t , and so Definition 3 is satisfied with M = 100, = 49, and any T . Hence, f (t) is of exponential order 49. (c) Since f (t) 3 2 = lim et t = lim e(t )t = , t t e t t t we see that f (t) grows faster than e for any . Thus f (t) is not of exponential order. lim (d) Similarly to (a), for any > 0, we get t ln t t ln t ln t + 1 1/t = lim t = lim = lim 2 t = 0 , t t t e t e t e t e lim and so f (t) is of exponential order for any positive . (e) Since,
2 1 2 et + et > et f (t) = cosh t = 2 2 t2 t and e grows faster than e for any fixed (see page 357 in the text), we conclude that cosh (t2 ) is not of exponential order. (f) This function is bounded: f (t) = t2 1 1 = 1, = f r1t2 + 1 +1 0+1 2 2 and so Definition 3 is satisfied with M = 1 and = 0. Hence, f (t) is of exponential order 0. (g) The function sin (t2 ) is bounded, namely, sin (t2 ) 1. For any fixed > 0, the limit of t4 /et , as t , is 0, which implies that t4 et for all t > T = T (). Thus, sin t2 + t4 e6t 1 + et e6t = 2e+6t . This means that f (t) is of exponential order for any > 6. 393 Chapter 7
(h) The function 3 + cos 4t is bounded because 3 + cos 4t 3 +  cos 4t 4. Therefore, by the triangle inequality, f (t) et  3 + cos 4t et  4, and, therefore, for any fixed , f (t) grows faster than et (because et does, and the other term is bounded). So, f (t) is not of exponential order. (i) Clearly, for any t > 0, t t2 = t < (1)t = t. t+1 t+1 et
2 /(t+1) 2 2 2 Therefore, < et , and Definition 3 holds with M = 1, = 1, and T = 0. (j) Since, for any x, 1 sin x 1, the given function is bounded. Indeed, sin et
2 + esin t sin et 2 + esin t 1 + e Thus it is of exponential order 0. 31. (a) N L e (a+ib)t (s) :=
0 e st (a+ib)t e dt =
0 N 0 e (a+ibs)t dt = lim N e(a+ibs)t dt
0 = Since N lim e a + ib  s (a+ibs)t = 1 lim e(as+ib)N  1 . (7.1) a + ib  s N e(as+ib)x = e(as)x eibx , where the first factor vanishes at if a  s < 0 while the second factor is a bounded ( eibx 1) and periodic function, the limit in (7.1) exists if and only if a  s < 0. Assuming that s > a, we get 1 1 1 lim e(as+ib)N  1 = (0  1) = . N a + ib  s a + ib  s s  (a + ib) 394 Exercises 7.2
(b) Note that s  (a + ib) = (s  a)  ib. Multiplying the result in (a) by the complex conjugate of the denominator, that is, (s  a) + bi, we get 1 (s  a) + ib (s  a) + ib , = = s  (a + ib) [(s  a)  ib] [(s  a) + ib] (s  a)2 + b2 where we used the fact that, for any complex number z, zz = z2 . (c) From (a) and (b) we klnow that L e(a+ib)t (s) = Writing (s  a) + ib . (s  a)2 + b2 (s  a) + ib sa b = + i, 2 + b2 2 + b2 (s  a) (s  a) (s  a)2 + b2 sa b sa + i = , (7.2) (s  a)2 + b2 (s  a)2 + b2 (s  a)2 + b2 sa b b (s) = Im + i = . (7.3) (s  a)2 + b2 (s  a)2 + b2 (s  a)2 + b2 we see that Re L e(a+ib)t (s) = Re Im L e(a+ib)t On the other hand, by Euler's formulas, Re est e(a+ib)t = est Re eat (cos bt + i sin bt) = est eat cos bt and so Re L e(a+ib)t (s) = Re es e(a+ib)t dt est e(a+ib)t dt = Re 0 0 =
0 Re es e(a+ib)t dt =
0 est eat cos bt dt = L eat cos bt (s), which together with (7.2) gives the last entry in Table 7.1. Similarly, Im L e(a+ib)t (s) = L eat sin bt (s), and so (7.3) gives the Laplace transform of eat sin bt. 395 Chapter 7
33. Let f (t) be a piecewise continuous function on [a, b], and let a function g(t) be continuous on [a, b]. At any point of continuity of f (t), the function (f g)(t) is continuous as the product of two continuous functions at this point. Suppose now that c is a point of discontinuity of f (t). Then onesided limits lim f (t) = L and lim f (t) = L+ tc tc+ exist. At the same time, continuity of g(t) yields lim g(t) = lim g(t) = lim g(t) = g(c). +
tc tc tc Thus, the product rule implies that onesided limits lim (f g)(t) = lim f (t) lim g(t) = L g(c)  
tc tc tc tc+ lim (f g)(t) = lim f (t) lim g(t) = L+ g(c) + +
tc tc exist. So, f g has a jump (even removable if g(c) = 0) discontinuity at t = c. Therefore, the product (f g)(t) is continuous at any point on [a, b] except possibly a finite number of points (namely, points of discontinuity of f (t)). EXERCISES 7.3: Properties of the Laplace Transform, page 365 1. Using the linearity of the Laplace transform we get L t2 + et sin 2t (s) = L t2 (s) + L et sin 2t (s). From Table 7.1 in Section 7.2 we know that L t2 (s) = Thus L t2 + et sin 2t (s) = 396 2 2 . + 3 s (s  1)2 + 4 2 2! = 3, 3 s s L et sin 2t (s) = 2 2 . = 2 + 22 (s  1) (s  1)2 + 4 Exercises 7.3
3. By the linearity of the Laplace transform, L et cos 3t + e6t  1 (s) = L et cos 3t (s) + L e6t (s)  L {1} (s). From Table 7.1 of the text we see that L et cos 3t (s) = L e6t (s) = L {1} (s) = s  (1) s+1 , = 2 + 32 [s  (1)] (s + 1)2 + 9 s > 6; s > 1; (7.4) (7.5) (7.6) 1 , s6 1 , s s > 0. Since (7.4), (7.5), and (7.6) all hold for s > 6, we see that our answer, L et cos 3t + e6t  1 (s) = 1 1 s+1 +  , 2+9 (s + 1) s6 s is valid for s > 6. Note that (7.4) and (7.5) could also be obtained from the Laplace transforms for cos 3t and 1, respectively, by applying the translation Theorem 3. 5. We use the linearity of the Laplace transform and Table 7.1 to get L 2t2 et  t + cos 4t (s) = 2L t2 et (s)  L {t} (s) + L {cos 4t} (s) 1 s 4 1 s 2  2+ 2 =  2+ 2 = 2 , 3 2 3 (s + 1) s s +4 (s + 1) s s + 16 which is valid for s > 0. 7. Since (t  1)4 = t4  4t3 + 6t2  4t + 1, we have from the linearity of the Laplace transform that L (t  1)4 (s) = L t4 (s)  4L t3 (s) + 6L t2 (s)  4L {t} (s) + L {1} (s). From Table 7.1 of the text, we get that, for s > 0, L t4 (s) = L t3 4! 24 = 5, 5 s s 3! 6 (s) = 4 = 4 , s s 397 Chapter 7
2 2! = 3, 3 s s 1! 1 L {t} (s) = 2 = 2 , s s 1 L {1} (s) = . s L t2 (s) = Thus L (t  1)4 (s) = 9. Since L et sin 2t (s) = we use Theorem 6 to get L et t sin 2t (s) = L t et sin 2t (s) =  L et sin 2t (s) = 
2 24 24 12 4 1  4 + 3  2+ , 5 s s s s s 2 , (s + 1)2 + 4 s > 0. = 2(1) (s + 1)2 + 4 2 (s + 1)2 + 4 4(s + 1) (s + 1)2 + 4 = . [(s + 1)2 + 4]2 11. We use the definition of cosh x and the linear property of the Laplace transform. L {cosh bt} (s) = L = ebt + ebt 2 (s) 1 1 1 1 s L ebt (s) + L ebt (s) = + . = 2 2 2 sb s+b s  b2 13. In this problem, we need the trigonometric identity sin2 t = (1  cos 2t)/2 and the linearity of the Laplace transform. L sin2 t (s) = L = 1  cos 2t 2 (s) 1 1 s 1 2 [L {1} (s)  L {cos 2t} (s)] =  2 . = 2 + 4) 2 2 s s +4 s(s 15. From the trigonometric identity cos2 t = (1 + cos 2t)/2, we find that cos3 t = cos t cos2 t = 398 1 1 cos t + cos t cos 2t . 2 2 Exercises 7.3
Next we write cos t cos 2t = Thus, 1 1 3 1 1 cos t + cos 3t + cos t = cos t + cos 3t. 2 4 4 4 4 We now use the linearity of the Laplace transform and Table 7.1 to find that cos3 t = L cos3 t (s) = which holds for s > 0. 17. Since sin A sin B = [cos(A  B)  cos(A + B)]/2, we get L {sin 2t sin 5t} (s) = L = 1 2 cos 3t  cos 7t 1 (s) = [L {cos 3t} (s)  L {cos 7t} (s)] 2 2 s s 20s  2 = 2 . 2+9 s s + 49 (s + 9)(s2 + 49) 1 3 s 1 s 3 L {cos t} (s) + L {cos 3t} (s) = + , 4 4 4 s2 + 1 4 s2 + 9 1 1 1 [cos(2t + t) + cos(2t  t)] = cos 3t + cos t. 2 2 2 19. Since sin A cos B = [sin(A + B) + sin(A  B)]/2, we get L {cos nt sin mt} (s) = L sin[(m + n)t] + sin[(m  n)t] (s) 2 m+n mn 1 1 + . = 2 + (m + n)2 2 + (m  n)2 2s 2s sa . (s  a)2 + b2 21. By the translation property of the Laplace transform (Theorem 3), L eat cos bt (s) = L {cos bt} (s  a) = 23. Clearly, (t sin bt) = (t) sin bt + t(sin bt) = sin bt + bt cos bt. Therefore, using Theorem 4 and the entry 30, that is, L {t sin bt} (s) = (2bs)/[(s2 + b2 )2 ], we obtain L {sin bt + bt cos bt} (s) = L {(t sin bt) } (s) = sL {t sin bt} (s)  (t sin bt) = s(2bs) 2bs 0 = 2 . 2 + b2 )2 (s (s + b2 )2 399
2 t=0 u2 u + b2 =
u=sa Chapter 7
25. (a) By property (6) on page 363 of the text, L {t cos bt} (s) =  [L {cos bt} (s)] =  (b) Again using the same property, we get L t2 cos bt (s) = L {t(t cos bt)} (s) =  [L {t cos bt} (s)] =  s2  b2 (s2 + b2 )2 = 2s3  6sb2 , (s2 + b2 )3 s > 0. s s2 + b2 = s2  b2 , (s2 + b2 )2 s > 0. 27. First observe that since f (t) is piecewise continuous on [0, ) and f (t)/t approaches a finite limit as t 0+ , we conclude that f (t)/t is also piecewise continuous on [0, ). Next, since for t 1 we have f (t)/t f (t), we see that f (t)/t is of exponential order since f (t) is. These observations and Theorem 2 on page 357 of the text show that L{f (t)/t} exists. When the results of Problem 26 are applied to f (t)/t, we see that
N lim L f (t) t (N) = 0. By Theorem 6, we have that F (s) =
0 est f (t) dt =
0 d test f (t) dt =  L t ds f (t) t (s) . Thus, s F (u) du =
s s d  L du f (t) t f (t) t (u) du = d L du f (t) t (u) du (s) . = L (s)  lim L
N f (t) t (N) = L f (t) t 29. From the linearity properties (2) and (3) on page 354 of the text we have L {g(t)} (s) = L {y (t) + 6y (t) + 10y(t)} (s) = L {y (t)} (s) + 6L {y (t)} (s) + 10L {y(t)} (s). Next, applying properties (2) and (4) on pages 361 and 362 yields L {g} (s) = s2 L {y} (s)  sy(0)  y (0) + 6 [sL {y} (s)  y(0)] + 10L {y} (s). 400 Exercises 7.3
Keeping in mind the fact that all initial conditions are zero the above becomes G(s) = s2 + 6s + 10 Y (s), Therefore, the transfer function H(s) is given by H(s) = 1 Y (s) = 2 . G(s) s + 6s + 10 where Y (s) = L {y} (s). 31. Using Definition 1 of the Laplace transform in Section 7.2, we obtain c L {g(t)} (s) =
0 e st g(t) dt =
0 (0) dt +
c est f (t  c) dt = t  c u, dt du esu f (u) du = ecs L {f (t)} (s). =
0 es(u+c) f (u) du = ecs
0 33. The graphs of the function f (t) = t and its translation g(t) to the right by c = 1 are shown in Figure 7A(a). We use the result of Problem 31 to find L{g(t)}. L {g(t)} (s) = e(1)s L {t} (s) = es . s2 35. The graphs of the function f (t) = sin t and its translation g(t) to the right by c = /2 units are shown in Figure 7A(b). We use the formula in Problem 31 to find L{g(t)}. L {g(t)} (s) = e(/2)s L {sin t} (s) = e(/2)s . s2 + 1 37. Since f (t) is of exponential order on [0, ), for some , M > 0, and T > 0, f (t) Met , for all t T. (7.7) On the other hand, piecewise continuity of f (t) on [0, ) implies that f (t) is bounded on any finite interval, in particular, on [0, T ]. That is, f (t) C, for all t in [0, T ]. (7.8) 401 Chapter 7
f (t)=t g(t)
2 g(t)
0 /2 f (t)=sin t
0 1 2 3 (a) (b) Figure 7A: Graphs of functions in Problems 33 and 35. From (7.7) and (7.8) it follows that, for s > , T T e
0 st f (t) dt =
0 e st f (t) dt +
T e st f (t) dt C
0 e st dt + M
T est et dt = Ce s st T + lim
0 N Me s (s)t N =
T C 1e s sT + Me(s)T  0 s as s . Therefore, (7) yields 0 sL {f } (s)  f (0) =
0 est f (t) dt 0 est f (t) dt  0 as s . Hence, by the squeeze theorem, lim sL {f } (s)  f (0) = 0 lim [sL {f } (s)  f (0)] = 0 lim sL {f } (s) = f (0). s s s EXERCISES 7.4: Inverse Laplace Transform, page 374 1. From Table 7.1, the function 6/(s  1)4 = (3!)/(s  1)4 is the Laplace transform of et tn with = 1 and n = 3. Therefore, L1 402 6 (s  1)4 (t) = et t3 . Exercises 7.4
3. Writing s2 s+1 s+1 s+1 , = 2 = + 2s + 10 (s + 2s + 1) + 9 (s + 1)2 + 32 we see that this function is the Laplace transform of et cos 3t (the last entry in Table 7.1 with = 1 and b = 3). Hence L1 s2 s+1 + 2s + 10 (t) = et cos 3t . 5. We complete the square in the denominator and use the linearity of the inverse Laplace transform to get L1 s2 1 + 4s + 8 (t) = L1 1 (s + 2)2 + 22 1 (t) = L1 2 2 (s + 2)2 + 22 (t) = 1 2t e sin 2t. 2 (See the Laplace transform formula for et sin bt in Table 7.1). 7. By completing the square in the denominator, we can rewrite (2s + 16)/(s2 + 4s + 13) as s2 2s + 16 2(s + 2) 4(3) 2s + 16 = + . = 2 + 32 2 + 32 + 4s + 4 + 9 (s + 2) (s + 2) (s + 2)2 + 32 Thus, by the linearity of the inverse Laplace transform, L1 s2 2s + 16 + 4s + 13 (t) = 2L1 s+2 (t) + 4L1 2 + 32 (s + 2) 2t = 2e cos 3t + 4e2t sin 3t . 3 (s + 2)2 + 32 (t) 9. We complete the square in the denominator, rewrite the given function as a sum of two entries in Table 7.1, and use the linearity of the inverse Laplace transform. This yields 3s  15 3 s5 = 2 2s2  4s + 10 2 s  2s + 5 3s  15 L1 = 2  4s + 10 2s (3/2)(s  1) 3(2) 3 (s  1)  4 =  2 (s  1)2 + 22 (s  1)2 + 22 (s  1)2 + 22 s1 2 3 1 L  3L1 2 + 22 2 (s  1) (s  1)2 + 22 3 = et cos 2t  3et sin 2t. 2 = 403 Chapter 7
11. In this problem, we use the partial fractions decomposition method. Since the denominator, (s  1)(s + 2)(s + 5), is a product of three nonrepeated linear factors, the expansion has the form s2  26s  47 A B C = + + (s  1)(s + 2)(s + 5) s1 s+2 s+5 A(s + 2)(s + 5) + B(s  1)(s + 5) + C(s  1)(s + 2) = . (s  1)(s + 2)(s + 5) Therefore, s2  26s  47 = A(s + 2)(s + 5) + B(s  1)(s + 5) + C(s  1)(s + 2). (7.9) Evaluating both sides of (7.9) for s = 1, s = 2, and s = 5, we find constants A, B, and C. s=1: (1)2  26(1)  47 = A(1 + 2)(1 + 5)
2 A = 4, B = 1, C = 6. s = 2 : (2)  26(2)  47 = B(2  1)(2 + 5) s = 5 : (5)2  26(5)  47 = C(5  1)(5 + 2) Hence, 6 1 4 s2  26s  47 =   . (s  1)(s + 2)(s + 5) s+5 s+2 s1 13. The denominator has a simple linear factor, s, and a double linear factor, s + 1. Thus, we look for the decomposition of the form A A(s + 1)2 + Bs(s + 1) + Cs B C 2s2  3s  2 = + = , + s(s + 1)2 s s + 1 (s + 1)2 s(s + 1)2 which yields 2s2  3s  2 = A(s + 1)2 + Bs(s + 1) + Cs. Evaluating this equality for s = 0 and s = 1, we find A and C, respectively. s=0: 2 = A(0 + 1)2 A = 2, C = 1. s = 1 : 2(1)2  3(1)  2 = C(1) To find B, we compare the coefficients at s2 in both sides of (7.10). 2 = A + B 404 B = 2  A = 0. (7.10) Exercises 7.4
Hence, 2s2  3s  2 1 2 =  . s(s + 1)2 (s + 1)2 s 15. First, we complete the square in the quadratic s2  2s + 5 to make sure that this polynomial is irreducible and to find the form of the decomposition. Since s2  2s + 5 = (s2  2s + 1) + 4 = (s  1)2 + 22 , we have 8s  2s2  14 A B(s  1) + C(2) A [(s  1)2 + 4] + [B(s  1) + 2C] (s + 1) = + = (s + 1)(s2  2s + 5) s + 1 (s  1)2 + 22 (s + 1) [(s  1)2 + 4] which implies that 8s  2s2  14 = A (s  1)2 + 4 + [B(s  1) + 2C] (s + 1). Taking s = 1, s = 1, and s = 0, we find A, B, and C, respectively. s = 1 : 8(1)  2(1)2  14 = A [(1  1)2 + 4] s=1: s=0: and so 8(1)  2(1)  14 = A [(1  1) + 4] + 2C(1 + 1)
2 2 2 2 A = 3, C = 1, B = 1, 8(0)  2(0)  14 = A [(0  1) + 4] + [B(0  1) + 2C] (0 + 1) 3 (s  1) + 2 8s  2s2  14 = + 2  2s + 5) (s + 1)(s s + 1 (s  1)2 + 4 17. First we need to completely factor the denominator. Since s2 + s  6 = (s  2)(s + 3), we have s(s2 3s + 5 3s + 5 = . + s  6) s(s  2)(s + 3) Since the denominator has only nonrepeated linear factors, we can write 3s + 5 A B C = + + s(s  2)(s + 3) s s2 s+3 for some choice of A, B and C. Clearing fractions gives us 3s + 5 = A(s  2)(s + 3) + Bs(s + 3) + Cs(s  2). 405 Chapter 7
With s = 0, this yields 5 = A(2)(3) so that A = 5/6. With s = 2, we get 11 = B(2)(5) so that B = 11/10. Finally, s = 3 yields 4 = C(3)(5) so that C = 4/15. Thus, s(s2 5 11 4 3s + 5 = +  . + s  6) 6s 10(s  2) 15(s + 3) 19. First observe that the quadratic polynomial s2 + 2s + 2 is irreducible because the discriminant 22  4(1)(2) = 4 is negative. Since the denominator has one nonrepeated linear factor and one nonrepeated quadratic factor, we can write (s  3)(s2 1 1 A B(s + 1) + C = = + , 2 + 1] + 2s + 2) (s  3)[(s + 1) s3 (s + 1)2 + 1 where we have chosen a form which is more convenient for taking the inverse Laplace transform. Clearing fractions gives us 1 = A (s + 1)2 + 1 + [B(s + 1) + C] (s  3). (7.11) With s = 3, this yields 1 = 17A so that A = 1/17. Substituting s = 1, we see that 1 = A(1) + C(4), or C = (A1)/4 = 4/17. Finally, the coefficient A+ B at s2 in the righthand side of (7.11) must be the same as in the lefthand side, that is, 0. So B = A = 1/17 and 1 s+1 4 1 1 =   . (s  3)(s2 + 2s + 2) 17 s  3 (s + 1)2 + 1 (s + 1)2 + 1 21. Since the denominator contains only nonrepeated linear factors, the partial fractions decomposition has the form A B C A(s  1)(s  6) + Bs(s  6) + Cs(s  1) 6s2  13s + 2 = + + = . s(s  1)(s  6) s s1 s6 s(s  1)(s  6) Therefore, 6s2  13s + 2 = A(s  1)(s  6) + Bs(s  6) + Cs(s  1). Evaluating both sides of this equation for s = 0, s = 1, and s = 6, we find constants A, B, and C. s = 0 : 2 = 6A s = 6 : 140 = 30C 406 A = 1/3, B = 1, C = 14/3. s = 1 : 5 = 5B Exercises 7.4
Hence, 1/3 1 14/3 6s2  13s + 2 = + + s(s  1)(s  6) s s1 s6 6s2  13s + 2 s(s  1)(s  6) and the linear property of the inverse Laplace transform yields L1 1 = L1 3 1 s + L1 1 s1 + 14 1 L 3 1 s6 = 14 6t 1 + et + e . 3 3 23. In this problem, the denominator of F (s) has a simple linear factor, s + 1, and a double linear factor, s + 3. Thus, the decomposition is the form A C A(s + 1) + B(s + 1)(s + 3) + C(s + 3)2 5s2 + 34s + 53 B = + = . + (s + 3)2 (s + 1) (s + 3)2 s + 3 s + 1 (s + 3)2 (s + 1) Therefore, we must have 5s2 + 34s + 53 = A(s + 1) + B(s + 1)(s + 3) + C(s + 3)2 . Substitutions s = 3 and s = 1 yield values of A and C, respectively. s = 3 : 4 = 2A s = 1 : 24 = 4C To find B, we take, say, s = 0 and get 53 = A + 3B + 9C Hence, L1 5s2 + 34s + 53 (s + 3)2 (s + 1) (t) = 2L1 1 (t)  L1 (s + 3)2 = 2te3t  e3t + 6et . 1 s+3 (t) + 6L1 1 s+1 (t) B= 53  A  9C = 1. 3 A = 2, C = 6. 25. Observing that the quadratic s2 + 2s + 5 = (s + 1)2 + 22 is irreducible, the partial fractions decomposition for F (s) has the form 7s2 + 23s + 30 A B(s + 1) + C(2) = + . 2 + 2s + 5) (s  2)(s s2 (s + 1)2 + 22 407 Chapter 7
Clearing fractions gives us 7s2 + 23s + 30 = A (s + 1)2 + 4 + [B(s + 1) + C(2)] (s  2). With s = 2, this yields 104 = 13A so that A = 8; s = 1 gives 14 = A(4) + C(6), or C = 3. Finally, the coefficient A + B at s2 in the righthand side must match the one in the lefthand side, which is 7. So B = 7  A = 1. Therefore, 8 (s + 1) + 3(2) 7s2 + 23s + 30 = + , 2 + 2s + 5) (s  2)(s s2 (s + 1)2 + 22 which yields L1 7s2 + 23s + 30 1 s+1 2 + 3L1 = 8L1  L1 2 + 2s + 5) 2 + 22 (s  2)(s s2 (s + 1) (s + 1)2 + 22 = 8e2t  et cos 2t + 3et sin 2t . 27. First, we find F (s). F (s) s2  4 = 5 s+1 F (s) = 5 5 = . (s + 1)(s2  4) (s + 1)(s  2)(s + 2) The partial fractions expansion yields A B C 5 = + + . (s + 1)(s  2)(s + 2) s+1 s2 s+2 Clearing fractions gives us 5 = A(s  2)(s + 2) + B(s + 1)(s + 2) + C(s + 1)(s  2). With s = 1, s = 2, and s = 2 this yields A = 5/3, B = 5/12, and C = 5/4. So, 1 5 5 (t) + L1 L1 {F (s)} (t) =  L1 3 s+1 12 5 2t 5 2t 5 e + e . =  et + 3 12 4 29. Solving for F (s) yields F (s) = 408 10s2 + 12s + 14 10s2 + 12s + 14 = . (s + 2)(s2  2s + 2) (s + 2)[(s  1)2 + 1] 1 s2 5 (t) + L1 4 1 s+2 (t) Exercises 7.4
Since, in the denominator, we have nonrepeated linear and quadratic factors, we seek for the decomposition 10s2 + 12s + 14 A B(s  1) + C(1) = + . (s + 2)[(s  1)2 + 1] s+2 (s  1)2 + 1 Clearing fractions, we conclude that 10s2 + 12s + 14 = A[(s  1)2 + 1] + [B(s  1) + C] (s + 2). Substitution s = 2 into this equation yields 30 = 10A or A = 3. With s = 1, we get 36 = A+3C and so C = (36A)/3 = 11. Finally, substitution s = 0 results 14 = 2A+2(CB) or B = A + C  7 = 7. Now we apply the linearity of the inverse Laplace transform and obtain L1 {F (s)} (t) = 3L1 1 s1 (t) + 7L1 s+2 (s  1)2 + 1 = 3e2t + 7et cos t + 11et sin t . (t) + 11L1 1 (s  1)2 + 1 (t) 31. Functions f1 (t), f2 (t), and f3 (t) coincide for all t in [0, ) except a finite number of points. Since the Laplace transform a function is a definite integral, it does not depend on values of the function at finite number of points. Therefore, in (a), (b), and (c) we have one and the same answer, that is L {f1 (t)} (s) = L {f2 (t)} (s) = L {f3 (t)} (s) = L {t} (s) = 1 . s2 By Definition 4, the inverse Laplace transform is a continuous function on [0, ). f3 (t) = t clearly satisfies this condition while f1 (t) and f2 (t) have removable discontinuities at t = 2 and t = 1, 6, respectively. Therefore, L1 1 s2 (t) = f3 (t) = t. 33. We are looking for L1 {F (s)} (t) = f (t). According to the formula given just before this problem, f (t) = 1 1 L t dF ds (t) 409 Chapter 7
(take n = 1 in the formula). Since F (s) = ln we have dF (s) d 1 1 = (ln(s + 2)  ln(s  5)) =  ds ds s+2 s5 dF 1 1 (t) = L1  (t) = e2t  e5t L1 ds s+2 s5 e5t  e2t 1 2t e  e5t = L1 {F (s)} (t) = . t t 35. Taking the derivative of F (s), we get d s2 + 9 d 2s 2s dF (s) = ln 2 = ln(s2 + 9)  ln(s2 + 1) = 2  2 . ds ds s +1 ds s +9 s +1 So, using the linear property of the inverse Laplace transform, we obtain L1 Thus L1 {F (s)} (t) = dF (s) ds (t) = 2L1 s2 s +9 (t)  2L1 s2 s +1 (t) = 2(cos 3t  cos t). s+2 s5 = ln(s + 2)  ln(s  5), 1 1 L t dF (s) ds (t) = 2(cos t  cos 3t) . t 37. By the definition, both, L1 {F1 } (t) and L1 {F2 } (t), are continuous functions on [0, ). Therefore, their sum, (L1 {F1 } + L1 {F2 }) (t), is also continuous on [0, ). Furthermore, the linearity of the Laplace transform yields L L1 {F1 } + L1 {F2 } (s) = L L1 {F1 } (s) + L L1 {F2 } (s) = F1 (s) + F2 (s). Therefore, L1 {F1 } + L1 {F2 } is a continuous function on [0, ) whose Laplace transform is F1 + F2 . By the definition of the inverse Laplace transform, this function is the inverse Laplace transform of F1 + F2 , that is, L1 {F1 } (t) + L1 {F2 } (t) = L1 {F1 + F2 } (t), 410 Exercises 7.4
and (3) in Theorem 7 is proved. To show (4), we use the continuity of L1 {F } to conclude that cL1 {F } is a continuous function. Since the linearity of the Laplace transform yields L cL1 {F } (s) = cL L1 {F } (s) = cF (s), we have cL1 {F } (t) = L1 {cF } (t). 39. In this problem, the denominator Q(s) := s(s  1)(s + 2) has only nonrepeated linear factors, and so the partial fractions decomposition has the form F (s) := 2s + 1 A B C = + + . s(s  1)(s + 2) s s1 s+2 To find A, B, and C, we use the residue formula in Problem 38. This yields A = lim sF (s) = lim 2(0) + 1 1 2s + 1 = = , s0 s0 (s  1)(s + 2) (0  1)(0 + 2) 2 2(1) + 1 2s + 1 = = 1, B = lim(s  1)F (s) = lim s1 s1 s(s + 2) (1)(1 + 2) 2(2) + 1 1 2s + 1 = = . C = lim (s + 2)F (s) = lim s2 s2 s(s  1) (2)(2  1) 2 2s + 1 1/2 1 1/2 = +  . s(s  1)(s + 2) s s1 s+2 41. In notation of Problem 40, P (s) = 3s2  16s + 5, Q(s) = (s + 1)(s  3)(s  2). Therefore, We can apply the Heaviside's expansion formula because Q(s) has only nonrepeated linear factors. We need the values of P (s) and Q (s) at the points r1 = 1, r2 = 3, and r3 = 2. Using the product rule, we find that Q (s) = (s  3)(s  2) + (s + 1)(s  2) + (s + 1)(s  3), and so Q (1) = (1  3)(1  2) = 12, Q (3) = (3 + 1)(3  2) = 4, Q (2) = (2 + 1)(2  3) = 3. 411 Chapter 7
Also, we compute P (1) = 24, Therefore, L1 3s2  16s + 5 (s + 1)(s  3)(s  2) (t) = P (1) (1)t P (3) (3)t P (2) (2)t e e + e = 2et 4e3t + 5e2t . + Q (1) Q (3) Q (2) P (3) = 16, P (2) = 15. 43. Since s2  2s + 5 = (s  1)2 + 22 , we see that the denominator of F (s) has nonrepeated linear factor s + 2 and nonrepeated irreducible quadratic factor s2  2s + 5 with = 1 and = 2 (in notation of Problem 40). Thus the partial fractions decomposition has the form F (s) = A(s  1) + 2B 6s2 + 28 C = . + 2  2s + 5)(s + 2) 2 + 22 (s (s  1) s+2 We find C by applying the real residue formula derived in Problem 38. C = lim 52 (s + 2)(6s2 + 28) 6s2 + 28 = lim 2 = = 4. 2  2s + 5)(s + 2) s2 (s s2 s  2s + 5 13 Next, we use the complex residue formula given in Problem 42, to find A and B. Since = 1 and = 2, the formula becomes 2B + i2A = lim Dividing we get 2B + i2A = 78 + 52i (10 + 24i)(3  2i) = = 6 + 4i. (3 + 2i)(3  2i) 13 (s2  2s + 5)(6s2 + 28) 6s2 + 28 6(1 + 2i)2 + 28 10 + 24i = lim = = . s1+2i (s2  2s + 5)(s + 2) s1+2i s + 2 (1 + 2i) + 2 3 + 2i Taking the real and imaginary parts yields 2B = 6, 2A = 4 Therefore, 2(s  1) + 2(3) 4 6s2 + 28 = . + 2  2s + 5)(s + 2) 2 + 22 (s (s  1) s+2 412 B = 3, A = 2. Exercises 7.5
EXERCISES 7.5: Solving Initial Value Problems, page 383 1. Let Y (s) := L {y} (s). Taking the Laplace transform of both sides of the given differential equation and using its linearity, we obtain L {y } (s)  2L {y } (s) + 5Y (s) = L {0} (s) = 0. (7.12) We can express L {y } (s) and L {y } (s) in terms of Y (s) using the initial conditions and Theorem 5 in Section 7.3. L {y } (s) = sY (s)  y(0) = sY (s)  2, L {y } (s) = s2 Y (s)  sy(0)  y (0) = s2 Y (s)  2s  4. Substituting back into (7.12) and solving for Y (s) yield s2 Y (s)  2s  4  2 [sY (s)  2] + 5Y (s) = 0 Y (s) s2  2s + 5 = 2s 2s 2(s  1) 2 2s = = + . Y (s) = 2 2 + 22 2 + 22 s  2s + 5 (s  1) (s  1) (s  1)2 + 22 s1 (s  1)2 + 22 2 (s  1)2 + 22 Applying now the inverse Laplace transform to both sides, we obtain y(t) = 2L1 (t) + L1 (t) = 2et cos 2t + et sin 2t. 3. Let Y (s) := L {y} (s). Taking the Laplace transform of both sides of the given differential equation, y + 6y + 9y = 0, and using the linearity of the Laplace transform, we obtain L {y } (s) + 6L {y } (s) + 9Y (s) = 0. We use formula (4), page 362, to express L {y } (s) and L {y } (s) in terms of Y (s). L {y } (s) = sY (s)  y(0) = sY (s) + 1, L {y } (s) = s2 Y (s)  sy(0)  y (0) = s2 Y (s) + s  6. Therefore, s2 Y (s) + s  6 + 6 [sY (s) + 1] + 9Y (s) = 0 413 Chapter 7 Y (s) s2 + 6s + 9 = s s 3 1 s = , =  Y (s) = 2 s + 6s + 9 (s + 3)2 (s + 3)2 s + 3 where the last equality comes from the partial fraction expansion of s/(s + 32 ). We apply the inverse Laplace transform to both sides and use Table 7.1 to obtain y(t) = 3L1 1 (s + 3)2 (t)  L1 1 s+3 (t) = 3te3t  e3t . 5. Let W (s) = L {w} (s). Then taking the Laplace transform of the equation and using linearity yield L {w } (s) + W (s) = L t2 + 2 (s) = L t2 (s) + 2L {1} (s) = Since L {w } (s) = s2 W (s)  sw(0)  w (0) = s2 W (s)  s + 1, we have s2 W (s)  s + 1 + W (s) = 2 2 + 3 s s 2(s2 + 1) 2 s + 1 W (s) = s  1 + s3 2 2 + . s3 s W (s) = s2 s 1 2  2 + 3. +1 s +1 s Now, taking the inverse Laplace transform, we obtain w = L1 s2 s +1  L1 s2 1 +1 + L1 2 s3 = cos t  sin t + t2 . 7. Let Y (s) := L {y} (s). Using the initial conditions and Theorem 5 in Section 7.3 we can express L {y } (s) and L {y } (s) in terms of Y (s), namely, L {y } (s) = sY (s)  y(0) = sY (s)  5, L {y } (s) = s2 Y (s)  sy(0)  y (0) = s2 Y (s)  5s + 4. Taking the Laplace transform of both sides of the given differential equation and using its linearity, we obtain L {y  7y + 10y} (s) = L {9 cos t + 7 sin t} (s) 414 s2 Y (s)  5s + 4  7 [sY (s)  5] + 10Y (s) = s2 7 9s + 2 +1 s +1 Exercises 7.5 5s3  39s2 + 14s  32 9s + 7 + 5s  39 = s2 + 1 s2 + 1 3 2 9s + 7 5s  39s + 14s  32 5s3  39s2 + 14s  32 Y (s) = 2 + 5s  39 = 2 = 2 . s +1 (s + 1)(s2  7s + 10) (s + 1)(s  5)(s  2) s2  7s + 10 Y (s) = The partial fractions decomposition of Y (s) has the form As + B C D 5s3  39s2 + 14s  32 = 2 + + . 2 + 1)(s  5)(s  2) (s s +1 s5 s2 Clearing fractions yields 5s3  39s2 + 14s  32 = (As + B)(s  5)(s  2) + C(s2 + 1)(s  2) + D(s2 + 1)(s  5). We substitute s = 5 and s = 2 to find C and D, resprectively, and then s = 0 to find B. s = 5 : 312 = 78C s = 2 : 120 = 15D s = 0 : 32 = 10B  2C  5D C = 4, D = 8, B = 0. Equating the coefficients at s3 , we also get A + C + D = 5, which implies that A = 1. Thus Y (s) = s2 s 4 8  + +1 s5 s2 y(t) = L1 {Y (s)} (t) = cos t  4e5t + 8e2t . 9. First, note that the initial conditions are given at t = 1. Thus, to use the method of Laplace transform, we make a shift in t and move the initial conditions to t = 0. z (t) + 5z (t)  6z(t) = 21et1 z (t + 1) + 5z (t + 1)  6z(t + 1) = 21e(t+1)1 = 21et . (7.13) Now, let y(t) := z(t + 1). Then the chain rule yields y (t) = z (t + 1)(t + 1) = z (t + 1), y (t) = [y (t)] = z (t + 1)(t + 1) = z (t + 1), and (7.13) becomes y (t) + 5y (t)  6y(t) = 21et (7.14) 415 Chapter 7
with initial conditions y(0) = z(0 + 1) = z(1) = 1, y (0) = z (0 + 1) = z (1) = 9. With Y (s) := L {y(t)} (s), we apply the Laplace transform to both sides of (7.14) and obtain L {y } (s) + 5L {y } (s)  6Y (s) = L 21et (s) = By Theorem 5, Section 7.3, L {y } (s) = sY (s)  y(0) = sY (s) + 1, L {y } (s) = s2 Y (s)  sy(0)  y (0) = s2 Y (s) + s  9. Substituting these expressions back into (7.15) and solving for Y (s) yield s2 Y (s) + s  9 + 5 [sY (s) + 1]  6Y (s) = 21 s1 s2 + 5s + 17 21 s+4 = s2 + 5s  6 Y (s) = s1 s1 s2 + 5s + 17 s2 + 5s + 17 s2 + 5s + 17 = = . Y (s) = (s  1)(s2 + 5s  6) (s  1)(s  1)(s + 6) (s  1)2 (s + 6) 21 . s1 (7.15) The partial fractions decomposition for Y (s) has the form A C B s2 + 5s + 17 = + . + (s  1)2 (s + 6) (s  1)2 s  1 s + 6 Clearing fractions yields s2 + 5s + 17 = A(s + 6) + B(s  1)(s + 6) + C(s  1)2 . Substitutions s = 1 and s = 6 give A = 3 and C = 1. Also, with s = 0, we have 17 = 6A  6B + C or B = 0. Therefore, Y (s) = 3 1  2 (s  1) s+6 y(t) = L1 3 1  2 (s  1) s+6 (t) = 3tet  e6t . Finally, shifting the argument back, we obtain z(t) = y(t  1) = 3(t  1)et1  e6(t1) . 416 Exercises 7.5
11. As in the previous problem (and in Example 3 in the text), we first need to shift the initial conditions to 0. If we set v(t) = y(t + 2), the initial value problem for v(t) becomes v (t)  v(t) = (t + 2)  2 = t, v(0) = y(2) = 3, v (0) = y (2) = 0. Taking the Laplace transform of both sides of this new differential equation gives us L {v } (s)  L {v} (s) = L {t} (s) = 1 . s2 If we denote V (s) := L {v} (s) and express L {v } (s) in terms of V (s) using (4) in Section 4.3 (with n = 2), that is, L {v } (s) = s2 V (s)  3s, we obtain s2 V (s)  3s  V (s) = Hence, v(t) = L1 {V (s)} (t) = L1  1 2 1 + + 2 s s+1 s1 (t) = t + et + 2et . V (s) = 1 s2 3s3 + 1 1 2 3s3 + 1 1 = 2 = 2 + + . s2 (s2  1) s (s + 1)(s  1) s s+1 s1 Since v(t) = y(t + 2), we have y(t) = v(t  2) and so y(t) = (t  2) + e(t2) + 2et2 = 2  t + e2t + 2et2 . 13. To shift the initial conditions to t = 0, we make the substitution x(t) := y(t + /2) in the original equation and use the fact that x (t) := y (t + /2), This yields y (t)  y (t)  2y(t) = 8 cos t  2 sin t 8 cos t +  2 sin t + = 8 cos t +  2 sin t + 2 2 2 2 x(0) = 1, x (0) = 0. x (t)  x (t)  2x(t) = 8 sin t  2 cos t, x (t) := y (t + /2). = 8 sin t  2 cos t 417 Chapter 7
Taking the Laplace transform of both sides in this last differential equation and using the fact that, with X(s) := L {x} (s), L {x } (s) = sX(s)  1 and L {x } (s) = s2 X(s)  s (which comes from the initial conditions and (4) in Section 7.3), we obtain s2 X(s)  s  [sX(s)  1]  2X(s) = L {8 sin t  2 cos t} (s) = 8  2s s3  s2  s + 7 +s1= s2 + 1 s2 + 1 3 2 3 2 s s s+7 s s s+7 = 2 . X(s) = 2 (s + 1)(s2  s  2) (s + 1)(s  2)(s + 1) s2  s  2 X(s) = s2 8 2s  2 +1 s +1 We seek for the partial fractions decomposition of X(s) in the form s 3  s2  s + 7 As + B C D = 2 + + . 2 + 1)(s  2)(s + 1) (s s +1 (s  2) s + 1 Solving yields A= Therefore, X(s) = (3/5) 1 (7/5)s (11/5) + 2 +  2+1 s s +1 (s  2) s + 1 11 3 7 sin t + e2t  et . x(t) = L1 {X(s)} (t) = cos t  5 5 5 7 , 5 B= 11 , 5 C= 3 , 5 D = 1. Finally, since y(t) = x(t  /2), we obtain the solution y(t) = 11 3 7 cos t   sin t  + e2(t/2)  e(t/2) 5 2 5 2 5 11 3 2t 7 (/2)t) sin t + cos t + e = e 5 5 5 15. Taking the Laplace transform of y  3y + 2y = cos t and applying the linearity of the Laplace transform yields L {y } (s)  3L {y } (s) + 2L {y} (s) = L {cos t} (s) = 418 s2 s . +1 (7.16) Exercises 7.5
If we put Y (s) = L {y} (s) and apply the property (4), page 362 of the text, we get L {y } (s) = sY (s), Substitution back into (7.16) yields s2 Y (s) + 1  3 [sY (s)] + 2Y (s) = s2  3s + 2 Y (s) = s2 s +1 L {y } (s) = s2 Y (s) + 1. s s2 + s  1 1 = s2 + 1 s2 + 1 s2 + s  1 s2 + s  1 Y (s) = 2 = 2 . (s + 1)(s2  3s + 2) (s + 1)(s  1)(s  2) 17. With Y (s) := L {y} (s), we find that L {y } (s) = sY (s)  y(0) = sY (s)  1, L {y } (s) = s2 Y (s)  sy(0)  y (0) = s2 Y (s)  s, and so the Laplace transform of both sides of the original equation yields L {y + y  y} (s) = L t3 (s) s2 Y (s)  s + [sY (s)  1]  Y (s) = Y (s) = s2 1 +s1 6 +s+1 s4 6 s4 s5 + s4 + 6 . = 4 2 s (s + s  1) 19. Let us denote Y (s) := L {y} (s). From the initial conditions and formula (4) on page 362 of the text we get L {y } (s) = sY (s)  y(0) = sY (s)  1, L {y } (s) = s2 Y (s)  sy(0)  y (0) = s2 Y (s)  s  1. The Laplace transform, applied to both sides of the given equation, yields s2 Y (s)  s  1 + 5 [sY (s)  1]  Y (s) = L et (s)  L {1} (s) = s3 + 5s2  6s + 1 1 +s+6= s(s  1) s(s  1) 3 2 s + 5s  6s + 1 Y (s) = . s(s  1)(s2 + 5s  1) s2 + 5s  1 Y (s) = 419 1 1 1  = s1 s s(s  1) Chapter 7
21. Applying the Laplace transform to both sides of the given equation yields L {y } (s)  2L {y } (s) + L {t} (s) = L {cos t} (s)  L {sin t} (s) = s1 . s2 + 1 If L {y} (s) =: Y (s), then it follows from the initial conditions and (4) on page 362 of the text that L {y } (s) = sY (s)  1, Therefore, Y (s) satisfies s2 Y (s)  s  3  2 [sY (s)  1] + Y (s) = Solving for Y (s) gives us s2  2s + 1 Y (s) = s3 + s2 + 2s s1 +s+1= s2 + 1 s2 + 1 3 2 s3 + s2 + 2s s + s + 2s = 2 Y (s) = 2 . (s + 1)(s2  2s + 1) (s + 1)(s  1)2 s1 . s2 + 1 L {y } (s) = s2 Y (s)  s  3. 23. In this equation, the righthand side is a piecewise defined function. Let us find its Laplace transform first. 2 L {g(t)} (s) =
0 e st g(t) dt =
0 2 e st t dt +
2 est 5 dt
N 2 = test s 2 
0 0 est 5est dt + lim N s s =  5e2s e2s 1 + 3se2s  e2s 2e2s 1  = + 2 + , s s2 s s s2 where we used integration by parts integrating est t. Using this formula and applying the Laplace transform to the given equation yields L {y } (s) + 4L {y} (s) = L {g(t)} (s) 420 s2 L {y} (s) + s + 4L {y} (s) = L {g(t)} (s) s3 + 1 + 3se2s  e2s s2 + 4 L {y} (s) = L {g(t)} (s)  s = s2 Exercises 7.5 L {y} (s) = s3 + 1 + 3se2s  e2s . s2 (s2 + 4) 25. Taking the Laplace transform of y  y + y  y = 0 and applying the linearity of the Laplace transform yields L {y } (s)  L {y } (s) + L {y } (s)  L {y} (s) = L {0} (s) = 0. (7.17) If we denote Y (s) := L {y} (s) and and apply property (4) on page 362 of the text, we get L {y } (s) = sY (s)  1, L {y } (s) = s2 Y (s)  s  1, LT y = s3 Y (s)  s2  s  3. Combining these equations with (7.17) gives us s3 Y (s)  s2  s  3  s2 Y (s)  s  1 + [sY (s)  1]  Y (s) = 0 s3  s2 + s  1 Y (s) = s2 + 3 s2 + 3 s2 + 3 Y (s) = 3 = . s  s2 + s  1 (s  1)(s2 + 1) Expanding Y (s) by partial fractions results Y (s) = s+1 2 s 1 2  2 =  2  2 . s1 s +1 s1 s +1 s +1 From Table 7.1 on page 358 of the text, we see that y(t) = L1 {Y (s)} (t) = 2et  cos t  sin t. 27. Let Y (s) := L {y} (s). Then, by Theorem 5 in Section 7.3, L {y } (s) = sY (s)  y(0) = sY (s) + 4, L {y } (s) = s2 Y (s)  sy(0)  y (0) = s2 Y (s) + 4s  4, L {y } (s) = s3 Y (s)  s2 y(0)  sy (0)  y (0) = s3 Y (s) + 4s2  4s + 2. Using these equations and applying the Laplace transform to both sides of the given differential equation, we get s3 Y (s) + 4s2  4s + 2 + 3 s2 Y (s) + 4s  4 + 3 [sY (s) + 4] + Y (s) = 0 421 Chapter 7 s3 + 3s2 + 3s + 1 Y (s) + 4s2 + 8s + 2 = 0 4s2 + 8s + 2 4s2 + 8s + 2 . Y (s) =  3 = s + 3s2 + 3s + 1 (s + 1)3 Therefore, the partial fractions decomposition of Y (s) has the form  A + B(s + 1) + C(s + 1)2 4s2 + 8s + 2 A B C = = + + (s + 1)3 (s + 1)3 (s + 1)2 s + 1 (s + 1)3 (4s2 + 8s + 2) = A + B(s + 1) + C(s + 1)2 . Substitution s = 1 yields A = 2. Equating coefficients at s2 , we get C = 4. At last, substituting s = 0 we obtain 2 = A + B + C Therefore, Y (s) = 4 2 + 3 (s + 1) s+1 y(t) = L1 {Y } (t) = t2 et  4et = t2  4 et . B = 2  A  C = 0. 29. Using the initial conditions, y(0) = a and y (0) = b, and formula (4) on page 362 of the text, we conclude that L {y } (s) = sY (s)  y(0) = sY (s)  a, L {y } (s) = s2 Y (s)  sy(0)  y (0) = s2 Y (s)  as  b, where Y (s) = L {y} (s). Applying the Laplace transform to the original equation yields s2 Y (s)  as  b  4 [sY (s)  a] + 3Y (s) = L {0} (s) = 0 s2  4s + 3 Y (s) = as + b  4a as + b  4a A B as + b  4a = = + . Y (s) = 2 s  4s + 3 (s  1)(s  3) s1 s3 Solving for A and B, we find that A = (3a  b)/2, B = (b  a)/2. Hence Y (s) = (3a  b)/2 (b  a)/2 + s1 s3 1 3a  b 1 b  a 1 L L y(t) = L1 {Y } (t) = (t) + 2 s1 2 3a  b t b  a 3t e + e . = 2 2 1 s3 (t) 422 Exercises 7.5
31. Similarly to Problem 29, we have L {y } (s) = sY (s)  y(0) = sY (s)  a, L {y } (s) = s2 Y (s)  sy(0)  y (0) = s2 Y (s)  as  b, with Y (s) := L {y} (s). Thus the Laplace transform of both sides of the the given equation yields L {y + 2y + 2y} (s) = L {5} (s) 5 s 2 as + (2a + b)s + 5 5 s2 + 2s + 2 Y (s) = + as + 2a + b = s s as2 + (2a + b)s + 5 as2 + (2a + b)s + 5 Y (s) = = . s(s2 + 2s + 2) s[(s + 1)2 + 1] s2 Y (s)  as  b + 2 [sY (s)  a] + 2Y (s) = We seek for an expansion of Y (s) of the form A B(s + 1) + C as2 + (2a + b)s + 5 = + . 2 + 1] s[(s + 1) s (s + 1)2 + 1 Clearing fractions, we obtain as2 + (2a + b)s + 5 = A (s + 1)2 + 1 + [B(s + 1) + C] s . Substitutions s = 0 and s = 1 give us s=0: 5 = 2A A = 5/2, C = A + a + b  5 = a + b  5/2. s = 1 : 5  a  b = A  C To find B, we can compare coefficients at s2 : a=A+B So, Y (s) = B = a  A = a  5/2. 5/2 (a  5/2)(s + 1) a + b  5/2 + + s (s + 1)2 + 1 (s + 1)2 + 1 5 t 5 5 y(t) = L1 {Y } (t) = + a  e cos t + a + b  2 2 2 et sin t . 423 Chapter 7
33. By Theorem 6 in Section 7.3, L t2 y (t) (s) = (1)2 d2 d2 [L {y (t)} (s)] = 2 [L {y (t)} (s)] . ds2 ds (7.18) On the other hand, equation (4) on page 362 says that L {y (t)} (s) = sY (s)  y(0), Substitution back into (7.18) yields L t2 y (t) (s) = d d2 d [sY (s)  y(0)] [sY (s)  y(0)] = 2 ds ds ds d = [sY (s) + Y (s)] = (sY (s) + Y (s)) + Y (s) = sY (s) + 2Y (s). ds Y (s) := L {y} (s). 35. Taking the Laplace transform of y + 3ty  6y = 1 and applying the linearity of the Laplace transform yields L {y } (s) + 3L {ty } (s)  6L {y} (s) = L {1} (s) = 1 . s (7.19) If we put Y (s) = L {y} (s) and apply property (4) on page 362 of the text with n = 2, we get L {y } (s) = s2 Y (s)  sy(0)  y (0) = s2 Y (s). Furthermore, as it was shown in Example 4, Section 4.5, L {ty } (s) = sY (s)  Y (s). Substitution (7.20) and (7.21) back into (7.19) yields s2 Y (s) + 3 [sY (s)  Y (s)]  6Y (s) = 3sY (s) + s2  9 Y (s) = Y (s) + 3 s  s 3 1 s 1 Y (s) =  2 . 3s 1 s (7.21) (7.20) This is a first order linear differential equation in Y (s), which can be solved by the techniques of Section 2.3. Namely, it has the integrating factor (s) = exp 424 3 s  s 3 ds = exp 3 ln s  s2 2 = s3 es /6 . 6 Exercises 7.5
Thus Y (s) = s s2 /6 1 1 1 (s)  2 ds = 3 s2 /6 e ds (s) 3s se 3 1 1 2 2 = 3 s2 /6 es /6 + C = 3 1 + Ces /6 . s se Just as in Example 4 on page 380 of the text, C must be zero in order to ensure that Y (s) 0 as s . Thus Y (s) = 1/s3 , and from Table 7.1 on page 358 of the text we get y(t) = L1 1 s3 1 (t) = L1 2 2 s3 (t) = t2 . 2 37. We apply the Laplace transform to the given equation and obtain L {ty } (s)  2L {y } (s) + L {ty} (s) = 0. (7.22) Using Theorem 5 (Section 7.3) and the initial conditions, we express L {y } (s) and L {y } (s) in terms of Y (s) := L {y} (s). L {y } (s) = sY (s)  y(0) = sY (s)  1, L {y } (s) = s2 Y (s)  sy(0)  y (0) = s2 Y (s)  s. We now involve Theorem 6 in Section 7.3 to get L {ty} (s) =  d [L {y} (s)] = Y (s). ds (7.25) (7.23) (7.24) Also, Theorem 6 and equation (7.24) yield L {ty } (s) =  d d 2 [L {y } (s)] =  s Y (s)  s = 1  2sY (s)  s2 Y (s). ds ds (7.26) Substituting (7.23), (7.25), and (7.26) into (7.22), we obtain 1  2sY (s)  s2 Y (s)  2 [sY (s)  1] + [Y (s)] = 0  s2 + 1 Y (s)  4sY (s) + 3 = 0 3 4s Y (s) = 2 . Y (s) + 2 s +1 s +1 425 Chapter 7
The integrating factor of this first order linear differential equation is (s) = exp Hence Y (s) = 3 1 1 3 s2 + 1 ds (s) 2 ds = 2 (s) s +1 (s + 1)2 (s3 + s) + (2s + C) s C 1 2s s3 + 3s + C = = 2 + 2 , = + 2 2 + 1)2 2 + 1)2 2 (s (s s + 1 (s + 1) (s + 1)2 s +1 2s + 1)2 C 1 L 2 2 + 1)2 s2 4s ds = exp 2 ln s2 + 1 +1 = s2 + 1
2 . where C is an arbitrary constant. Therefore, y(t) = L1 {Y } (t) = L1 s2 (t) + L1 (s2 (t) + (s2 (t) . Using formulas (24), (29) and (30) on the inside back cover of the text, we finally get y(t) = cos t + t sin t + c(sin t  t cos t), where c := C/2 is an arbitrary constant. 39. Similarly to Example 5, we have the initial value problem (18), namely, Iy (t) = ke(t), y(0) = 0, y (0) = 0, for the model of the mechanism. This equation leads to equation (19) for the Laplace transforms Y (s) := L {y(t)} (s) and E(s) := L {e(t)} (s): s2 IY (s) = kE(s). But, this time, e(t) = y(t)  a and so E(s) = L {y(t)  a} (s) = Y (s)  Substituting this relation into (7.27) yields s2 IE(s) + aIs = kE(s) E(s) =  as aIs = 2 . 2I + k s s + (k/I) a s Y (s) = E(s) + a . s (7.27) Taking the inverse Laplace transform, we obtain e(t) = L1 {E(s)} (t) = aL1 s s2 +( k/I)2 (t) = a cos k/It . 426 Exercises 7.5
41. As in Problem 40, the differential equation modeling the automatic pilot is Iy (t) = ke(t)  e (t) , but now the error e(t) is given by e(t) = y(t)  at. Let Y (s) := L {y(t)} (s), E(s) := L {e(t)} (s). Notice that, as in Example 5 on page 382, we have y(0) = y (0) = 0, and so e(0) = 0. Using these initial conditions and Theorem 5 in Section 7.3, we obtain L {y (t)} (s) = s2 Y (s) and L {e (t)} (s) = sE(s). (7.28) Applying the Laplace transform to both sides of (7.28) we then conclude that IL {y (t)} (s) = kL {e(t)} (s)  L {e (t)} (s) Since e(t) = y(t)  at, E(s) = L {e(t)} (s) = L {y(t)  at} (s) = Y (s)  aL {t} (s) = Y (s)  or Y (s) = E(s) + a/s2 . Substitution back into (7.29) yields Is2 E(s) + a = (k + s)E(s) s2 Is2 + s + k E(s) = aI a aI = 2 . E(s) = 2 Is + s + k s + (/I)s + (k/I) a s2 Is2 Y (s) = kE(s)  sE(s) = (k + s)E(s). (7.29) Completing the square in the denominator, we write E(s) in the form suitable for inverse Laplace transform. E(s) = = /(2I)]2 a + (k/I)  2 /(4I 2 ) 4kI  2 /(2I) . 4kI  2 [s + /(2I)]2 + (4kI  2 )/(4I 2) 2Ia 427 [s + a = 2 + (4kI  2 )/(4I 2 ) [s + /(2I)] Chapter 7
Thus, using Table 7.1 on page 358 of the text, we find that e(t) = L1 {E(s)} (t) = 2Ia 4kI  2 et/(2I) sin 4kI  2 t . 2I Compare this with Example 5 of the text and observe, how for moderate damping with < 2 kI, the oscillations of Example 5 die out exponentially. EXERCISES 7.6: Transforms of Discontinuous and Periodic Functions, page 395 1. To find the Laplace transform of g(t) = (t  1)2 u(t  1) we apply formula (5) on page 387 of the text with a = 1 and f (t) = t2 . This yields L (t  1)2 u(t  1) (s) = es L t2 (s) = The graph of g(t) = (t  1)2 u(t  1) is shown in Figure 7B(a). 3. The graph of the function y = t2 u(t  2) is shown in Figure 7B(b). For this function, formula (8) on page 387 is more convenient. To apply the shifting property, we observe that g(t) = t2 and a = 2. Hence g(t + a) = g(t + 2) = (t + 2)2 = t2 + 4t + 4. Now the Laplace transform of g(t + 2) is L t2 + 4t + 4 (s) = L t2 (s) + 4L {t} (s) + 4L {1} (s) = Hence, by formula (8), we have L t2 u(t  2) (s) = e2s L {g(t + 2)} (s) = e2s 4 4 2 + 2+ s3 s s = e2s (4s2 + 4s + 2) . s3 2 4 4 + 2+ . s3 s s 2es . s3 5. The function g(t) equals zero until t reaches 1, at which point g(t) jumps to 2. We can express this jump by (2  0)u(t  1). At t = 2 the function g(t) jumps from the value 2 to the value 1. This can be expressed by adding the term (1  2)u(t  2). Finally, the jump at t = 3 from 1 to 3 can be accomplished by the function (3  1)u(t  3). Hence g(t) = 0 + (2  0)u(t  1) + (1  2)u(t  2) + (3  1)u(t  3) = 2u(t  1)  u(t  2) + 2u(t  3) 428 Exercises 7.6
y=(t1)2 u(t1) y=t2 u(t2) 4 5 2 0 1 2 3 0 2 (a) (b) Figure 7B: Graphs of functions in Problems 1 and 3. and, by the linearity of the Laplace transform, L {g(t)} (s) = 2L {u(t  1)} (s)  L {u(t  2)} (s) + 2L {u(t  3)} (s) es e2s e3s = 2  +2 s s s s 2s 3s + 2e e e . = s 7. Observe from the graph that g(t) is given by 0, t < 1, t, 1 < t < 2, 1, 2 < t. The function g(t) equals zero until t reaches 1, at which point g(t) jumps to the function t. We can express this jump by tu(t  1). At t = 2 the function g(t) jumps from the function t to the value 1. This can be expressed by adding the term (1  t)u(t  2). Hence g(t) = 0 + tu(t  1) + (1  t)u(t  2) = tu(t  1)  (t  1)u(t  2). 429 Chapter 7
Taking the Laplace transform of both sides and using formula (8) on page 387, we find that the Laplace transform of the function g(t) is given by L {g(t)} (s) = L {tu(t  1)} (s)  L {(t  1)u(t  2)} (s) = es L {(t + 1)} (s)  e2s L {(t  1) + 2} (s) = es  e2s L {t + 1} (s) = es  e2s 1 1 + 2 s s = (es  e2s )(s + 1) . s2 9. First, we find the formula for g(t) from the picture given. 0, t < 1, t  1, 1 < t < 2, 3  t, 2 < t < 3, 0, 3 < t. Thus, this function jumps from 0 to t  1 at t = 1, from t  1 to 3  t at t = 2, and from 3  t to 0 at t = 3. Since the function u(t  a) has the unit jump from 0 to 1 at t = a, we can express g(t) as g(t) = [(t  1)  0]u(t  1) + [(3  t)  (t  1)]u(t  2) + [0  (3  t)]u(t  3) = (t  1)u(t  1) + (4  2t)u(t  2) + (t  3)u(t  3). Therefore, L {g(t)} (s) = L {(t  1)u(t  1)} (s) + L {(4  2t)u(t  2)} (s) + L {(t  3)u(t  3)} (s) = es L {(t + 1)  1} (s) + e2s L {4  2(t + 2)} (s) + e3s L {(t + 3)  3} (s) es  2e2s + e3s = es L {t} (s)  2e2s L {t} (s) + e3s L {t} (s) = . s2 11. We use formula (6) on page 387 of the text with a = 2 and F (s) = 1/(s  1). Since f (t) = L1 {F (s)} (t) = L1 we get L 430
1 1 s1 (t) = et f (t  2) = et2 , e2s s1 (t) = f (t  2)u(t  2) = et2 u(t  2). Exercises 7.6
13. Using the linear property of the inverse Laplace transform, we obtain L1 e2s  3e4s s+2 (t) = L1 e2s s+2 (t)  3L1 e4s s+2 (t) . To each term in the above equation, we can apply now formula (6), page 387 of the text with F (s) = 1/(s + 2) and a = 2 and a = 4, respectively. Since f (t) := L1 {F (s)} (t) = L1 {1/(s + 2)} (t) = e2t , we get L1 e2s s+2 (t)  3L1 e4s s+2 (t) = f (t  2)u(t  2)  3f (t  4)u(t  4) = e2(t2) u(t  2)  3e2(t4) u(t  4) . 15. Since F (s) := s s s+2 1 = = 2 2 + 12 2 + 12 + 4s + 5 (s + 2) (s + 2) (s + 2)2 + 12 f (t) := L1 {F (s)} (t) = e2t (cos t  2 sin t) , s2 applying Theorem 8 we get L1 se3s s2 + 4s + 5 (t) = f (t  3)u(t  3) = e2(t3) [cos(t  3)  2 sin(t  3)] u(t  3). 17. By partial fractions, 6 7 s5 = + (s + 1)(s + 2) s+1 s+2 so that L1 e3s (s  5) (s + 1)(s + 2) (t) = 6L1 = 6L1 = e3s s+1 1 s+1 (t) + 7L1 e3s s+2 (t) 1 s+2 (t  3)u(t  3) (t  3)u(t  3) + 7L1 6e(t3) + 7e2(t3) u(t  3) = 7e62t  6e3t u(t  3). 431 Chapter 7
19. In this problem, we apply methods of Section 7.5 of solving initial value problems using the Laplace transform. Taking the Laplace transform of both sides of the given equation and using the linear property of the Laplace transform, we get L {I } (s) + 2L {I } (s) + 2L {I} (s) = L {g(t)} (s). Let us denote I(s) := L {I} (s). By Theorem 5, Section 7.3, L {I } (s) = sI(s)  I(0) = sI(s)  10, L {I } (s) = s2 I(s)  sI(0)  I (0) = s2 I(s)  10s. (7.31) (7.30) To find the Laplace transform of g(t), we express this function using the unit step function u(t). Since g(t) identically equals to 20 for 0 < t < 3, jumps from 20 to 0 at t = 3 and then jumps from 0 to 20 at t = 4, we can write g(t) = 20 + (0  20)u(t  3) + (20  0)u(t  4) = 20  20u(t  3) + 20u(t  4). Therefore, L {g(t)} (s) = L {20  20u(t  3) + 20u(t  4)} (s) = 20L {1  u(t  3) + u(t  4)} (s) = 20 Substituting this equation and (7.31) into (7.30) yields s2 I(s)  10s + 2 [sI(s)  10] + 2I(s) = 20 e3s + e4s 1 . I(s) = 10 + 20 s s[(s + 1)2 + 1] 1 e3s e4s  + s s s (7.32) 1  e3s + e4s . s Since L1 {1/s} (t) = 1 and L1 1 s[(s + 1)2 + 1] (t) = L1 = 1 1 s+1  1s  2+1 2 (s + 1) (s + 1)2 + 1 (t) 1 1  et (cos t + sin t) , 2 applying the inverse Laplace transform to both sides of (7.32) yields e3s e4s 1 + 20 I(t) = L1 10 + 20 s s[(s + 1)2 + 1] s[(s + 1)2 + 1] 432 (t) Exercises 7.6
y=I(t) 10 3 4 8 0 Figure 7C: The graph of the function y = I(t) in Problem 19. = 10  10u(t  3) 1  e(t3) (cos(t  3) + sin(t  3)) +10u(t  4) 1  e(t4) (cos(t  4) + sin(t  4)) = 10  10u(t  3) 1 + e(t3) (cos t + sin t) +10u(t  4) 1  e(t4) (cos t + sin t) . The graph of the solution, y = I(t), 0 < t < 8, is depicted in Figure 7C. 21. In the windowed version (11) of f (t), fT (t) = t and T = 2. Thus 2 FT (s) :=
0 est fT (t) dt =
0 est t dt =  test est  2 s s 2 0 =  2e2s e2s 1 1  2se2s  e2s  2 + 2 = . s s s s2 From Theorem 9 on page 391 of the text, we obtain L {f (t)} (s) = FT (s) 1  2se2s  e2s . = 1  e2s s2 (1  e2s ) The graph of the function y = f (t) is given in Figure B.45 in the answers of the text. 23. We use formula (12) on page 391 of the text. With the period T = 2, the windowed version 433 Chapter 7
fT (t) of f (t) is f (t), 0 < t < 2, 0, otherwise et , 0 < t < 1, 1, 1 < t < 2, 0, otherwise. fT (t) = = Therefore, 1 2 FT (s) =
0 e st fT (t) dt =
1 e st t e dt +
1 (s+1) est dt es  e2s s = and, by (12), e (s + 1) (s+1)t +
0 e s 0 st 2 =
1 1e s+1 + L {f (t)} (s) = 1  e(s+1) es  e2s 1 + . 1  e2s s+1 s The graph of f (t) is shown in Figure B.46 in the answers of the text. 25. Similarly to Example 6 on page 392 of the text, f (t) is a periodic function with period T = 2a, whose windowed version has the form f2a (t) = 1  u(t  a), 0 < t < 2a. Thus, using the linearity of the Laplace transform and formula (4) on page 386 for the Laplace transform of the unit step function, we have F2a (s) = L {f2a (t)} (s) = L {1} (s)  L {u(t  a)} (s) = Applying now Theorem 9 yields L {f (t)} (s) = 1 1  eas 1 1  eas 1 = = . 2as as )(1 + eas ) 1e s (1  e s s(1 + eas ) 1  eas 1 eas  = . s s s 27. Observe that if we let f2a (t) = 434 f (t), 0 < t < 2a, 0, otherwise, Exercises 7.6
denote the windowed version of f (t), then from formula (12) on page 391 of the text we have L {f (t)} (s) = Now f2a (t) = t t t t + 2  u(t  a) + 0  2  a a a a 2(t  a)u(t  a) (t  2a)u(t  2a) t  + . = a a a u(t  2a) L {f2a (t)} (s) L {f2a (t)} (s) . = 1  e2as (1  eas )(1 + eas ) Hence, L {f2a (t)} (s) = 2 1 1 L {t} (s)  L {(t  a)u(t  a)} (s) + L {(t  2a)u(t  2a)} (s) a a a 2 as as 11 (1  eas ) 2e 1e 1 =  + = 2 1  2eas + e2as = a s2 a s2 a s2 as as2 (1  eas ) /(as2 ) 1  eas L {f (t)} (s) = = 2 . (1  eas )(1 + eas ) as (1 + eas ) e3s L {y } (s) + L {y} (s) = L {u(t  3)} (s) = . s Since L {y } (s) = s2 L {y} (s)  sy(0)  y (0) = s2 L {y} (s)  1, substitution yields s2 L {y} (s)  1 + L {y} (s) = e3s s e3s 1 s 1 1 + = 2 + e3s  2 . L {y} (s) = 2 s + 1 s(s2 + 1) s +1 s s +1 1 s  2 s s +1
2 and 29. Applying the Laplace transform to both sides of the given differential equation, we obtain By formula (6) on page 387 of the text, L1 e3s Hence y(t) = L1 s2 1 1 s + e3s  2 +1 s s +1 (t) = sin t + [1  cos(t  3)]u(t  3) 1 s  2 s s +1 (t) = L1 (t  3)u(t  3) = [1  cos(t  3)]u(t  3). The graph of the solution is shown in Figure B.47 in the answers of the text. 435 Chapter 7
31. We apply the Laplace transform to both sides of the differential equation and get L {y } (s) + L {y} (s) = L {t  (t  4)u(t  2)} (s) = 1  L {(t  4)u(t  2)} (s) . s2 (7.33) Since (t  4)u(t  2) = [(t  2)  2]u(t  2), we can use formula (5) from Theorem 8 to find its Laplace transform. With f (t) = t  2 and a = 2, this formula yields L {(t  4)u(t  2)} (s) = e2s L {t  2} (s) = e2s Also, L {y } (s) = s2 L {y} (s)  sy(0)  y (0) = s2 L {y} (s)  1. Substitution back into (7.33) yields s2 L {y} (s)  1 + L {y} (s) = 1  e2s s2 1  2s 1 = L {y} (s) = 2  e2s 2 2 s s (s + 1) 1 2  s2 s 1 2s 1 1 2  2 .  e2s 2  + 2 s2 s s s +1 s +1 1 2 .  2 s s Applying now the inverse Laplace transform and using formula (6) on page 387 of the text, we obtain y(t) = L1 1 1 2s 1 2  2  e2s 2  + 2 (t) 2 s s s s +1 s +1 2s 1 2 1  2  + 2 (t  2)u(t  2) = t  L1 2 s s s +1 s +1 = t  [(t  2)  2 + 2 cos(t  2)  sin(t  2)] u(t  2) = t + [4  t + sin(t  2)  2 cos(t  2)] u(t  2). See Figure B.48 in the answers of the text. 33. By formula (4) on page 386 of the text, L {u(t  2)  u(t  4)} (s) = e2s e4s  . s s Thus, taking the Laplace transform of y + 2y + 2y = u(t  2)  u(t  4) and applying the initial conditions y(0) = y (0) gives us s2 Y (s)  s  1 + 2 [sY (s)  1] + 2Y (s) = 436 e2s  e4s , s Exercises 7.6
where Y (s) is the Laplace transform of y(t). Solving for Y (s) yields Y (s) = s+3 e2s  e4s + s2 + 2s + 2 s(s2 + 2s + 2) s+1 e4s 2(1) e2s =  . (7.34) + + (s + 1)2 + 12 (s + 1)2 + 12 s[(s + 1)2 + 12 ] s[(s + 1)2 + 12 ] Since 1 1 1 (s2 + 2s + 2)  (s2 + 2s) 1 1 s+1  , = =  2 + 12 ] 2 + 12 ] 2 + 12 s[(s + 1) 2 s[(s + 1) 2 s (s + 1) (s + 1)2 + 12 we have L1 1 s[(s + 1)2 + 12 ] (t) = L1 = 1 1 1 s+1   2 + 12 2 s (s + 1) (s + 1)2 + 12 (t) 1 1  et cos t  et sin t 2 and, by formula (6) on page 387 of the text, L1 e2s s[(s + 1)2 + 12 ] 1 2 1 = 2 1 (t) = 2 1 = 2 (t) = 1  e(t2) cos(t  2)  e(t2) sin(t  2) u(t  2) 1  e2t (cos t + sin t) u(t  2) 1  e(t4) cos(t  4)  e(t4) sin(t  4) u(t  4) 1  e4t (cos t + sin t) u(t  4). L1 e4s s[(s + 1)2 + 12 ] Finally, taking the inverse Laplace transform in (7.34) yields y(t) = et cos t + 2et sin t + 1 1  e2t (cos t + sin t) u(t  2) 2 1  1  e4t (cos t + sin t) u(t  4) . 2 35. We take the Laplace transform of the both sides of the given equation and obtain L {z } (s) + 3L {z } (s) + 2L {z} (s) = L e3t u(t  2) (s). (7.35) 437 Chapter 7
We use the initial conditions, z(0) = 2 and z (0) = 3, and formula (4) from Section 7.3 to express L {z } (s) and L {z } (s) in terms of Z(s) := L {z} (s). That is, L {z } (s) = sZ(s)  z(0) = sZ(s)  2, L {z } (s) = s2 Z(s)  sz(0)  z (0) = s2 Z(s)  2s + 3. In the righthand side of (7.35), we can use, say, the translation property of the Laplace transform (Theorem 3, Section 7.3) and the Laplace transform of the unit step function (formula (4), Section 7.6). L e3t u(t  2) (s) = L {u(t  2)} (s + 3) = Therefore, (7.35) becomes s2 Z(s)  2s + 3 + 3 [sZ(s)  2] + 2Z(s) = s2 + 3s + 2 Z(s) = 2s + 3 + Z(s) = e2(s+3) s+3 e2(s+3) . s+3 e2(s+3) s+3 1 2s + 3 + e2s6 s2 + 3s + 2 (s + 3)(s2 + 3s + 2) 1/2 1 1 1/2 1 + + e2s6  + . = s+1 s+2 s+3 s+2 s+1 Hence, z(t) = L1 1 1 1/2 1 1/2 + + e6 e2s  + (t) s+1 s+2 s+3 s+2 s+1 1 1 (t) + L1 (t) = L1 s+1 s+2 1 1 1 e6 L1  2L1 + L1 + 2 s+3 s+2 s+1 6 e e3(t2)  2e2(t2) + e(t2) u(t  2) = et + e2t + 2 1 = et + e2t + e3t  2e2(t+1) + e(t+4) u(t  2) 2 (t  2)u(t  2) 37. Since 2 L {g(t)} (s) =
0 e st g(t) dt =
0 est sin t dt = est (s sin t  cos t) s2 + 1 2 =
0 1  e2s , s2 + 1 438 Exercises 7.6
applying the Laplace transform to the original equation yields L {y } (s) + 4L {y} (s) = L {g(t)} (s) 1  e2s s2 + 1 s+3 1 e2s L {y} (s) = 2 +  . s + 4 (s2 + 1)(s2 + 4) (s2 + 1)(s2 + 4) s2 L {y} (s)  s  3 + 4L {y} (s) = Using the partial fractions decomposition 1 (s2 + 4)  (s2 + 1) 1 1 1 2 1 = =  2 , 2 + 1)(s2 + 4) 2 + 1)(s2 + 4) 2 +1 (s 3 (s 3 s 6s +4 we conclude that L {y} (s) = and so y(t) = L1 2 1 1 (t) + L1 (t) +4 3 s2 + 1 1 1 1 2  2 L1 (t  2)u(t  2) 2 +1 3s 6s +4 1 1 1 4 sin(t  2)  sin 2(t  2) u(t  2) = cos 2t + sin 2t + sin t  3 3 3 6 1 1 1 4 sin t  sin 2t u(t  2) = cos 2t + sin 2t + sin t  3 3 3 6 1 1 = cos 2t + [1  u(t  2)] sin t + [8 + u(t  2)] sin 2t . 3 6 s2 s +4 4 (t) + L1 3 s2 s2 1 1 4 2 1 1 1 2 s + 2 + 2  e2s  2 2+1 +4 3s +4 3s +1 3s 6s +4 39. We can express g(t) using the unit step function as g(t) = tu(t  1) + (1  t)u(t  5) = [(t  1) + 1]u(t  1)  [(t  5) + 4]u(t  5). Thus, formula (5) on page 387 of the text yields L {g(t)} (s) = es L {t + 1} (s)  e5s L {t + 4} (s) = es 1 1 + 2 s s  e5s 4 1 + 2 s s . 439 Chapter 7
Let Y (s) = L {y} (s). Applying the Laplace transform to the given equation and using the initial conditions, we obtain L {y } (s) + 5L {y } (s) + 6Y (s) = L {g(t)} (s) s2 Y (s)  2 + 5 [sY (s)] + 6Y (s) = L {g(t)} (s) 1 4 1 1 + s2 + 5s + 6 Y (s) = 2 + es 2 +  e5s s s s2 s s+1 4s + 1 2 + es 2 2  e5s 2 2 . Y (s) = 2 s + 5s + 6 s (s + 5s + 6) s (s + 5s + 6) (7.36) Using partial fractions decomposition, we can write 2 2 2 =  , + 5s + 6 s+2 s+3 s+1 1/36 1/6 2/9 1/4 = + 2  + , 2 (s2 + 5s + 6) s s s s+2 s+3 4s + 1 1/6 19/36 7/4 11/9 = 2 +  + . 2 (s2 + 5s + 6) s s s s+2 s+3 s2 Therefore, L1 L1 L1 2 (t) = 2e2t  2e3t , + 5s + 6 t e2t 2e3t s+1 1 +  + , (t) = s2 (s2 + 5s + 6) 36 6 4 9 4s + 1 19 t 7e2t 11e3t +  + . (t) = s2 (s2 + 5s + 6) 36 6 4 9 s2 Using these equations and taking the inverse Laplace transform in (7.36), we finally obtain y(t) = 2e2t  2e3t + t  1 e2(t1) 2e3(t1) 1 +  + u(t  1) 36 6 4 9 19 t  5 7e2(t5) 11e3(t5) +  + + u(t  5). 36 6 4 9 41. First observe that for s > 0, T > 0, we have 0 < eT s < 1 so that 1 = 1 + eT s + e2T s + e3T s + T s 1e 440 (7.37) Exercises 7.6
and the series converges for all s > 0. Thus, 1 1 1 1 = 1 + eT s + e2T s + e3T s + = T s ) T s (s + )(1  e s+ 1e s+ eT s e2T s 1 + + + , = s+ s+ s+ and so L1 1 (s + )(1  eT s ) (t) = L1 1 eT s e2T s + + + (t). s+ s+ s+ (7.38) Taking for granted that the linearity of the inverse Laplace transform extends to the infinite sum in (7.38) and ignoring convergence questions yields L
1 1 (s + )(1  eT s ) = L 1 1 s+ +L 1 eT s s+ +L 1 e2T s s+ + = et + e(tT ) u(t  T ) + e(t2T ) u(t  2T ) + as claimed. 43. Using the expansion (7.37) obtained in Problem 41, we can represent L {g} (s) as L {g} (s) = 1 = 2 1 + eT s + e2T s + e3T s + s2 + 2 1  eT s s + 2 + eT s 2 + e2T s 2 + . = 2 2 2 s + s + s + 2 Since L1 {/(s2 + 2 )} (t) = sin t, using the linearity of the inverse Laplace transform (extended to infinite series) and formula (6) in Theorem 8, we obtain g(t) = L1 + 2 (t) + L1 + 2 (t  T )u(t  T ) +L1 s2 s2 s2 (t  2T )u(t  2T ) + + 2 = sin t + [sin (t  T )]u(t  T ) + [sin (t  2T )]u(t  2T ) + as stated. 441 Chapter 7
45. In order to apply the method of Laplace transform to given initial value problem, let us find L {f } (s) first. Since the period of f (t) is T = 1 and f (t) = et on (0, 1), the windowed version of f (t) is f1 (t) = and so F1 (s) =
0 et , 0 < t < 1, 0,
1 otherwise, e(1s)t e dt = 1s
1 e st f1 (t) dt =
0 e st t =
0 1  e1s . s1 Hence, Theorem 9 yields the following formula for L {f } (s): L {f } (s) = 1  e1s . (s  1)(1  es ) 1  e1s (s  1)(1  es ) We can now apply the Laplace transform to the given differential equation and obtain L {y } (s) + 3L {y } (s) + 2L {y} (s) = 1  e1s (s  1)(1  es ) 1s 1e 1  e1s L {y} (s) = = (s  1)(s2 + 3s + 2)(1  es ) (s  1)(s + 1)(s + 2)(1  es ) 1e e 1 + . L {y} (s) = (s  1)(s + 1)(s + 2) 1  es (s  1)(s + 1)(s + 2) s2 L {y} (s) + 3 [sL {y} (s)] + 2L {y} (s) = Using the partial fractions decomposition 1 1/6 1/2 1/3 =  + (s  1)(s + 1)(s + 2) s1 s+1 s+2 we find that L {y} (s) = e/2 e/3 1e e/6 1  + + s1 s+1 s+2 6 (s  1)(1  es ) 1 1e 1 1e +  s ) 2 (s + 1)(1  e 3 (s + 2)(1  es ) e e e 1  e 1 1 y(t) = et  et + e2t + L (t) 6 2 3 6 (s  1)(1  es ) 1 1 1  e 1 1  e 1 L L (t) + (t). (7.39)  s ) 2 (s + 1)(1  e 3 (s + 2)(1  es ) 442 Exercises 7.6
To each of the three inverse Laplace transforms in the above formula we can apply results of Problem 42(a) with T = 1 and = 1, 1, and 2, respectively. Thus, for n < t < n + 1, we have L1 L1 L1 1 (s  1)(1  es ) 1 (s + 1)(1  es ) 1 (s + 2)(1  es ) e(n+1)  1 , e1  1 en+1  1 (t) = et , e1 e2(n+1)  1 (t) = e2t . e2  1 (t) = et Finally, substitution back into (7.39) yields y(t) = e t e t e 2t 1  e t e(n+1)  1 e  e + e + e 6 2 3 6 e1  1 1  e t en+1  1 1  e 2t e2(n+1)  1 e e  + 2 e1 3 e2  1 etn et (1 + e  en+1 ) e2t (1 + e + e2  e2n+2 )  + . = 6 2 3(e + 1) 47. Since et = k=0 tk k! k! sk+1 and L tk (s) = , using the linearity of the Laplace transform we have L e t (s) = L
k=0 tk k! (s) =
k=0 L tk (s) = k! k=0 1 k!/sk+1 = k! s k=0 1 s k . (7.40) We can apply now the summation formula for geometric series, that is, 1 + x + x2 + = 1 , 1x which is valid for x < 1. With x = 1/s, s > 1, (7.40) yields L et (s) = 1 1 1 = . s 1  (1/s) s1 443 Chapter 7
49. Recall that the Taylor's series for cos t about t = 0 is cos t = 1  so that t2 t4 t6 t2n +  + + (1)n + 2! 4! 6! (2n)! t2n1 1  cos t t t3 t5 =  + + + (1)n+1 + . t 2! 4! 6! (2n)! 1  cos t t (1)n+1 1 1 L {t} (s)  L t3 (s) + + L t2n1 (s) + 2! 4! (2n)! 11 11 (1)n+1 1 =  ++ + 2 s2 4 s4 2n s2n (1)n+1 1 (1)n+1 = . = 2n s2n 2ns2n n=1 n=1 Thus L (s) = To sum this series, recall that ln(1  x) =  n=1 xn . n Hence, ln 1 + Thus, we have 1 1 ln 1 + 2 2 s =
n=1 1 s2 =
n=1 (1)n = ns2n n=1 (1)n+1 . ns2n (1)n+1 =L 2ns2n 1  cos t t (s) . This formula can also be obtained by using the result of Problem 27 in Section 7.3 of the text. 51. We use formula (17) on page 394 of the text. (a) With r = 1/2, (17) yields L t
1/2 [(1/2) + 1] (1/2) (s) = = 1/2 = = (1/2)+1 s s s . s (b) This time, r = 7/2, and (17) becomes L t7/2 (s) = 444 [(7/2) + 1] (9/2) = 9/2 . (7/2)+1 s s Exercises 7.6
From the recursive formula (16) we find that 9 2 7 +1 = 2 7 7 = 2 2 5 75 = 22 2 L t
7/2 3 753 = 222 2 1 7531 = 2222 2 105 . = 16 Therefore, 105 . (s) = 16s9/2 53. According to the definition (11) of the function fT (t), fT (t  kT ) = 0 if the point t  kT does not belong to (0, T ). Therefore, fixed t, in the series (13) all the terms containing fT (t  kT ) with k's such that t  kT 0 or t  kT T vanish. In the remaining terms, k satisfies 0 < t  kT < T t t 1<k < . T T But, for any fixed t, there is at most one k satisfying this condition. 55. Recall that ex = 1 + x + Substituting 1/s for x above yields e1/s = 1  Thus, we have s
1/2 1/s xn x2 ++ + . 2! n! 1 1 1 (1)n +  ++ + . s 2!s2 3!s3 n!sn e = 1 s1/2  1 s3/2 1 (1)n + ++ + = 2!s5/2 n!sn+1/2 n=0 (1)n . n!sn+1/2 By Problem 52 of this section, L1 so that L
1 1 sn+(1/2) (t) = 2n tn(1/2) , 1 3 5 (2n  1) s 1/2 1/s e (t) = L = 1 n=0 (1)n n!sn+1/2 1 (t) n=0 (1)n 1 L n! sn+(1/2) (t) =
n=0 2n tn(1/2) (1)n . n! 1 3 5 (2n  1) 445 Chapter 7
Multiplying the nth term by [2 4 (2n)]/[2 4 (2n)], we obtain L
1 s 1/2 1/s e (1)n (2n )2 tn(1/2) (1)n (2n )2 tn = (t) = (2n)! (2n)! t n=0 n=0 1 1 (1)n (2 t)2n = cos 2 t . = (2n)! t n=0 t 57. Recall that the Maclaurin expansion of ln(1  x) is ln(1  x) = 
n=1 xn , n which converges for x < 1. Hence, substitution 1/s2 for x yields ln 1 + 1 s2 =
n=1 (1)n = ns2n n=1 (1)n+1 . ns2n Assuming that the inverse Laplace transform can be computed termwise, we obtain L
1 1 ln 1 + 2 s =L 1 n=1 (1)n+1 ns2n =
n=1 (1)n+1 1 L n 1 s2n . From Table 7.1 in Section 7.2, L tk = k!/sk+1, k = 1, 2, . . . . Thus L1 1/sk+1 = tk /k!. With k = 2n  1, this yields L1 and, therefore, L1 ln 1 + Since cos t =
n=0 1 s2n (t) = t2n1 , (2n  1)! n = 1, 2, . . . 1 s2 (t) =
n=1 2 (1)n+1 t2n1 = n (2n  1)! t n=1 (1)n 2n t . (2n)! (7.41) (1)n 2n (1)n 2n t =1+ t , (2n)! (2n)! n=1 2(1  cos t) 2 (cos t  1) = . t t (7.41) implies that L1 ln 1 + 446 1 s2 (t) =  Exercises 7.6
59. Applying the Laplace transform to both sides of the original equation and using its linearity, we obtain L {y } (s)  L {y} (s) = L {G3 (t  1)} (s). Initial conditions, y(0) = 0 and y (0)=2, and Theorem 5 in Section 7.3 imply that L {y } (s) = s2 L {y} (s)  sy(0)  y (0) = s2 L {y} (s)  2. In the righthand side of (7.42), we can apply the result of Problem 58(c) with a = 3 and b = 1 to get L {G3 (t  1)} (s) = Thus (7.42) becomes es  e4s s L {y} (s)  2  L {y} (s) = s es  e4s 2 + . L {y} (s) = 2 s 1 s(s2  1)
2 (7.42) es  e4s . s Substituting partial fraction decompositions s2 yields L {y} (s) = 1 1  + es  e4s s1 s+1 1/2 1 1  + es + = s1 s+1 s1 L
1 1 1 2 =  , 1 s1 s+1 s(s2 1 1/2 1/2 1 = +   1) s1 s+1 s 1/2 1 1/2 +  s1 s+1 s 1/2 1/2 1 1/2 1   e4s +  . s+1 s s1 s+1 s et + et  2 , (t) = 2 (7.43) Since 1/2 1/2 1 +  s1 s+1 s formula (6) on page 387 of the text gives us L1 es 1/2 1/2 1 +  s1 s+1 s (t) = L1 1/2 1/2 1 +  s1 s+1 s t1 e + e1t  2 u(t  1), = 2 (t  1)u(t  1) 447 Chapter 7
L1 e4s 1/2 1 1/2 +  s1 s+1 s (t) = L1 1/2 1 1/2 +  s1 s+1 s t4 e + e4t  2 = u(t  4). 2 (t  4)u(t  4) Taking the inverse Laplace transform in (7.43) yields y(t) = et  et + et4 + e4t  2 et1 + e1t  2 u(t  1)  u(t  4). 2 2 61. In this problem, we use the method of solving "mixing problems" discussed in Section 3.2. So, let x(t) denote the mass of salt in the tank at time t with t = 0 denoting the moment when the process started. Thus, using the formula mass = volume concentration , we have the initial condition x(0) = 500 (L) 0.2 (kg/L) = 100 (kg). For the rate of change of x(t), that is, x (t), we use then relation x (t) = input rate  output rate . While the output rate (through the exit valve C) can be computed as output rate = 3x(t) x(t) (kg/L) 12 (L/min) = (kg/min) 500 125 (7.44) for all t, the input rate has different formulas for the first 10 minute and after that. Namely, 0 < t < 10 (valve A) : 10 < t (valve B) : input rate = 12 (L/min) 0.4 (kg/L) = 4.8 (kg/min); input rate = 12 (L/min) 0.6 (kg/L) = 7.2 (kg/min). In other words, the input rate is a function of t, which can be written as input rate = g(t) = 448 4.8, 0 < t < 10, 7.2, 10 < t. Exercises 7.6
Using the unit step function, we can express g(t) = 4.8 + 2.4u(t  10) (kg/min). Therefore (7.44) becomes x (t) = g(t)  3x(t) 125 x (t) + 3 x(t) = 4.8 + 2.4u(t  10) 125 (7.45) with the initial condition x(0) = 100. Taking the Laplace transform of both sides yields L {x } (s) + Since 2.4 = 100 s[s + (3/125)] 100s + 4.8 = 100 s[s + (3/125)] 1 1  , s s + (3/125) 2 1  , s s + (3/125) 4.8 2.4e10s 3 L {x} (s) = L {4.8 + 2.4u(t  10)} (s) = + 125 s s 4.8 2.4e10s 3 L {x} (s) = + [sL {x} (s)  100] + 125 s s 100s + 4.8 2.4 10s . L {x} (s) = + e s[s + (3/125)] s[s + (3/125)] (7.46) applying the inverse Laplace transform in (7.46), we get x(t) = 100 2  e3t/125 + 100 1  e3(t10)/125 u(t  10). Finally, dividing by the volume of the solution in the tank, which constantly equals to 500 L, we conclude that concentration = 0.4  0.2e3t/125 + 0.2 1  e3(t10)/125 u(t  10). 63. In this problem, the solution still enters the tank at the rate 12 L/min, but leaves the tank at the rate only 6 L/min. Thus, every minute, the volume of the solution in the tank increases by 12  6 = 6 L. Therefore, the volume, as a function of t, is given by 500 + 6t and so output rate = 3x(t) x(t) (kg/L) 6 (L/min) = (kg/min). 500 + 6t 250 + 3t Instead of equation (7.45) in Problem 61, we now have x (t) = g(t)  3x(t) 250 + 3t (250 + 3t)x (t) + 3x(t) = (250 + 3t)[48 + 24u(t  10)]. 449 Chapter 7
This equation has polynomial coefficients and can also be solved using the Laplace transform method. (See the discussion in Section 7.5, page 380, and Example 4.) But, as an intermediate step, one will obtain a first order linear differential equation for L {x} (s). EXERCISES 7.7: Convolution, page 405 1. Let Y (s) := L {y} (s), G(s) := L {g} (s). Taking the Laplace transform of both sides of the given differential equation and using the linear property of the Laplace transform, we obtain L {y } (s)  2L {y } (s) + Y (s) = G(s). The initial conditions and Theorem 5, Section 7.3, imply that L {y } (s) = sY (s) + 1, L {y } (s) = s2 Y (s) + s  1. Thus, substitution yields s2 Y (s) + s  1  2 [sY (s) + 1] + Y (s) = G(s) s2  2s + 1 Y (s) = 3  s + G(s) 3s 1 G(s) 2 G(s) Y (s) = 2  . + 2 = + 2 s  2s + 1 s  2s + 1 (s  1) s  1 (s  1)2 Taking now the inverse Laplace transform, we obtain y(t) = 2L1 1 (s  1)2 (t)  L1 1 s1 (t) + L1 G(s) (s  1)2 (t) . Using Table 7.1, we find that L1 1 s1 (t) = et , L1 1 (s  1)2 (t) = tet , and, by the convolution theorem,
t L 1 G(s) (s  1)2 (t) = L 1 1 G(s) (t) = tet g(t) = (s  1)2 (t  v)etv g(v) dv.
0 450 Exercises 7.7
Thus y(t) = 2tet  et +
0 t (t  v)etv g(v) dv. 3. Taking the Laplace transform of y + 4y + 5y = g(t) and applying the initial conditions y(0) = y (0) = 1 gives us s2 Y (s)  s  1 + 4 [sY (s)  1] + 5Y (s) = G(s), where Y (s) := L {y} (s), G(s) := L {g} (s). Thus Y (s) = s2 s+5 G(s) s+2 3 G(s) + 2 = + + . 2+1 2+1 + 4s + 5 s + 4s + 5 (s + 2) (s + 2) (s + 2)2 + 1 Taking the inverse Laplace transform of Y (s) with the help of the convolution theorem yields
t y(t) = e2t cos t + 3e2t sin t +
0 e2(tv) sin(t  v)g(v) dv.. 5. Since L1 {1/s} (t) = 1 and L1 {1/(s2 + 1)} (t) = sin t, writing s(s2 1 1 1 = 2 + 1) s s +1 and using the convolution theorem, we obtain
t L1 1 2 + 1) s(s (t) = 1 sin t =
0 sin v dv =  cos v t = 0 1  cos t. 7. From Table 7.1, L1 {1/(s  a)} (t) = eat . Therefore, using the linearity of the inverse Laplace transform and the convolution theorem, we have
t L1 14 (s + 2)(s  5) (t) = 14L1 1 1 s+2 s5
t (t) = 14e2t e5t = 14
0 e2(tv) e5v dv = 14e2t
0 e7v dv = 2e2t e7t  1 = 2 e5t  e2t . 451 Chapter 7
9. Since s/(s2 + 1)2 = [s/(s2 + 1)] [1/(s2 + 1)] the convolution theorem tells us that
t L1 s 2 + 1)2 (s (t) = L1 s s 2 + 1 s2 + 1 s (t) = cos t sin t =
0 cos(t  v) sin v dv. Using the identity sin cos = [sin( + ) + sin(  )]/2, we get
t L1 s 2 + 1)2 (s 1 (t) = 2 1 = 2 [sin t + sin(t  2v)] dv
0 cos(t  2v) v sin t + 2 t =
0 t sin t . 2 11. Using the hint, we can write s 1 1 = + , (s  1)(s + 2) s + 2 (s  1)(s + 2) so that by the convolution theorem, Theorem 11 on page 400 of the text, L1 s (s  1)(s + 2) (t) = L1 = e
2t 1 s+2 +e e
t t (t) + L1
2t 1 (s  1)(s + 2)
t (t) =e 2t +
0 etv e2v dv et 3t 2e2t et e 1 = + . 3 3 3 = e2t + et
0 e3v dv = e2t  13. Note that f (t) = t e3t . Hence, by (8) on page 400 of the text, L {f (t)} (s) = L {t} (s)L e3t (s) = 15. Note that 1 1 1 = 2 . 2 s3 s s (s  3) t y(v) sin(t  v) dv = sin t y(t).
0 Let Y (s) := L {y} (s). Taking the Laplace transform of the original equation, we obtain Y (s) + 3L {sin t y(t)} (s) = L {t} (s) 452 Exercises 7.7 1 1 3 Y (s) = 2 Y (s) + 2 s2 s +1 s 2 s +1 (3/8)2 (1/4) Y (s) = 2 2 + 2 = 2 s (s + 4) s s + 22 (3/8)2 (1/4) t 3 sin 2t y(t) = L1 + 2 (t) = + . 2 2 s s +2 4 8 Y (s) + 3L {sin t} (s)Y (s) = 17. We use the convolution Theorem 11 to find the Laplace transform of the integral term. t Y (s) (t  v)y(v) dv (s) = L {t y(t)} (s) = L {t} (s)L {y(t)} (s) = 2 , L s
0 where Y (s) denotes the Laplace transform of y(t). Thus taking the Laplace transform of both sides of the given equation yields Y (s) + 1 Y (s) = 2 s s Y (s) = s2 s +1 y(t) = L1 s2 s +1 (t) = cos t . 19. By the convolution theorem, t 2Y (s) 2 (t  v) y(v) dv (s) = L t2 y(t) (s) = L t2 (s)L {y(t)} (s) = . L s3
0 Hence, applying the Laplace transform to the original equation yields Y (s) + 2Y (s) 6 3 = L t3 + 3 (s) = 4 + 3 s s s 6 + 3s3 s3 3 Y (s) = 3 = s +2 s4 s 3 y(t) = L1 (t) = 3. s 21. As in Example 3 on page 402 of the text, we first rewrite the integrodifferential equation as y (t) + y(t)  y(t) sin t =  sin t , We now take the Laplace transform of (7.47) to obtain [sY (s)  1] + Y (s)  s2 1 1 Y (s) =  2 , +1 s +1 453 y(0) = 1. (7.47) Chapter 7
where Y (s) = L {y} (s). Thus, Y (s) = s s s2 = 2 = 3 + s2 + s s s +s+1 (s + 1/2)2 + 3/4 s + 1/2 (1/ 3)( 3/2) =  . 2 + 3/4 (s + 1/2) (s + 1/2)2 + 3/4 Taking the inverse Laplace transform yields 3t t/2 cos y(t) = e 2 1  et/2 sin 3 3t 2 . 23. Taking the Laplace transform of the differential equation, and assuming zero initial conditions, we obtain s2 Y (s) + 9Y (s) = G(s), where Y = L {y}, G = L {g}. Thus, H(s) = The impulse response function is then h(t) = L1 {H(s)} (t) = L1 1 s2 + 9 1 (t) = L1 3 3 s2 + 3 2 (t) = sin 3t . 3 Y (s) 1 = 2 . G(s) s +9 To solve the initial value problem, we need the solution to the corresponding homogeneous problem. The auxiliary equation, r 2 + 9 = 0, has roots, r = 3i. Thus, a general solution to the homogeneous equation is yh (t) = C1 cos 3t + C2 sin 3t. Applying the initial conditions y(0) = 2 and y (0) = 3, we obtain 2 = y(0) = (C1 cos 3t + C2 sin 3t)
t=0 = C1 ,
t=0 3 = y (0) = (3C1 sin 3t + 3C2 cos 3t So = 3C2 C1 = 2 , C2 = 1 . yk (t) = 2 cos 3t  sin 3t, 454 Exercises 7.7
and the formula for the solution to the original initial value problem is
t 1 y = (h g)(t) + yk (t) = 3 g(v) sin 3(t  v) dv + 2 cos 3t  sin 3t.
0 25. Taking the Laplace transform of both sides of the given equation and assuming zero initial conditions, we get L {y  y  6y} (s) = L {g(t)} (s) Thus, H(s) = Y (s) 1 1 = 2 = G(s) s s6 (s  3)(s + 2) s2 Y (s)  sY (s)  6Y (s) = G(s). is the transfer function. The impulse response function h(t) is then given by
t h(t) = L 1 1 (s  3)(s + 2) (t) = e e 3t 2t =
0 e 3(tv) 2v e dv = e 3t e5v 5 t =
0 e3t  e2t . 5 To solve the given initial value problem, we use Theorem 12. To this end, we need the solution yk (t) to the corresponding initial value problem for the homogeneous equation. That is, y  y  6y = 0, y(0) = 1, y (0) = 8 (see (19) in the text). Applying the Laplace transform yields s2 Yk (s)  s  8  [sYk (s)  1]  6Yk (s) = 0 s+7 s+7 2 1 Yk (s) = 2 = =  s s6 (s  3)(s + 2) s3 s+2 1 2  yk (t) = L1 {Yk (s)} (t) = L1 (t) = 2e3t  e2t . s3 s+2 So,
t 1 y(t) = (h g)(t) + yk (t) = 5 e3(tv)  e2(tv) g(v) dv + 2e3t  e2t .
0 455 Chapter 7
27. Taking the Laplace transform and assuming zero initial conditions, we find the transfer function H(s). s2 Y (s)  2sY (s) + 5Y (s) = G(s) Therefore, the impulse response function is h(t) = L1 {H(s)} (t) = L1 1 (s  1)2 + 22 (t) = 1 1 L 2 2 (s  1)2 + 22 (t) = 1 t e sin 2t . 2 H(s) = Y (s) 1 = 2 . G(s) s  2s + 5 Next, we find the solution yk (t) to the corresponding initial value problem for the homogeneous equation, y  2y + 5y = 0, y(0) = 0, y (0) = 2. Since the associated equation, r 2  2r + 5 = 0, has roots r = 1 2i, a general solution to the homogeneous equations is yh (t) = et (C1 cos 2t + C2 sin 2t) . We satisfy the initial conditions by solving 0 = y(0) = C1 2 = y (0) = C1 + 2C2 Hence, yk (t) = et sin 2t and
t C1 = 0, C2 = 1. 1 y(t) = (h g)(t) + yk (t) = 2 is the desired solution. etv sin 2(t  v)g(v) dv + et sin 2t
0 29. With given data, the initial value problem becomes 5I (t) + 20I (t) + 1 I(t) = e(t), 0.005 I(0) = 1, I (0) = 8. Using formula (15) on page 403 of the text, we find the transfer function H(s) = 456 5s2 1 1 1 = . + 20s + 200 5 (s + 2)2 + 62 Exercises 7.7
Therefore, h(t) = L1 1 1 5 (s + 2)2 + 62 (t) = 1 1 L 30 6 (s + 2)2 + 62 (t) = 1 2t e sin 6t. 30 Next, we consider the initial value problem 5I (t) + 20I (t) + 200I(t) = 0, I(0) = 1, I (0) = 8 for the corresponding homogeneous equation. Its characteristic equation, 5r 2 + 20r + 200 = 0, has roots r = 2 6i, which yield a general solution Ih (t) = e2t (C1 cos 6t + C2 sin 6t) . We find constants C1 and C2 so that the solution satisfies the initial conditions. Thus we have 1 = I(0) = C1 , 8 = I (0) = 2C1 + 6C2 and so Ik (t) = e2t (sin 6t  cos 6t). Finally,
t C1 = 1 , C2 = 1 , 1 I(t) = h(t) e(t) + Ik (t) = 30 31. By the convolution theorem, we get e(v)e2(tv) sin 6(t  v) dv + e2t (sin 6t  cos 6t) .
0 L {1 1 1} (s) = L {1} (s)L {1 1} (s) = L {1} (s)L {1} (s)L {1} (s) = Therefore, the definition of the inverse Laplace transform yields 1 1 1 = L1 1 s3 (t) = 1 1 L 2 2 s3 (t) = 1 2 t . 2 1 s 3 = 1 . s3 33. Using the linear property of integrals, we have
t t f (g + h) =
0 t f (t  v)[g + h](v) dv =
0 t f (t  v)[g(v) + h(v)] dv =
0 f (t  v)g(v) dv +
0 f (t  v)h(v) dv = f g + f h. 457 Chapter 7
35. Since
t t f (v) dv =
0 0 1 f (v) dv = 1 f (t), we conclude that t 1 L f (v) dv (s) = L {1 f (t)} (s) = L {1} (s)L {f (t)} (s) = F (s). s
0 Hence, by the definition of the inverse Laplace transform,
t f (v) dv = L1
0 1 F (s) (t). s (Note that the integral in the lefthand side is a continuous function.) 37. Actually, this statement holds for any continuously differentiable function h(t) on [0, ) satisfying h(0) = 0. Indeed, first of all,
t 0 (h g)(0) =
0 h(t  v)g(v) dv
t=0 =
0 h(v)g(v) dv = 0 since the interval of integration has zero length. Next, we apply the Leibniz's rule to find the derivative of (h g)(t). (h g) (t) = 0 t t h(t  v)g(v) dv =
0 t h(t  v)g(v) dv + h(t  v)g(v) t
t v=t =
0 h (t  v)g(v) dv + h(0)g(t) =
0 h (t  v)g(v) dv since h(0) = 0. Therefore,
0 (h g) (0) =
0 h (v)g(v) dv = 0, again as a definite integral with equal limits of integration. 458 Exercises 7.8
EXERCISES 7.8: Impulses and the Dirac Delta Function, page 412 1. By equation (3) on page 407 of the text, (t2  1)(t) dt = t2  1
 t=0 = 1. 3. By equation (3) on page 407 of the text, (sin 3t) t 
 dt = sin 3 2 2 = 1. 5. Formula (6) of the Laplace transform of the Dirac delta function yields e2t (t  1) dt = L {(t  1)} (2) = es
0 s=2 = e2 . 7. Using the linearity of the Laplace transform and (6) on page 409 of the text, we get L {(t  1)  (t  3)} (s) = L {(t  1)} (s)  L {(t  3)} (s) = es  e3s . 9. Since (t  1) = 0 for t < 1, L {t(t  1)} (s) :=
0 e st t(t  1) dt =
 est t(t  1) dt = est t t=1 = es by equation (3) on page 407 of the text. Another way to solve this problem is to use Theorem 6 inj Section 7.3. This yields L {t(t  1)} (s) =  d (es ) d L {(t  1)} (s) =  = es . ds ds 11. Since (t  ) = 0 for t < , we use the definition of the Laplace transform and formula (3), page 407 of the text, to conclude that L {(sin t)(t  )} (s) :=
0 e st (sin t)(t  ) dt =
 est (sin t)(t  ) dt = et sin = 0. 459 Chapter 7
13. Let W (s) := L {w} (s). Using the initial conditions and Theorem 5 in Section 7.3, we find that L {w } (s) = s2 W (s)  sw(0)  w (0) = s2 W (s). Thus, applying the Laplace transform to both sides of the given equation yields s2 W (s) + W (s) = L {(t  )} (s) = es W (s) = es . s2 + 1 Taking the inverse Laplace transform of both sides of the last equation and using Theorem 8 in Section 7.6, we get w(t) = L
1 es s2 + 1 (t) = L1 s2 1 +1 (t)u(t) = sin(t)u(t) = (sin t)u(t). 15. Let Y := L {y}. Taking the Laplace transform of y + 2y  3y = (t  1)  (t  2) and applying the initial conditions y(0) = 2, y (0) = 2, we obtain s2 Y (s)  2s + 2 + 2 [sY (s)  2]  3Y (s) = L {(t  1)  (t  2)} (s) = es  e2s 2s + 2 es e2s 2s + 2 + es  e2s = +  Y (s) = s2 + 2s  3 (s + 3)(s  1) (s + 3)(s  1) (s + 3)(s  1) s 1 1 e 1 1 e2s 1 1 = + +    , s1 s+3 4 s1 s+3 4 s1 s+3 so that by Theorem 8 on page 387 of the text we get y(t) = et + e3t + 1 t2 1 t1 e  e3(t1) u(t  1)  e  e3(t2) u(t  2). 4 4 17. Let Y := L {y}. We use the initial conditions to find that L {y } (s) = s2 Y (s)  sy(0)  y (0) = s2 Y (s)  2. Thus taking the Laplace transform of both sides of the given equation and using formula (6) on page 409, we get s2 Y (s)  2  Y (s) = 4L {(t  2)} (s) + L t2 (s) = 4e2s + 460 Y (s) = 2(s3 + 1) 4e2s + 3 2 = 2e2s s2  1 s (s  1) 1 1  s1 s+1 + 2 s3 2 2 2  3 . s1 s s Exercises 7.8
Now we can apply the inverse Laplace transform. y(t) = L1 2e2s = 2 L1 1 1  s1 s+1  L1 + 2 2 2  3 s1 s s (t) 1 s1 1 s+1 +2L1 (t  2)u(t  2) (t)  L1 2 s3 (t)  2L1 1 s (t) = 2 et2  e2t 1 s1 t 2 u(t  2) + 2e  t  2. 19. Let W (s) := L {w} (s). We apply the Laplace transform to the given equation and obtain L {w } (s) + 6L {w } (s) + 5W (s) = L et (t  1) (s). From formula (4) on page 362 of the text we see that L {w } (s) = sW (s)  w(0) = sW (s), L {w } (s) = s2 W (s)  sw(0)  w (0) = s2 W (s)  4. Also, the translation property (1), Section 7.3, of the Laplace transform yields L et (t  1) (s) = L {(t  1)} (s  1) = e(s1) = e1s . Substituting (7.49) and (7.50) back into (7.48), we obtain s2 W (s)  4 + 6 [sW (s)] + 5W (s) = e1s 4 + e1s 1 1 e 4 + e1s = =  + es W (s) = 2 s + 6s + 5 (s + 1)(s + 5) s+1 s+5 4 Finally, the inverse Laplace transform of both sides of this equation yields w(t) = et  e5t + e (t1) e  e5(t1) u(t  1) . 4 (7.50) (7.49) (7.48) 1 1  s+1 s+5 . 21. We apply the Laplace transform to the given equation, solve the resulting equation for L {y} (s), and then use the inverse Laplace transforms. This yields L {y } (s) + L {y} (s) = L {(t  2)} (s) 461 Chapter 7 s2 L {y} (s)  1 + L {y} (s) = e2s y(t) = L1 L {y} (s) = 1 + e2s s2 + 1 1 1 (t) + L1 (t  2)u(t  2) s2 + 1 s2 + 1 = sin t + [sin(t  2)]u(t  2) = [1 + u(t  2)] sin t. The graph of the solution is shown in Figure B.49 in the answers of the text. 23. The solution to the initial value problem y + y = (t  2), is given in Problem 21, that is y1 (t) = [1 + u(t  2)] sin t. Thus, if y2 (t) is the solution to the initial value problem y + y = (t  ), y(0) = 0, y (0) = 0, (7.51) y(0) = 0, y (0) = 1 then, by the superposition principle (see Section 4.5), y(t) = y1 (t) + y2 (t) is the desired solution. The Laplace transform of both sides in (7.51) yields s2 L {y} (s) + L {y} (s) = es y2 (t) = L1 s2 1 +1 L {y} (s) =  es s2 + 1 (t  )u(t  ) = [sin(t  )]u(t  ) = u(t  ) sin t. (We have used zero initial conditions to express L {y } in terms of L {y}.) Therefore, the answer is y(t) = y1 (t) + y2 (t) = [1 + u(t  2)] sin t + u(t  ) sin t = [1 + u(t  ) + u(t  2)] sin t. The sketch of this curve is given in Figure B.50 . 25. Taking the Laplace transform of y + 4y + 8y = (t) with zero initial conditions yields s2 Y (s) + 4sY (s) + 8Y (s) = L {(t)} (s) = 1. 462 Exercises 7.8
Solving for Y (s), we obtain Y (s) = so that h(t) = L1 1 2 2 (s + 2)2 + 22 (t) = 1 2t e sin 2t . 2 s2 1 1 2 1 = = 2+4 + 4s + 8 (s + 2) 2 (s + 2)2 + 22 Notice that H(s) for y +4y +8y = g(t) with y(0) = y (0) = 0 is given by H(s) = 1/(s2 +4s+8), so that again h(t) = L1 {H(s)} (t) = 1 2t e sin 2t . 2 27. The Laplace transform of both sides of the given equation, with zero initial conditions and g(t) = (t), gives us s2 L {y} (s)  2sL {y} (s) + 5L {y} (s) = L {(t)} (s) 1 1 . L {y} (s) = 2 = s  2s + 5 (s  1)2 + 22 The inverse Laplace transform now yields h(t) = L1 1 (s  1)2 + 22 (t) = 1 1 L 2 2 (s  1)2 + 22 (t) = 1 t e sin 2t . 2 29. We solve the given initial value problem to find the displacement x(t). Let X(s) := L {x} (s). Applying the Laplace transform to the differential equation yields L {x } (s) + 9X(s) = L 3 t  Since L {x } (s) = s2 X(s)  sx(0)  x (0) = s2 X(s)  s, the above equation becomes s2 X(s)  s + 9X(s) = 3es/2 Therefore, x(t) = L1 s2 s 3  es/2 2 2 +3 s + 32 (t) 463 X(s) = s  3es/2 s 3  es/2 2 . = 2 2+9 2 s s +3 s + 32 2 (s) = 3es/2 . Chapter 7
= cos 3t  sin 3 t  2 u t 2 = 1u t 2 cos 3t . Since, for t > /2, u(t  /2) 1, we conclude that x(t) 0 for t > . 2 This means that the mass stops after the hit and remains in the equilibrium position thereafter. 31. By taking the Laplace transform of ay + by = cy = (t), y(0) = y (0) = 0, and solving for Y := L {y}, we find that the transfer function is given by H(s) = as2 1 . + bs + c If the roots of the polynomial as2 + bs + c are real and distinct, say r1 , r2 , then H(s) = Thus 1 er1 t  er2 t r1  r2 and clearly h(t) is bounded as t if and only if r1 and r2 are less than or equal to zero. h(t) = If the roots of as2 + bs + c are complex, then, by the quadratic formula, they are given by 4ac  b2 b i  2a 2a so that the real part of the roots is b/(2a). Now H(s) = 1 1 1 1 1 = 2 = 2 + (4ac  b2 )/(4a2 ) + bs + c a s + (b/a)s + (c/a) a [s + b/(2a)] 4ac  b2 /(2a) 2 = 4ac  b2 [s + b/(2a)]2 + [ 4ac  b2 /(2a)]2 as2 1/(r1  r2 ) 1/(r1  r2 ) 1 =  . (s  r1 )(s  r2 ) s  r1 s  r2 4ac  b2 2 t , h(t) = e(b/2a)t sin 2a 4ac  b2 and again it is clear that h(t) is bounded if and only if b/(2a), the real part of the roots of so that as2 + bs + c, is less than or equal to zero. 464 Exercises 7.8
33. Let a function f (t) be defined on (, ) and continuous in a neighborhood of the origin, t = 0. Since (t) = 0 for any t = 0, so does the product f (t)(t). Therefore, f (t)(t) dt =
  f (t)(t) dt for any > 0. (7.52) By the mean value theorem, for any small enough (so that f (t) is continuous on (, )) there exists a point in (, ) such that f (t)(t) dt = f ( )
  (t) dt = f ( )
 (t) dt = f ( ) . Together with (7.52) this yields f (t)(t) dt = f ( ) ,
 for any > 0. Now we take limit, as 0, in both sides. 0 0 lim  f (t)(t) dt = lim [f ( )] . Note that the integral in the lefthand side does not depend on , and so the limit equals to the integral itself. In the righthand side, since belongs to (, ), 0 as 0, and the continuity of f (t) implies that f ( ) converges to f (0), as 0. Combining these observations, we get the required. 35. Following the hint, we solve the initial value problem EIy (4) (x) = L(x  ), y(0) = y (0) = 0, y (0) = A, y (0) = B. Using these initial conditions and Theorem 5 in Section 7.3 with n = 4, we obtain L y (4) (x) (s) = s4 L {y(x)} (s)  sA  B, 465 Chapter 7
and so the Laplace transform of the given equation yields EI s4 L {y(x)} (s)  sA  B = LL {(x  )} (s) = Les . Therefore, L {y(x)} (s) = A B L es + 3+ 4 4 EI s s s s L e A B y(x) = L1 + 3 + 4 (x) 4 EI s s s 3! A L (x  )u(x  ) + L1 = L1 4 EI3! s 2! L A B = (x  )3 u(x  ) + x2 + x3 . 6EI 2 6 2! s3 (x) + B 1 L 3! 3! s4 (x) (7.53) Next, we are looking for A and B such that y (2) = y (2) = 0. Note that, for x > , u(x  ) 1 and so (7.53) becomes y(x) = Differentiating we get y (x) = L (x  ) + A + Bx EI and y (x) = L + B. EI A B L (x  )3 + x2 + x3 . 6EI 2 6 Hence, A and B must satisfy 0 = y (2) = [L/(EI)](2  ) + A + 2B, 0 = y (2) = L/(EI) + B Substitution back into (7.53) yields the solution y(x) = EXERCISES 7.9: L (x  )3 u(x  ) + 3x2  x3 . 6EI A = L/(EI), B = L/(EI). Solving Linear Systems with Laplace Transforms, page 416 1. Let X(s) = L {x} (s), Y (s) = L {y} (s). Applying the Laplace transform to both sides of the given equations yields L {x } (s) = 3X(s)  2Y (s), L {y } (s) = 3Y (s)  2X(s). 466 (7.54) Exercises 7.9
Since L {x } (s) = sX(s)  x(0) = sX(s)  1, L {y } (s) = sY (s)  y(0) = sY (s)  1, the system (7.54) becomes sX(s)  1 = 3X(s)  2Y (s), sY (s)  1 = 3Y (s)  2X(s) (s  3)X(s) + 2Y (s) = 1, 2X(s) + (s  3)Y (s) = 1. (7.55) Subtracting the second equation from the first equation yields (s  5)X(s) + (5  s)Y (s) = 0 So, from the first equation in (7.55) we get (s  3)X(s) + 2X(s) = 1 X(s) = 1 s1 x(t) = L1 1 s1 (t) = et . X(s) = Y (s). Since Y (s) = X(s), y(t) = x(t) = et . 3. Let Z(s) = L {z} (s), W (s) = L {w} (s). Using the initial conditions we conclude that L {z } (s) = sZ(s)  z(0) = sZ(s)  1, L {w } (s) = sW (s)  w(0) = sW (s). Using these equations and taking the Laplace transform of the equations in the given system, we obtain [sZ(s)  1] + [sW (s)] = Z(s)  W (s), sZ(s)  1]  [sW (s)] = Z(s)  W (s) Subtracting equations yields 2sW (s) = 0 W (s) = 0 w(t) = L1 {0} (t) 0. (s  1)W (s) + (s + 1)W (s) = 1, (s  1)W (s)  (s  1)W (s) = 1. (7.56) Substituting W (s) into either equation in (7.56), we obtain (s  1)Z(s) = 1 Z(s) = 1 s1 z(t) = L1 1 s1 (t) = et . 467 Chapter 7
5. Denote X(s) = L {x} (s), Y (s) = L {y} (s). The Laplace transform of the given equations yields L {x } (s) = Y (s) + L {sin t} (s), L {y } (s) = X(s) + 2L {cos t} (s), which becomes sX(s)  2 = Y (s) + 1/(s2 + 1), sY (s) = X(s) + 2s/(s2 + 1) sX(s)  Y (s) = (2s2 + 3)/(s2 + 1), X(s) + sY (s) = 2s/(s2 + 1) after expressing L {x } and L {y } in terms of X(s) and Y (s). Multiplying the second equation by s and adding the result to the first equation, we get s2  1 Y (s) = 4s2 + 3 s2 + 1 Y (s) = 4s2 + 3 . (s  1)(s + 1)(s2 + 1) Since the partial fractions decomposition for Y (s) is 4s2 + 3 7/4 7/4 1/2 =  + 2 , (s  1)(s + 1)(s2 + 1) s1 s+1 s +1 taking the inverse Laplace transform yields y(t) = L1 7/4 1/2 7/4  + 2 s1 s+1 s +1 (t) = 7 t 7 t 1 e  e + sin t . 4 4 2 From the second equation in the original system, x(t) = y  2 cos t = 7 t 7 t 3 e + e  cos t . 4 4 2 7. We will first write this system without using operator notation. Thus, we have x  4x + 6y = 9e3t , x  y + y = 5e3t . (7.57) By taking the Laplace transform of both sides of both of these differential equations and using the linearity of the Laplace transform, we obtain L {x } (s)  4X(s) + 6Y (s) = 9/(s + 3) , X(s)  L {y } (s) + Y (s) = 5/(s + 3) , 468 (7.58) Exercises 7.9
where X(s) and Y (s) are the Laplace transforms of x(t) and y(t), respectively. Using the initial conditions x(0) = 9 and y(0) = 4, we can express L {x } (s) = sX(s)  x(0) = sX(s) + 9, L {y } (s) = sY (s)  y(0) = sY (s)  4. Substituting these expressions into the system given in (7.58) and simplifying yields (s  4)X(s) + 6Y (s) = 9 + 9s  18 9 = , s+3 s+3 5 4s  7 X(s) + (s + 1)Y (s) = 4 + = . s+3 s+3 By multiplying the second equation above by (s  4), adding the resulting equations, and simplifying, we obtain s2  5s + 10 Y (s) = 4s2  18s  46 (4s + 7)(s  4) 9s  18 + = s+3 s+3 s+3 2 4s  18s  46 . Y (s) = (s + 3)(s2  5s + 10) Note that the quadratic s2  5s + 10 = (s  5/2)2 + 15/4 is irreducible. The partial fractions decomposition yields Y (s) = 46s  334 22 1 + 2 + 15/4 17 (s  5/2) s+3 s  5/2 (s  5/2)2 + 15/4 1 46 = 17 and so 146 15  5 15/2 (s  5/2)2 + 15/4 + 22 1 , s+3 46 5t/2 e y(t) = L1 {Y (s)} (t) = cos 17 15t 2 146 15 5t/2 e sin  85 15t 2 + 22 3t e . 17 From the second equation in the system (7.57) above, we find that 15t 115 5t/2 3t 3t e x(t) = 5e + y (t)  y(t) = 5e + cos 17 2 15t 23 15 73 15 219 5t/2 5t/2 + e sin cos  e  17 17 2 17 15t 2  66 3t e 17 469 Chapter 7
150 5t/2 e =  cos 17 15t 2 334 15 5t/2 e sin  85 15t 2  3 3t e . 17 9. Taking the Laplace transform of both sides of both of these differential equations yields the system L {x } (s) + X(s) + 2L {y } (s) = 0, 3L {x } (s)  3X(s) + 2L {y } (s) + 4Y (s) = 0, where X(s) = L {x} (s), Y (s) = L {y} (s). Using the initial conditions x(0) = 2, x (0) = 7 and y(0) = 4, y (0) = 9, we see that L {x } (s) = s2 X(s)  sx(0)  x (0) = s2 X(s)  2s + 7, L {y } (s) = sY (s)  y(0) = sY (s)  4, L {y } (s) = s2 Y (s)  sy(0)  y (0) = s2 Y (s)  4s + 9. Substituting these expressions into the system given above yields [s2 X(s)  2s + 7] + X(s) + 2 [sY (s)  4] = 0, 3 [s2 X(s)  2s + 7]  3X(s) + 2 [s2 Y (s)  4s + 9] + 4Y (s) = 0, which simplifies to (s2 + 1) X(s) + 2sY (s) = 2s + 1, 3 (s2 + 1) X(s) + 2 (s2 + 2) Y (s) = 2s + 3. function X(s). Thus, we obtain 2s2 + 6s + 4 Y (s) = 8s + 6 Y (s) = 5 1 4s + 3 =  , (s + 2)(s + 1) s+2 s+1 (7.59) Multiplying the first equation by 3 and adding the two resulting equations eliminates the where we have factored the expression 2s2 + 6s + 4 and used the partial fractions expansion. Taking the inverse Laplace transform, we obtain y(t) = L1 {Y (s)} (t) = 5L1 1 s+2 (t)  L1 1 s+1 (t) = 5e2t  et . To find the solution x(t), we again examine the system given in (7.59) above. This time we will eliminate the function Y (s) by multiplying the first equation by s2 + 2 and the second 470 Exercises 7.9
equation by s and adding the resulting equations. Thus, we have s2 + 3s + 2 s2 + 1 X(s) = 2s3  s2 + s + 2 2s3  s2 + s + 2 X(s) = . (s + 2)(s + 1)(s2 + 1) Expressing X(s) in a partial fractions expansion, we find that X(s) = and so 4 1 s   2 s+2 s+1 s +1 Hence, the solution to this initial value problem is x(t) = L1 (t) = 4e2t  et  cos t. 1 s 4   2 s+2 s+1 s +1 x(t) = 4e2t  et  cos t and y(t) = 5e2t  et . 11. Since L {x } (s) = sX(s)  x(0) = sX(s) L {y } (s) = sY (s)  y(0) = sY (s) , applying the Laplace transform to the given equations yields sX(s) + Y (s) = L {1  u(t  2)} (s) = X(s) + sY (s) = L {0} (s) = 0 . From the second equation, X(s) = sY (s). Substituting this into the first equation, we eliminate X(s) and obtain s2 Y (s) + Y (s) = 1  e2s s 2s 1e = 1  e2s Y (s) = s(1  s2 ) 1  e2s 1 e2s  = , s s s and 1/2 1/2 1   . s s1 s+1 Using now the linear property of the inverse Laplace transform and formula (6) on page 387, we get y(t) = L1 1/2 1/2 1   s s1 s+1 (t)  L1 1/2 1/2 1   s s1 s+1 (t  2)u(t  2) 471 Chapter 7
= 1 et + et et2 + e(t2) u(t  2).  1 2 2 Since, from the second equation in the original system, x = y , we have x(t) =  1  et + et et2 + e(t2)  1 u(t  2) 2 2 et2  e(t2) et  et  = u(t  2). 2 2 13. Since, by formula (8) on page 387 of the text, L {(sin t)u(t  )} (s) = es L {sin(t + )} (s) = es L { sin t} (s) =  applying the Laplace transform to the given system yields L {x } (s)  L {y } (s) = L {(sin t)u(t  )} (s), L {x} (s) + L {y } (s) = L {0} (s) [sX(s)  1]  [sY (s)  1] =  X(s) + [sY (s)  1] = 0, es , s2 + 1 es , s2 + 1 where we have used the initial conditions, x(0) = 1 and y(0) = 1, and Theorem 4, Section 7.3, to express L {x } (s) and L {y } (s) in terms of X(s) = L {x} (s) and Y (s) = L {y} (s). The above system simplifies to X(s)  Y (s) =  es , s(s2 + 1) X(s) + sY (s) = 1. From the second equation, X(s) = 1  sY (s), and with this substitution the first equation becomes 1  sY (s)  Y (s) =  es 1 es es 1 Y (s) = 1 + = + . s(s2 + 1) s(s2 + 1) s + 1 s + 1 s(s + 1)(s2 + 1) Using partial fractions we express Y (s) = 472 1 1/2 (1/2)s 1/2 1 + es   2  2 s+1 s s+1 s +1 s +1 Exercises 7.9
and so y(t) = et + 1  1 (t) e  2 1 = et + 1  e(t) + 2 1 1 cos(t  )  sin(t  ) u(t  ) 2 2 1 1 cos t + sin t u(t  ). 2 2 Finally, x(t) = y (t) = et  1 (t) 1 1  sin t + cos t u(t  ). e 2 2 2 15. First, note that the initial conditions are given at the point t = 1. Thus, for the Laplace transform method, we have to shift the argument to get zero initial point. Let us denote u(t) := x(t + 1) The chain rule yields u (t) = x (t + 1)(t + 1) = x (t + 1), v (t) = y (t + 1)(t + 1) = y (t + 1). and v(t) := y(t + 1). In the original system, we substitute t + 1 for t to get x (t + 1)  2y(t + 1) = 2, x (t + 1) + x(t + 1)  y (t + 1) = (t + 1)2 + 2(t + 1)  1, and make u and v substitution. This yields u (t)  2v(t) = 2, u (t) + u(t)  v (t) = (t + 1)2 + 2(t + 1)  1 = t2 + 4t + 2 with initial conditions u(0) = 1, v(0) = 0. Taking the Laplace transform and using formula (2) on page 361 of the text, we obtain the system [sU(s)  1]  2V (s) = 2 , s 4 2 2 + 2+ , 3 s s s [sU(s)  1] + U(s)  sV (s) = where U(s) = L {u} (s), V (s) = L {v} (s). Expressing U(s) = 2 2V (s) 1 + 2+ s s s 473 Chapter 7
from the first equation and substituting this into the second equation, we obtain 2V (s) 2 2 1 4 2 2 + 2V (s) + + 2+  sV (s) = 3 + 2 + , s s s s s s s which yields 1 2 2 1 U(s) = 3 + 2 + . 2 s s s s Applying now inverse Laplace transforms yields V (s) = u(t) = t2 + 2t + 1 = (t + 1)2 , Finally, x(t) = u(t  1) = t2 and y(t) = v(t  1) = t  1. v(t) = L1 1 s2 (t) = t. 17. As in Problem 15, first we make a shift in t to move the initial conditions to t = 0. Let u(t) := x(t + 2) and v(t) := y(t + 2). With t replaced by t + 2, the original system becomes x (t + 2) + x(t + 2)  y (t + 2) = 2tet , x (t + 2)  x (t + 2)  2y(t + 2) = et or u (t) + u(t)  v (t) = 2tet , u (t)  u (t)  2v(t) = et , u(0) = 0, with u (0) = 1, v(0) = 1. Applying the Laplace transform to these equations and expressing L {u }, L {u }, and L {v } in terms of U = L {u} and V = L {v} (see formula (4) on page 362 of the text, we obtain [sU(s)] + U(s)  [sV (s)  1] = 2L tet (s) = s2 U(s)  1  [sU(s)]  2V (s) =  1 . s1 2 , (s  1)2 We multiply the first equation by 2, the second equation by s, and subtract the resulting equations in order to eliminate V (s). Thus we get s(s2  s)  2(s + 1) U(s) = s  474 4 s  +2 s  1 (s  1)2 Exercises 7.9 s3  s2  2s  2 U(s) = s3  s2  2s  2 (s  1)2 U(s) = 1 . (s  1)2 The inverse Laplace transform then yields u(t) = L1 1 (s  1)2 (t) = tet x(t) = u(t  2) = (t  2)et2 . We find y(t) from the second equation in the original system. tet2  (t  1)et2 + et2 x (t)  x (t) + et2 = = et2 . y(t) = 2 2 19. We first take the Laplace transform of both sides of all three of these equations and use the initial conditions to obtain a system of equations for the Laplace transforms of the solution functions: sX(s) + 6 = 3X(s) + Y (s)  2Z(s), sY (s)  2 = X(s) + 2Y (s) + Z(s), sZ(s) + 12 = 4X(s) + Y (s)  3Z(s). Simplifying yields (s  3)X(s)  Y (s) + 2Z(s) = 6, X(s) + (s  2)Y (s)  Z(s) = 2, 4X(s)  Y (s) + (s + 3)Z(s) = 12. To solve this system, we will use substitution to eliminate the function Y (s). Therefore, we solve for Y (s) in the first equation in (7.60) to obtain Y (s) = (s  3)X(s) + 2Z(s) + 6. Substituting this expression into the two remaining equations in (7.60) and simplifying yields (s2  5s + 7)X(s) + (2s  5)Z(s) = 6s + 14, (s + 1)X(s) + (s + 1)Z(s) = 6. (7.61) (7.60) Next we will eliminate the function X(s) from the system given in (7.61). To do this we can either multiply the first equation by (s + 1) and the second by (s2  5s + 7) and add, or we can solve the last equation given in (7.61) for X(s) to obtain X(s) = Z(s) + 6 , s+1 (7.62) 475 Chapter 7
and substitute this into the first equation in (7.61). By either method we see that Z(s) = 12s2 + 38s  28 12s2 + 38s  28 = . (s + 1)(s2  3s + 2) (s + 1)(s  2)(s  1) Now, Z(s) has the partial fraction expansion Z(s) = 1 13 + . s+1 s1 Therefore, by taking inverse Laplace transforms of both sides of this equation, we obtain z(t) = L1 {Z(s)} (t) = L1 1 13 + s+1 s1 (t) = 13et + et . To find X(s), we will use equation (7.62) and the expression found above for Z(s). Thus, we have X(s) = Z(s) + 13 1 6 7 1 6 = + + = + s+1 s+1 s1 s+1 s+1 s1 7 1 x(t) = L1 {X(s)} (t) = L1 + (t) = 7et + et . s+1 s1 To find y(t), we could substitute the expressions that we have already found for X(s) and Z(s) into the Y (s) = (s  3)X(s) + 2Z(s) + 6, which we found above, or we could return to the original system of differential equations and use x(t) and z(t) to solve for y(t). For the latter method, we solve the first equation in the original system for y(t) to obtain y(t) = x (t)  3x(t) + 2z(t) = 7et + et + 21et  3et  26et + 2et = 2et . Therefore, the solution to the initial value problem is x(t) = 7et + et , y(t) = 2et , z(t) = 13et + et . 21. We refer the reader to the discussion in Section 5.1 in obtaining the system (1) on page 242 of the text governing interconnected tanks. All the arguments provided remain in force except for the one affected by the new "valve condition", which the formula for the input rate for 476 Exercises 7.9
the tank A. In Section 5.1, just fresh water was pumped into the tank A and so there was no salt coming from outside of the system into the tank A . Now we have more complicated rule: the incoming liquid is fresh water for the first 5 min, but then it changes to a solution having a concentration 2 kg/L. This solution contributes additional 2 (kg/L) 6 (L/min) = 12 (kg/min) to the input rate into the tank A. Thus, from the valve, we have 0, t < 5, 12, t>5 = 12u(t  5) (kg/min) of salt coming to the tank A. With this change, the system (1) in the text becomes x = x/3 + y/12 + 12u(t  5), y = x/3  y/3. (7.63) Also, we have the initial conditions x(0) = x0 = 0, y(0) = y0 = 4. Let X := L {x} and Y := L {y}. Taking the Laplace transform of both equations in the system above, we get 1 1 Y (s) + 12L {u(t  5)} (s), L {x } (s) =  X(s) + 3 12 1 1 L {y } (s) = X(s)  Y (s). 3 3 Since L {u(t  5)} (s) = e5s /s and L {x } (s) = sX(s)  x(0) = sX(s), L {y } (s) = sY (s)  y(0) = sY (s)  4, we obtain 1 12e5s 1 Y (s) + , sX(s) =  X(s) + 3 12 s 1 1 sY (s)  4 = X(s)  Y (s) 3 3 which simplifies to 144e5s , s X(s) + (3s + 1)Y (s) = 12. 4(3s + 1)X(s)  Y (s) = 477 Chapter 7
From the second equation in this system, we have X(s) = (3s + 1)Y (s)  12. Substitution into the first equation yields 144e5s 4(3s + 1) [(3s + 1)Y (s)  12]  Y (s) = s 144e5s . 4(3s + 1)2  1 Y (s) = 48(3s + 1) + s Note that 4(3s + 1)2  1 = [2(3s + 1) + 1] [2(3s + 1)  1] = (6s + 3)(6s + 1) = 36 s + Therefore, Y (s) = 4e5s 4(3s + 1) + 3(s + 1/2)(s + 1/6) s(s + 1/2)(s + 1/6) 48 2 24 72 2 + + e5s +  = , (s + 1/2) (s + 1/6) s s + 1/2 s + 1/6 1 2 s+ 1 . 6 where we have applied the partial fractions decomposition. Taking the inverse Laplace transform and using Theorem 8 in Section 7.6 for the inverse Laplace transform of the term having the exponential factor, we get y(t) = 2L1 1 (s + 1/2) + 48L1 (t) + 2L1 1 s 1 (t) (s + 1/6) 1 + 24L1  72L1 s + 1/2 1 s + 1/6 (t  5)u(t  5) = 2et/2 + 2et/6 + 48 + 24e(t5)/2  72e(t5)/6 u(t  5). From the second equation in (7.63), after some algebra, we find x(t). x(t) = 3y (t) + y = et/2 + et/6 + 48  12e(t5)/2  36e(t5)/6 u(t  5). 23. Recall that Kirchhoff's voltage law says that, in an electrical circuit consisting of an inductor of L H, a resistor of R , a capacitor of C F, and a voltage source of E V, EL + ER + EC = E, 478 (7.64) Exercises 7.9
where EL , ER , and EC denote the voltage drops across the inductor, resistor, and capacitor, respectively. These voltage grops are given by EL = L dI , dt ER := RI, EC := q , C (7.65) where I denotes the current passing through the correspondent element. Also, Kirchhoff's current law states that the algebraic sum of currents passing through any point in an electrical network equals to zero. The electrical network shown in Figure 7.28 consists of three closed circuits: loop 1 through the battery, R1 = 2 resistor, L1 = 0.1 H inductor, and L2 = 0.2 H inductor; loop 2 through the inductor L1 and R2 = 1 resistor; loop 3 through the battery, resistors R1 and R2 , and inductor L2 . We apply Kirchhoff's voltage law (7.64) to two of these loops, say, the loop 1 and the loop 2, and (since the equation obtained from Kirchhoff's voltage law for the third loop is a linear combination of the other two) Kirchhoff's current law to one of the junction points, say, the upper one. Thus, choosing the clockwise direction in the loops and using formulas (7.65), we obtain Loop 1: ER1 + EL1 + EL2 = E Loop 2: EL1 + ER2 = 0 0.1I3  I2 = 0 2I1 + 0.1I3 + 0.2I1 = 6; with the negative sign due to the counterclockwise direction of the current I2 in this loop; Upper junction point: I1  I2  I3 = 0. Therefore, we have the following system for the currents I1 , I2 , and I3 : 2I1 + 0.1I3 + 0.2I1 = 6, 0.1I3  I2 = 0, I1  I2  I3 = 0 479 (7.66) Chapter 7
with initial conditions I1 (0) = I2 (0) = I3 (0) = 0. Let I1 (s) := L {I1 } (s), I2 (s) := L {I2 } (s), and I3 (s) := L {I3 } (s). Using the initial conditions, we conclude that L {I1 } (s) = sI1 (s)  I1 (0) = sI1 (s), L {I3 } (s) = sI3 (s)  I3 (0) = sI3 (s). Using these equations and taking the Laplace transform of the equations in (7.66), we come up with (0.2s + 2)I1 (s) + 0.1sI3(s) = 0.1sI3 (s)  I2 (s) = 0, I1 (s)  I2 (s)  I3 (s) = 0 Expressing I2 (s) = 0.1sI3 (s) from the second equation and substituting this into the third equation, we get I1 (s)  0.1sI3(s)  I3 (s) = 0 I1 (s) = (0.1s + 1)I3 (s). 6 , s The latter, when substituted into the first equation, yields (0.2s + 2)(0.1s + 1)I3 (s) + 0.1sI3 (s) = 2(0.1s + 1)2 + 0.1s I3 (s) = I3 (s) = 6 s 6 s 6 300 = . s[2(0.1s + 1)2 + 0.1s] s(s + 20)(s + 5) We use the partial fractions decomposition to find that 1 4 3  I3 (s) = + s s + 20 s + 5 and so 3 1 4 I3 (t) = L1 +  (t) = 3 + e20t  4e5t . s s + 20 s + 5 Now we can find I2 (t) using the second equation in (7.66). I2 (t) = 0.1I3 (t) = 0.1 3 + e20t  4e5t Finally, the third equation in (7.66) yields I1 (t) = I2 (t) + I3 (t) = 3  e20t  2e5t . 480 = 2e20t + 2e5t . Review Problems
REVIEW PROBLEMS: page 418 1. By the definition of Laplace transform, 2 L {f } (s) =
0 est f (t) dt =
0 est (3) dt +
2 est (6  t) dt. For the first integral, we have
2 est (3) dt =
0 3est s t=2 =
t=0 3(1  e2s ) . s The second integral is an improper integral. Using integration by parts, we obtain M M st t=M st e e est (6  t) dt = lim est (6  t) dt = lim (6  t)  (1)dt M M s t=2 s
2 2 2 = = Thus M lim 4e2s (6  M)esM e  + 2 s s s 4e s
2s st t=M t=2 M lim  (6  M)e s sM + e sM s2  e2s 4e2s e2s =  2 . s2 s s 1 1  2 s s L {f } (s) = 3(1  e2s ) 4e2s e2s 3 +  2 = + e2s s s s s . 3. From Table 7.1 on page 358 of the text, using the formula for the Laplace transform of eat tn with n = 2 and a = 9, we get L t2 e9t (s) = 2! 2 = . 3 [s  (9)] (s + 9)3 5. We use the linearity of the Laplace transform and Table 7.1 to obtain L e2t  t3 + t2  sin 5t (s) = L e2t (s)  L t3 (s) + L t2 (s)  L {sin 5t} (s) 3! 2! 6 5 1 2 5 1  4+ 3 2  4+ 3 2 . = = 2 s2 s s s +5 s2 s s s + 25 481 Chapter 7
7. We apply Theorem 6 in Section 7.3 and obtain L {t cos 6t} (s) =  s2  36 d (s2 + 36)  s(2s) d s = 2 . = L {cos 6t} (s) =  ds ds s2 + 62 (s2 + 36)2 (s + 36)2 9. We apply formula (8), Section 7.6, on page 387 of the text and the linear property of the Laplace transform to get L t2 u(t  4) (s) = e4s L (t + 4)2 (s) = e4s L t2 + 8s + 16 (s) 8 16 4 8 2 1 + 2+ = e4s = 2e4s 3 + 2 + . 3 s s s s s s 11. Using the linearity of the inverse Laplace transform and Table 7.1 we find L1 7 (s + 3)3 (t) = 7 1 L 2! 2! [s  (3)]3 (t) = 7 2 3t te . 2 13. We apply partial fractions to find the inverse Laplace transform. Since the quadratic polynomial s2 + 4s + 13 = (s + 2)2 + 32 is irreducible, the partial fraction decomposition for the given function has the form 4s2 + 13s + 19 A B(s + 2) + C(3) . = + (s  1)(s2 + 4s + 13) s1 (s + 2)2 + 32 Clearing fractions yields 4s2 + 13s + 19 = A[(s + 2)2 + 32 ] + [B(s + 2) + C(3)](s  1) . With s = 1, this gives 36 = 18A or A = 2. Substituting s = 2, we get 9 = 9A  9C Finally, with s = 0, we compute 19 = 13A + (2B + 3C)(1) Thus B = 2. C = A  1 = 1. 2 2(s + 2) + (1)(3) 4s2 + 13s + 19 = + , 2 + 4s + 13) (s  1)(s s1 (s + 2)2 + 32 482 Review Problems
and so L1 4s2 + 13s + 19 (s  1)(s2 + 4s + 13) (t) = 2L1 s+2 (t) (s + 2)2 + 32 3 (t) +L1 (s + 2)2 + 32 = 2et + 2e2t cos 3t + e2t sin 3t . (t) + 2L1 1 s1 15. The partial fraction decomposition for the given function has the form A C A(s + 2) + B(s + 1)(s + 2) + C(s + 1)2 B 2s2 + 3s  1 = + = . + (s + 1)2 (s + 2) (s + 1)2 s + 1 s + 2 (s + 1)2 (s + 2) Thus 2s2 + 3s  1 = A(s + 2) + B(s + 1)(s + 2) + C(s + 1)2 . We evaluate both sides of this equation at s = 2, 1, and 0. This yields s = 2 : 2(2)2 + 3(2)  1 = C(2 + 1)2 s = 1 : 2(1)2 + 3(1)  1 = A(1 + 2) s=0: Therefore, L1 2s2 + 3s  1 (s + 1)2 (s + 2) (t) = L1 2 1 1 + + 2 (s + 1) s+1 s+2 (t) = 2tet + et + e2t . 1 = 2A + 2B + C C = 1, A = 2, B = (1  2A  C)/2 = 1. 17. First we apply Theorem 8 in Section 7.6 to get L
1 e2s (4s + 2) (s  1)(s + 2) (t) = L1 4s + 2 (s  1)(s + 2) (t  2)u(t  2). (7.67) Applying partial fractions yields 2 2 4s + 2 = + (s  1)(s + 2) s1 s+2 Therefore, it follows from (7.67) that L
1 L1 4s + 2 (s  1)(s + 2) (t) = 2et + 2e2t . e2s (4s + 2) (s  1)(s + 2) (t) = 2et2 + 2e2(t2) u(t  2) = 2et2 + 2e42t u(t  2). 483 Chapter 7
19. Applying the Laplace transform to both sides of the given equation and using the linearity of the Laplace transform yields L {y  7y + 10y} (s) = L {y } (s)  7L {y } (s) + 10L {y} (s) = 0. By Theorem 5 in Section 7.3, L {y } (s) = sL {y} (s)  y(0) = sL {y} (s), L {y } (s) = s2 L {y} (s)  sy(0)  y (0) = s2 L {y} (s) + 3, where we have used the initial conditions, y(0) = 0 and y (0) = 3. Substituting these expressions into (7.68), we get s2 L {y} (s) + 3  7 [sL {y} (s)] + 10L {y} (s) = 0 Thus y(t) = L1 1 1  s2 s5 (t) = L1 1 s2 (t)  L1 1 s5 (t) = e2t  e5t . (s2  7s + 10)L {y} (s) + 3 = 0 3 3 1 1 L {y} (s) = 2 = =  . s  7s + 10 (s  2)(s  5) s2 s5 (7.68) 21. Let Y (s) := L {y} (s). Taking the Laplace transform of the given equation and using properties of the Laplace transform, we obtain L {y + 2y + 2y} (s) = L t2 + 4t (s) = Since L {y } (s) = sY (s)  y(0) = sY (s), we have s2 Y (s) + 1 + 2 [sY (s)] + 2Y (s) = 484 2 + 4s s3 2 + 4s 2 + 4s  s3 (s2 + 2s + 2)Y (s) = 1= s3 s3 L {y } (s) = s2 Y (s)  sy(0)  y (0) = s2 Y (s) + 1, 2 4 2 + 4s + 2 = . 3 s s s3 Review Problems Y (s) = 2 + 4s  s3 2 + 4s  s3 = 3 . s3 (s2 + 2s + 2) s [(s + 1)2 + 12 ] The partial fraction decomposition for Y (s) has the form 2 + 4s  s3 B C D(s + 1) + E(1) A . = 3+ 2+ + 3 [(s + 1)2 + 12 ] s s s s (s + 1)2 + 12 Clearing fractions, we obtain 2 + 4s  s3 = A[(s + 1)2 + 1] + Bs[(s + 1)2 + 1] + Cs2 [(s + 1)2 + 1] + [D(s + 1) + E]s3 . Comparing coefficients at the corresponding power of s in both sides of this equation yields s0 : 2 = 2A s : 4 = 2A + 2B s2 : 0 = A + 2B + 2C s4 : 0 = C + D
3 1 A = 1, B = (4  2A)/2 = 1, C = (A + 2B)/2 = 3/2, D = C = 3/2, E = 1  B  2C  D = 1/2. s : 1 = B + 2C + D + E Therefore, Y (s) = (3/2)(s + 1) 1 3/2 (1/2)(1) 1 + + 2  3 2 + 12 s s s (s + 1) (s + 1)2 + 12 3 3 t2 1 y(t) = L1 {Y (s)} (t) = + t  + et cos t  et sin t . 2 2 2 2 es . s 23. By formula (4) in Section 7.6, L {u(t  1)} (s) = Thus, applying the Laplace transform to both sides of the given equation and using the initial conditions, we get L {y + 3y + 4y} (s) = es s s2 Y (s)  1 + 3 [sY (s)] + 4Y (s) = Y (s) = 1 es + s2 + 3s + 4 s(s2 + 3s + 4) es s 485 Chapter 7 Y (s) = 1 1 , + es (s + 3/2)2 + ( 7/2)2 s[(s + 3/2)2 + ( 7/2)2 ] where Y (s) := L {y} (s). To apply the inverse Laplace transform, we need the partial fraction decomposition of the last fraction above. A B(s + 3/2) + C( 7/2) = + . s s[(s + 3/2)2 + ( 7/2)2 ] (s + 3/2)2 + ( 7/2)2 1 Solving for A, B, and C yields A= Therefore, (1/4)(s + 3/2) (3/4 7)( 7/2) 1 s 1/4  +e  Y (s) = s (s + 3/2)2 + ( 7/2)2 (s + 3/2)2 + ( 7/2)2 (s + 3/2)2 + ( 7/2)2 and the inverse Laplace transform gives y(t) = L1 1 (s + 3/2)2 + ( 7/2)2 +L1 2 = e3t/2 sin 7 (t) (t  1)u(t  1) 1 , 4 1 B= , 4 3 C= . 4 7 (1/4)(s + 3/2) (3/4 7)( 7/2) 1/4   s (s + 3/2)2 + (7/4) (s + 3/2)2 + (7/4) 7t 2 1 1 3(t1)/2  e cos + 4 4 7(t  1) 2 3  e3(t1)/2 sin 4 7
7(t  1) 2 u(t  1). 25. Let Y (s) := L {y} (s). Then, from the initial conditions, we have L {y } (s) = sY (s)  y(0) = sY (s), Moreover, Theorem 6 in Section 7.3 yields d d L {y } (s) =  [sY (s)] = sY (s)  Y (s), ds ds d d 2 s Y (s) = s2 Y (s)  2sY (s). L {ty } (s) =  L {y } (s) =  ds ds L {ty } (s) =  486 L {y } (s) = s2 Y (s)  sy(0)  y (0) = s2 Y (s). Review Problems
Hence, applying the Laplace transform to the given equation and using the linearity of the Laplace transform, we obtain L {ty + 2(t  1)y  2y} (s) = L {ty } (s) + 2L {ty } (s)  2L {y } (s)  2L {y} (s) = 0 s2 Y (s)  2sY (s) + 2 [sY (s)  Y (s)]  2 [sY (s)]  2Y (s) = 0 4(s + 1) Y (s) + s(s + 2)Y (s)  4(s + 1)Y (s) = 0 Y (s) = 0. s(s + 2) Separating variables and integrating yields 1 dY 4(s + 1) 1 = ds = 2 + ds Y s(s + 2) s s+2 ln Y  = 2(ln s + ln s + 2) + C c1 eC = 2 , Y (s) = 2 2 s (s + 2) s (s + 2)2 where c1 = 0 is an arbitrary constant. Allowing c1 = 0, we also get the solution Y (s) 0, which was lost in separation of variables. Thus Y (s) = and so y(t) = L1 {Y (s)} (t) = c1 t  1 + te2t + e2t = c t  1 + te2t + e2t , 4 s2 (s 1 c1 c1 1 1 1 =  + + 2 2 2 + 2) 4 s s (s + 2) s+2 where c = c1 /4 is an arbitrary constant. 27. Note that the original equation can be written in the form y(t) + t y(t) = e3t . Let Y (s) := L {y} (s). Applying the Laplace transform to both sides of this equation and using Theorem 11 in Section 7.7, we obtain L {y(t) + t y(t)} (s) = Y (s) + L {t} (s)Y (s) = L e3t (s) 1 s2 1 Y (s) + 2 Y (s) = Y (s) = . s s+3 (s + 3)(s2 + 1) 487 Chapter 7
The partial fraction decomposition for Y (s) has the form A Bs + C A(s2 + 1) + (Bs + C)(s + 3) s2 = + 2 = . (s + 3)(s2 + 1) s+3 s +1 (s + 3)(s2 + 1) Thus s2 = A(s2 + 1) + (Bs + C)(s + 3). Evaluating both sides of this equation at s = 3, 0, and 2 yields s = 3 : s=0: s = 2 : Therefore, Y (s) = 3/10 9/10 (1/10)s + 2  2 s+3 s +1 s +1 9 3t 1 3 e + cos t  sin t . y(t) = L1 {Y (s)} (t) = 10 10 10 9 = A(10) 0 = A + 3C A = 9/10, C = A/3 = 3/10, B = (5A + C  4)/2 = 1/10. 4 = 5A  2B + C 29. To find the transfer function, we use formula (15) on page 403 of the text. Comparing given equation with (14), we find that a = 1, b = 5, and c = 6. Thus (15) yields H(s) = as2 1 1 = 2 . + bs + c s  5s + 6 The impulse response function h(t) is defined as L1 {H} (t). Using partial fractions, we see that H(s) = 1 1 1 1 = =   5s + 6 (s  3)(s  2) s3 s2 1 1  h(t) = L1 (t) = e3t  e2t . s3 s2 s2 31. Let X(s) := L {x} (s), Y (s) := L {y} (s). Using the initial condition, we obtain L {x } (s) = sX(s)  x(0) = sX(s), 488 L {y } (s) = sY (s)  y(0) = sY (s). Review Problems
Therefore, applying the Laplace transform to both sides of the equations in the given system yields sX(s) + Y (s) = L {0} (s) = 0, 1  e2s 1 e2s  = . X(s) + sY (s) = L {1  u(t  2)} (s) = s s s Expressing Y (s) = sX(s) from the first equation and substituting this into the second equation, we eliminate Y (s): X(s)  s2 X(s) = Since  1 1 1/2 1/2 =   , s(s  1)(s + 1) s s1 s+1 1  e2s s 1  e2s 1  e2s = . X(s) =  2 s(s  1) s(s  1)(s + 1) the inverse Laplace transform yields x(t) = L1 1 1/2 1/2   (t) s s1 s+1 1/2 1/2 1/2 1/2 1 1   (t)  L1   = L1 s s1 s+1 s s1 s+1 t t t2 (t2) e +e e +e = 1  1 u(t  2) . 2 2 1  e2s (t  2)u(t  2) We now find y(t) from the first equation in the original system. y(t) = x (t) = et  et et2  e(t2)  u(t  2) . 2 2 489 Chapter 7 490 CHAPTER 8: Series Solutions of Differential Equations
EXERCISES 8.1: Introduction: The Taylor Polynomial Approximation, page 430 1. To find Taylor approximations y(0) + y (0) 2 y (0) 3 y (0) x+ x + x + , 1! 2! 3! we need the values of y(0), y (0), y (0), etc. y(0) is provided by the initial condition, y(0) = 1. Substituting x = 0 into the given differential equation, y (x) = x2 + y(x)2 , we obtain y (0) = 02 + y(0)2 = 0 + 12 = 1. Differentiating both sides of (8.1) yields y (x) = 2x + 2y(x)y (x), and so y (0) = 2(0) + 2y(0)y (0) = 0 + 2(1)(1) = 2. Hence y(x) = 1 + 2 1 x + x2 + = 1 + x + x2 + . 1! 2! (8.1) 3. Using the initial condition, y(0) = 0 we substitute x = 0 and y = 0 into the given equation and find y (0). y (0) = sin(0) + e0 = 1. 491 Chapter 8
To determine y (0), we differentiate the given equation with respect to x and substitute x = 0, y = 0, and y = 1 in the formula obtained: y (0) = (sin y + ex ) = (sin y) + (ex ) = y cos y + ex , y (0) = 1 cos 0 + e0 = 2. Similarly, differentiating y (x) and substituting, we obtain y = (y cos y + ex ) = (y cos y) + (ex ) = y cos y + (y ) ( sin y) + ex , y (0) = y (0) cos 0 + (y (0)) ( sin y(0)) + e0 = 2 cos 0 + (1)2 ( sin 0) + 1 = 3. Thus the first three nonzero terms in the Taylor polynomial approximations to the solution of the given initial value problem are y(x) = y(0) + y (0) y (0) 2 y (0) 3 x+ x + x + 1! 2! 3! 1 2 3 1 = 0 + x + x2 + x3 + = x + x2 + x3 + . 1 2 6 2
2 2 5. We need the values of x(0), x (0), x (0), etc. The first two are given by the initial conditions: x(0) = 1, Writing the given equation in the form x (t) = tx(t) we find that x (0) = 0 x(0) = 0 1 = 0. Differentiating (8.2) and substituting t = 0 we conclude that x (t) =  [tx (t) + x(t)] x (t) =  [tx (t) + 2x (t)] x(5) (t) =  [tx (t) + 3x (t)] x(6) (t) =  tx(4) (t) + 4x (t) Therefore, x(t) = 1  492
(4) x (0) = 0. (8.2) x (0) =  [0 x (0) + x(0)] = 1, x(4) (0) =  [0 x (0) + 2x (0)] = 0, x(5) (0) =  [0 x (0) + 3x (0)] = 0, x(6) (0) =  0 x(4) (0) + 4x (0) = 4. 1 3 4 6 t6 t3 t + t + = 1 + + . 3! 6! 6 180 Exercises 8.1
7. We use the initial conditions to find y (0). Writing the given equation in the form y () = y()3 + sin and substituting = 0, y(0) = 0 we get y (0) = y(0)3 + sin 0 = 0. Differentiating the given equation we obtain y = (y ) =  y 3 + (sin ) = 3y 2 y + cos Similarly, we get y (4) = (y ) = 3y 2 y  6y (y )  sin y (4) (0) = 3y(0)2 y (0)  6y(0) (y (0))  sin 0 = 0.
2 2 y (0) = 3y(0)2y (0) + cos 0 = 3(0)2 (0) + 1 = 1. To simplify further computations we observe that since the Taylor expansion for y() has the form 1 3 + , 3! then the Taylor expansion for y()3 must begin with the term (1/3!)3 9 , so that y() = y()3 Hence y (5) =  y 3 y (6) =  y 3 y (7) =  y 3
(3) (4) (5) (k) =0 =0 for k = 0, 1, . . . , 8 .  cos + sin + cos y (5) (0) =  y 3 y (6) (0) =  y 3 y (7) (0) =  y 3 (3) =0 (4) =0 (5) =0  cos 0 = 1,  sin 0 = 0, + cos 0 = 1. Thus, the first three nonzero terms of the Taylor approximations are y() = 1 1 1 1 5 1 7 1 3  5 + 7 + = 3  + + 3! 5! 7! 6 120 5040 493 Chapter 8
9. (a) To construct p3 (x) we need f (1), f (1), f (1), and f (1). Thus we have f (x) = ln x f (x) = x
1 f (1) = ln 1 = 0, f (1) = (1)1 = 1, f (1) = (1)2 = 1, f (1) = 2(1)3 = 2, f (x) = x2 f (x) = 2x3 and so p3 (x) = 0 + 1 2 1 (x  1) + (x  1)2 + (x  1)3 1! 2! 3! (x  1)2 (x  1)3 + . = x1 2 3 (b) To apply formula (6), we first compute f (4) (x) = [f (x)] = 2x3 Thus, the error formula (6) yields ln x  p3 (x) =: e3 (x) = f (4) () 6 4 (x  1)4 (x  x0 )4 = (x  1)4 =  4! 24 4 4 (1.5  1)4 (0.5)4 = ln(1.5)  p3 (1.5) =  4 4 4 4 1 (0.5)4 = = 0.015625 , ln(1.5)  p3 (1.5) 4 64 = 6x4 . where we have used the fact > 1. (c) Direct calculations yield ln(1.5)  p3 (1.5) 0.405465  0.5  (d) See Figure B.51 in the answers of the text. 11. First, we rewrite the given equation in the form y = py  qy + g. 494 (0.5)2 (0.5)3 + 2 3 0.011202 . Exercises 8.1
On the righthand side of this equation, the function y is differentiable (y exists) and the functions y, p, q, and g are differentiable (even twice). Thus we conclude that its lefthand side, y , is differentiable being the product, sum, and difference of differentiable functions. Therefore, y = (y ) exists and is given by y = (py  qy + g) = p y  py  q y  qy + g . Similarly, we conclude that the righthand side of the equation above is a differentiable function since all the functions involved are differentiable (notice that we have just proved the differentiability of y ). Hence, y , its lefthand side is differentiable as well, i.e., (y ) = y (4) does exist. 13. With form k = r = A = 1 and = 10, the Duffing's equation becomes y + y + y 3 = cos 10t or y = y  y 3 + cos 10t. Substituting the initial conditions, y(0) = 0 and y (0) = 1 into the latter equation yields y (0) = y(0)  y(0)3 + cos(10 0) = 0  (0)3 + cos 0 = 1. Differentiating the given equation, we conclude that y = y  y 3 + cos 10t = y  3y 2 y  10 sin 10t, which, at t = 0, gives y (0) = y (0)  3y(0)2y (0)  10 sin(10 0) = 1  3(0)2 (1)  10 sin 0 = 1. Thus, the Taylor polynomial approximations to the solution of the given initial value problem are y(t) = y(0) + y (0) 2 y (0) 3 1 1 y (0) t+ t + t + = t + t2  t3 + . 1! 2! 3! 2 6 15. For the Taylor polynomial p2 (x), we need y(0), y (0), and y (0). We already know y(0) and y (0) from the initial conditions: y(0) = 1 and y (0) = 0. 495 Chapter 8
Expressing y (x) from the given equation yields y (x) =  2y (x) + xy(x) . x (8.3) The formal substitution of x = 0 in (8.3) gives "0/0"indeterminate form. On the other hand, since the differentiability of a function implies its continuity, and we are given that y(x) has derivatives of all orders at x = 0, we conclude that all the derivatives of y(x) are continuous at x = 0. Therefore, y (0) = lim y (x),
x0 and we can find the above limit by applying L'Hospital's rule. Namely, y (0) = lim  2y (x) + xy(x) x0 x [2y (x) + xy(x)] =  lim [2y (x) + xy (x) + y(x)] , =  lim x0 x0 (x) and the last limit can be found by substitution due to the continuity of y(x) and its derivatives at x = 0. Hence, y (0) =  [2y (0) + 0 y (0) + y(0)] = 2y (0)  1. Solving for y (0) yields y (0) = 1/3, and so p2 (x) = y(0) + EXERCISES 8.2: y (0) 2 y (0) x2 x+ x =1 . 1! 2! 6 Power Series and Analytic Functions, page 438 1. Since an = 2n /(n + 1), the ratio test yields lim an+1 2(n+1) /(n + 2) 21 (n + 1) 1 = lim = lim = = L. n /(n + 1) n 2 n an n+2 2 1 = 2. L n So, the radius of convergence is = 496 Exercises 8.2
In this power series, x0 = 1. Hence, the endpoints of the interval of convergence are x1 = x0 + = 1 + 2 = 3, x2 = x0  = 1  2 = 1. At the point x1 , the series becomes n=0 2n (3  1)n = n+1 n=0 1 = n+1 (harmonic series); at the point x2 we have n=0 2n (1  1)n = n+1 n=0 (1)n < n+1 by alternating series test. Therefore, the set of convergence is [1, 3). 3. We will use the ratio test given in Theorem 2 on page 432 of the text to find the radius of convergence for this power series. Since an = n2 /2n , we see that (n + 1)2 /2n+1 (n + 1)2 an+1 = = . an n2 /2n 2n2 Therefore, we have lim 1 1 (n + 1)2 an+1 (n + 1)2 1 lim lim 1 + = = lim = 2 2 n an 2n 2 n n 2 n n
2 n = 1 . 2 Thus, the radius of convergence is = 2. Hence, this power series converges absolutely for x + 2 < 2. That is, for 2 < x + 2 < 2 or  4 < x < 0. We must now check the end points of this interval. We first check the end point 4 or x + 2 = 2 which yields the series n=0 n2 (2)n = 2n (1)n n2 . n=0 This series diverges since the nth term, an = (1)n n2 , does not approach zero as n goes to infinity. (Recall that it is necessary for the nth term of a convergent series to approach zero 497 Chapter 8
as n goes to infinity. But this fact in itself does not prove that a series converges.) Next, we check the end point x = 0 or x + 2 = 2 which yields the series n=0 n2 2n = 2n n2 . n=0 Again, as above, this series diverges. Therefore, this power series converges in the open interval (4, 0) and diverges outside of this interval. 5. With an = 3/n3 , the ratio test gives L = lim n 3/(n + 1)3 = lim 3 n n n + 1 3/n
3 = n n n + 1 lim 3 = 1. Therefore, the radius of convergence is = 1/L = 1. At the points x0 = 2 1, that is, x = 3 and x = 1, we have the series n=0 3 n3 and
n=0 3(1)n , n3 which are known to converge. Therefore, the set of convergence of the given series is the closed interval [1, 3]. 7. By writing a2k x2k = a2k x2 k =
k=0 bk z k , k=0 2 k=0 where bk := a2k and z := x , we obtain a power series centered at the origin. The ratio test then yields the radius of convergence to be 1/L, where L = lim So, the series k k=0 bk z k a2(k+1) bk+1 a2k+2 = lim = lim . k k bk a2k a2k converges for z < 1/L and diverges for z > 1/L. Since z = x2 , 1 L x2 < 1 L 1 x < . L z < Hence, the original series converges for x < 1/ L and diverges for x > 1/ L. By the definition, 1/ L is its radius of convergence. 498 Exercises 8.2
The second statement can be proved in a similar way, since a2k+1 x2k+1 = x a2k+1 x2 k =x
k=0 bk z k , k=0 k=0 where bk := a2k+1 and z := x2 . 9. Since the addition of power series reduces to the addition of the coefficients at the corresponding powers of the variable, we make the following changes in indices of summation. f (x) : n k f (x) = g(x) = k k=0 [1/(k + 1)] x (k+1) k x . k=0 2 , g(x) : n  1 k Therefore, f (x) + g(x) =
k=0 1 xk + k+1 2
k=0 (k+1) k x =
k=0 1 + 2k1 xk . k+1 11. We want to find the product f (x)g(x) of the two series f (x) =
n=0 x2 x3 x4 xn =1+x+ + + + , n! 2 6 24 and g(x) = sin x = k=0 (1)k x3 x5 x7 x2k+1 = x  +  + . (2k + 1)! 6 120 7! Therefore, we have f (x)g(x) = x3 x5 x7 x2 x3 x4 + + + x +  + 2 6 24 6 120 7! 1 1 1 1 1 1 1    + = x + x2 + x3 + x4 + x5 + 2 6 6 6 24 12 120 1 = x + x2 + + x3 + . 3 1+x+ Note that since the radius of convergence for both of the given series is = , the expansion of the product f (x)g(x) also converges for all values of x. 499 Chapter 8
13. Using formula (6) on page 434 of the text, we obtain f (x)g(x) =
n=0 (1)n n x n! (1)n xn
n=0 = 1x+ 1 2 1 3 x  x + 2 6 1  x + x2  x3 + 1 2 (1) x2 + = (1)(1) + [(1)(1) + (1)(1)] x + (1)(1) + (1)(1) + = 1  2x + 15. (a) Let q(x) = obtain 5 2 x + 2 n n=0 an x . Multiplying both sides of the given equation by n=0 xn /n!, we an x
n=0 n n=0 1 n x n! =
n=0 1 n x . 2n n n=0 x /n!. Thus, the righthand side, n=0 xn /2n , is the Cauchy product of q(x) and (b) With cn = 1/2n and bn = 1/n!, formula (6) on page 434 of the text yields: n=0: n=1: n=2: n=3: etc. (c) The system in (b) simplifies to 1 = a0 , 1/2 = a0 + a1 , 1/4 = a0 /2 + a1 + a2 , 1/8 = a0 /6 + a1 /2 + a2 + a3 , . . . 500 a0 = 1 , a1 = 1/2  a0 = 1/2 , a2 = 1/4  a0 /2  a1 = 1/4 , a3 = 1/8  a0 /6  a1 /2  a2 = 1/24 , . . . 1 20 1 21 1 22 1 23 1 = a0 ; 0! 1 1 = c1 = a0 b1 + a1 b0 = a0 + a1 = a0 + a1 ; 1! 0! a0 1 1 1 + a1 + a2 ; = c2 = a0 b2 + a1 b1 + a2 b0 = a0 + a1 + a2 = 2! 1! 0! 2 a0 a1 + + a2 + a3 ; = c3 = a0 b3 + a1 b2 + a2 b1 + a3 b0 = 6 2 = c0 = a0 b0 = a0 Exercises 8.2
Thus, q(x) = 1  17. Since
n 1 1 1 3 x + x2  x + . 2 4 24 lim an+1 (1)n+1 = lim = lim 1 = 1, n n an (1)n by the ratio test, we find the radius of convergence of the given series to be = 1/1 = 1 > 0. Therefore, Theorem 4 of page 434 of the text can be applied. This yields (1 + x)
1 =
n=1 (1) nx n n1 (1 + x) 2 =
n=1 (1)n nxn1 , and the radius of convergence of this series is also = 1. 19. Here we will assume that this series has a positive radius of convergence. Thus, since we have f (x) =
n=0 an xn = a0 + a1 x + a2 x2 + a3 x3 + + an xn + , we can differentiate term by term to obtain f (x) = 0 + a1 + a2 2x + a3 3x + + an nx
2 n1 + =
n=1 an nxn1 . Note that the summation for f (x) starts at zero while the summation for f (x) starts at one. 21. Using the ratio test, we find that the radius of convergence of the given series is = 1 1 = = 1 > 0. n+1 /(1)n  limn (1) 1 Thus, by Theorem 4 on page 434 of the text,
x x g(x) =
0 f (t) dt =
0 x n=0 (1)n tn dt =
n=0 (1) n 0 tn dt =
n=0 (1)n 1 tn+1 n+1 x =
0 n=0 (1)n n+1 x . n+1 501 Chapter 8
On the other hand,
x g(x) =
0 dt = ln(1 + t) 1+t x 0 = ln(1 + x), x (1, 1). 23. Setting k = n  1, we have n = k + 1. Note that k = 0 when n = 1. Hence, substitution into the given series yields n1 nan x
n=1 =
k=0 (k + 1)ak+1 xk . 25. We let n + 1 = k so that n = k  1; when n = 0, then k = 1. Thus, an x
n=0 n+1 =
k=1 ak1 xk . 27. Termwise multiplication yields x
2 n(n + 1)an x =
n=0 n=0 n n(n + 1)an x x =
n=0 n 2 n(n + 1)an xn+2 . Now we can shift the summation index by letting k = n + 2. Then we have n = k  2, n + 1 = k  1, k = 2 when n = 0, and so n(n + 1)an x
n=0 n+2 =
k=2 (k  2)(k  1)ak2xk . By replacing k by n, we obtain the desired form. 29. We need to determine the nth derivative of f (x) at the point x = . Thus, we observe that f (x) = f (0) (x) = cos x f (x) =  sin x f (x) =  cos x f (x) = sin x f (4) (x) = cos x f () = f (0) () = cos = 1, f () =  sin = 0, f () =  cos = 1, f () = sin = 0, f (4) () = cos = 1. Since f (4) (x) = cos x = f (x), the four derivatives given above will be repeated indefinitely. Thus, we see that f (n) () = 0 if n is odd and f (n) () = 1 if n is even (where the signs 502 Exercises 8.2
alternate starting at 1 when n = 0). Therefore, the Taylor series for f about the point x0 = is given by f (x) = 1 + 0 + 1 (1)n+1 (x  )2n 1 (x  )2 + 0  (x  )4 + + + 2! 4! (2n)! =
n=0 (1)n+1 (x  )2n . (2n)! 31. Writing f (x) = 1+x (1  x) + 2x 1 = = 1 + 2x , 1x 1x 1x we can use the power series expansion (3) on page 433 of the text (geometric series) to obtain the desired Taylor series. Thus we have 1 f (x) = 1 + 2x xk = 1 + 2xk+1 . = 1 + 2x 1x k=0 k=0 Shifting the summation index, that is, letting k + 1 = n, yields f (x) = 1 +
k=0 2x k+1 =1+
n=1 2xn . 33. Using the formula cj = f (j) (x0 ) j! for the coefficients of the Taylor series for f (x) about x0 , we find f (x0 ) = x3 + 3x  4 f (x0 ) = 3x + 3 f (x0 ) = 6x f (x) 6 f Therefore, x3 + 3x  4 = 6(x  1) + 3(x  1)2 + (x  1)3 . 503
(j) x=1 2 x=1 x=1 =0 c0 = 0, c1 = 6/1! = 6, c2 = 6/2! = 3, c3 = 6/3! = 1, cj = 0 for j 4. =6 =6 (x) 0 Chapter 8
35. (a) We have 1 1 1 = = , x 1 + (x  1) 1s Since 1/(1  s) = yields the expansion 1 1 = = x 1s which is valid for s = x  1 < 1 0 < x < 2. n=0 where s = (x  1). sn , the substitution s = (x  1) into both sides of this equality s =
n=0 n=0 n [(x  1)] =
n n=0 (1)n (x  1)n , (b) Since the above series has positive radius of convergence = 1, Theorem 4 on page 434 of the text can be applied. Hence, for 0 < x < 2,
x x ln x =
1 1 dt = t
n x (1) (t  1)
n 1 n=0 x n dt =
n=0 n (1) n 1 (t  1)n dt 1 (1) = (t  1)n+1 n+1 n=0 =
1 n=0 (1) (x  1)n+1 = n+1 k=1 (1)k1 (x  1)k . k 37. For n = 0, f (0) (0) := f (0) = 0 by the definition of f (x). To find f (0), we use the definition of the derivative. f (x)  f (0) e1/x = lim . f (0) = lim x0 x0 x0 x that t + when x 0+ and t  when x 0 . Thus we have e1/x t 1 2 = lim tet = lim t2 = lim lim 2 = 0, t t e t 2tet x0 x where we applied L'Hospital's rule to the indeterminate form /. Therefore, the limit in (8.4) exists and equals 0. For any x = 0, f (x) = e1/x 504
2 2 2 (8.4) We compute lefthand and righthand side limits by making the substitution t = 1/x. Note = e1/x 2  1 x2 = 2 1/x2 e . x3 Exercises 8.3
Next, we proceed by induction. Assuming that, for some n 1, f (n) (0) = 0 and f (n) (x) = p where p(t) is a polynomial in t, we show that f (n+1) (0) = 0 and f (n+1) (x) = q 1 x e1/x ,
2 1 x e1/x , 2 x = 0, x = 0, where q(t) is a polynomial in t. This will imply that f (n) (0) = 0 for all n 0. Indeed, the substitution t = 1/x in the onesided limits yields f (n) (x)  f (n) (0) p(1/x)e1/x tp(t) r(t) = lim = lim lim = lim t2 , t2 t e t e x0 x0 x0 x where r(t) = a0 tk + + ak is a polynomial. Applying the L'Hospital's rule k times, we obtain r(t) = 2 t et lim r (t) r (t) = lim 2 2 t (et ) t 2tet r (t) k!a0 = lim 2 = = lim 2 = 0. t (4t2 + 2)et t (2k tk + )et lim
2 Since both onesided limits exist and are equal, the regular limit exists and equals to the same number. That is, f (n+1) (0) = lim For any x = 0, f (n+1) (x) = = p p 1 x e1/x 1 x
2 f (n) (x)  f (n) (0) = 0. x0 x0 1 1 1 1 2 2 e1/x + p e1/x  2 x x x x 1 1/x2 2 1/x2 =q , e e x3 x = p 1 x 1 +p x2 where q(t) = p (t)t2 + p(t)2t3 . EXERCISES 8.3: Power Series Solutions to Linear Differential Equations, page 449 1. Dividing the given equation by (x + 1) yields y  x2 3 y + y = 0. x+1 x+1 505 Chapter 8
x2 3 , q(x) = . x+1 x+1 These are rational functions and so they are analytic everywhere except, perhaps, at zeros p(x) =  of their denominators. Solving x + 1 = 0, we find that x = 1, which is a point of infinite discontinuity for both functions. Consequently, x = 1 is the only singular point of the given equation. 3. Writing the equation in standard form yields y + The coefficients p() = are quotients of analytic functions, and so they are analytic everywhere except zeros = 2 of the denominator where they have infinite discontinuities. Hence, the given equation has two singular points, = 2. 5. In standard form, the equation becomes x + Hence p(t) = t2 t+1 t+1 = , t2 (t + 1)(t  2) q(t) =  t2 t2 t2 = . t2 (t + 1)(t  2) t2 t+1 t2 x  2 x = 0. t2 t t2 2 2 2 and q() = sin 2  2 2 2 sin y + 2 y = 0. 2 2 Thus we see that The point t = 1 is a removable singularity for p(t) since, for t = 1, we can cancel (t+1)term in the numerator and denominator, and so p(t) becomes analytic at t = 1 if we set p(1) := lim p(t) = lim
t1 t1 1 1 = . t2 3 At the point t = 2, p(t) has infinite discontinuity. Thus p(t) is analytic everywhere except t = 2. Similarly, q(t) is analytic everywhere except t = 1. Therefore, the given equation has two singular points, t = 1 and t = 2. 506 Exercises 8.3
7. In standard form, this equation becomes y + cos x y = 0. sin x Thus, p(x) = 0 and, hence, is analytic everywhere. We also see that q(x) = cos x = cot x. sin x Note that q(x) is the quotient of two functions (cos x and sin x) that each have a power series expansion with a positive radius of convergence about each real number x. Thus, according to page 434 of the text, we see that q(x) will also have a power series expansion with a positive radius of convergence about every real number x as long as the denominator, sin x, is not equal to zero. Since the cotangent function is at integer multiples of , we see that q(x) is not defined and, therefore, not analytic at n. Hence, the differential equation is singular only at the points n, where n is an integer. 9. Dividing the differential equation by sin , we get y  ln y = 0. sin Thus, p() 0 and q() =  ln / sin . The function q() is not defined for 0 because of the logarithmic term and has infinite discontinuities at positive zeros of the denominator. Namely, sin = 0 = k, k = 1, 2, 3, . . . . At all other points , q() is analytic as a quotient of two analytic functions. Hence, the singular points of the given equation are 0 and = k, k = 1, 2, 3, . . . . 11. The coefficient, x + 2, is a polynomial, and so it is analytic everywhere. Therefore, x = 0 is an ordinary point of the given equation. We seek a power series solution of the form y(x) =
n=0 an x n y (x) =
n=1 nan xn1 , 507 Chapter 8
where we have applied Theorem 4 on page 434 of the text to find the power series expansion of y (x). We now substitute the power series for y and y into the given differential equation and obtain nan x
n=1 n1 + (x + 2)
n=0 n1 an xn = 0 n=1 nan x +
n=0 2an x +
n=0 n an xn+1 = 0. (8.5) To sum these series, we make shifts in indices of summation so that they sum over the same power of x. In the first sum, we set k = n  1 so that n = k + 1 and k runs from 0 to ; in the second sum, we just replace n by k; in the third sum, we let k = n + 1 and so n = k  1, and the summation starts from 1. Thus the equation (8.5) becomes (k + 1)ak+1 x +
k=0 k=0 k 2ak x +
k=1 k k ak1 xk = 0 a1 +
k=1 (k + 1)ak+1 x + 2a0 +
k=1 2ak x k +
k=1 ak1 xk = 0 (a1 + 2a0 ) +
k=1 [(k + 1)ak+1 + 2ak + ak1 ] xk = 0. For the power series on the lefthand side to be identically zero, we must have all zero coefficients. Hence, a1 + 2a0 = 0 This yields a1 + 2a0 = 0 a1 = 2a0 , a2 = (2a1  a0 ) /2 = (4a0  a0 ) /2 = 3a0 /2 , a3 = (2a2  a1 ) /3 = (3a0 + 2a0 ) /3 = a0 /3 , k = 1 : 2a2 + 2a1 + a0 = 0 k = 2 : 3a3 + 2a2 + a1 = 0 . . . Therefore, 3a0 2 a0 3 3x2 x3 y(x) = a0  2a0 x + x  x + = a0 1  2x +  + , 2 3 2 3 where a0 is an arbitrary constant (which is, actually, y(0)). 508 and (k + 1)ak+1 + 2ak + ak1 = 0 for all k 1. Exercises 8.3
13. This equation has no singular points since the coefficients p(x) 0 and q(x) = x2 are analytic everywhere. So, let z(x) =
k=0 ak x k z (x) =
k=1 kak x k1 z (x) =
k=2 k(k  1)ak xk2 , where we used Theorem 4 on page 434 of the text differentiating the series termwise. Substitution z and z into the given equation yields z  x2 z = k(k  1)ak xk2  x2 ak xk =
k=0 k=2 k(k  1)ak xk2 
k=0 ak xk+2 . k=2 We now shift indices of summation so that they sum over the same power of x. For the first sum, we substitute n = k  2 so that k = n + 2, k  1 = n + 1, and the summation starts from n = 0. In the second summation, we let n = k + 2 which yields k = n  2 and n = 2 as the starting index. Thus we obtain z x z =
n=0 2 (n + 2)(n + 1)an+2 x 
n n=2 an2 xn . Next step in writing the righthand side as a single power series is to start both summations at the same point. To do this we observe that (n + 2)(n + 1)an+2 x 
n n=0 n=2 an2 x = 2a2 + 6a3 x +
n=2 n (n + 2)(n + 1)an+2 x 
n n=2 an2 xn = 2a2 + 6a3 x +
n=2 [(n + 2)(n + 1)an+2  an2 ] xn . In order for this power series to equal zero, each coefficient must be zero. Therefore, we obtain 2a2 = 0, 6a3 = 0 and (n + 2)(n + 1)an+2  an2 = 0, n 2. From the first two equations we find that a2 = 0 and a3 = 0. Next we take n = 2 and n = 3 in the above recurrence relation and get n = 2 : (4)(3)a4  a0 = 0 n = 3 : (5)(4)a5  a1 = 0 a4 = a0 /12 , a5 = a1 /20 . 509 Chapter 8
Hence, z(x) =
k=0 ak xk = a0 + a1 x + (0)x2 + (0)x3 + = a0 1 + a0 4 a1 5 x + x + 12 20 x4 x5 + + a1 x + + . 12 20 15. Zero is an ordinary point for this equation since the functions p(x) = x  1 and q(x) = 1 are both analytic everywhere and, hence, at the point x = 0. Thus, we can assume that the solution to this linear differential equation has a power series expansion with a positive radius of convergence about the point x = 0. That is, we assume that y(x) = a0 + a1 x + a2 x2 + a3 x3 + = an xn .
n=0 In order to solve the differential equation we must find the coefficients an . To do this, we must substitute y(x) and its derivatives into the given differential equation. Hence, we must find y (x) and y (x). Since y(x) has a power series expansion with a positive radius of convergence about the point x = 0, we can find its derivative by differentiating term by term. We can similarly differentiate y (x) to find y (x). Thus, we have y (x) = 0 + a1 + 2a2 x + 3a3 x + =
n=1 2 nan xn1 n(n  1)an xn2 . y (x) = 2a2 + 6a3 x + =
n=2 By substituting these expressions into the differential equation, we obtain y + (x  1)y + y =
n=2 n(n  1)an x n2 + (x  1)
n=1 nan x n1 +
n=0 an xn = 0. Simplifying yields n(n  1)an x
n=2 n2 +
n=1 nan x 
n n=1 nan x n1 +
n=0 an xn = 0. (8.6) We want to be able to write the lefthand side of this equation as a single power series. This will allow us to find expressions for the coefficient of each power of x. Therefore, we first need 510 Exercises 8.3
to shift the indices in each power series above so that they sum over the same powers of x. Thus, we let k = n  2 in the first summation and note that this means that n = k + 2 and that k = 0 when n = 2. This yields n(n  1)an x
n=2 n2 =
k=0 (k + 2)(k + 1)ak+2 xk . In the third power series, we let k = n  1 which implies that n = k + 1 and k = 0 when n = 1. Thus, we see that n1 nan x
n=1 =
k=0 (k + 1)ak+1 xk . For the second and last power series we need only to replace n with k. Substituting all of these expressions into their appropriate places in equation (8.6) above yields (k + 2)(k + 1)ak+2 x +
k=0 k=1 k kak x 
k k=0 (k + 1)ak+1 x +
k=0 k ak xk = 0. Our next step in writing the lefthand side as a single power series is to start all of the summations at the same point. To do this we observe that k (k + 2)(k + 1)ak+2 x = (2)(1)a2 x +
k=0 k=1 0 (k + 2)(k + 1)ak+2 xk , (k + 1)ak+1 xk = (1)a1 x0 + ak xk = a0 x0 + (k + 1)ak+1 xk ,
k=1 k=0 ak xk .
k=1 k=0 Thus, all of the summations now start at one. Therefore, we have (2)(1)a2 x +
k=1 0 (k + 2)(k + 1)ak+2 x +
k=1 k kak xk (1)a1 x0  (k + 1)ak+1xk + a0 x0 + ak xk = 0
k=1 k=1 2a2  a1 + a0 +
k=1 (k + 2)(k + 1)ak+2 xk + kak xk  (k + 1)ak+1 xk + ak xk = 0 511 Chapter 8 2a2  a1 + a0 +
k=1 ((k + 2)(k + 1)ak+2 + (k + 1)ak  (k + 1)ak+1 ) xk = 0. In order for this power series to equal zero, each coefficient must be zero. Therefore, we obtain 2a2  a1 + a0 = 0 and (k + 2)(k + 1)ak+2 + (k + 1)ak  (k + 1)ak+1 = 0, ak+1  ak ak+2 = , k 1, k+2 k1 a2 = a1  a0 , 2 where we have canceled the factor (k + 1) from the recurrence relation, the last equation obtained above. Note that in this recurrence relation we have solved for the coefficient with the largest subscript, namely ak+2 . Also, note that the first value for k in the recurrence relation is the same as the first value for k used in the summation notation. By using the recurrence relation with k = 1, we find that a2  a1 = a3 = 3 recurrence equation, we obtain a3  a2 = a4 = 4 (a1 + a0 ) a1  a0  2a1 + a0 ) 6 2 = , 4 12 a1  a0  a1 (a1 + a0 ) 2 = , 3 6 where we have plugged in the expression for a2 that we found above. By letting k = 2 in the where we have plugged in the values for a2 and a3 found above. Continuing this process will allow us to find as many coefficients for the power series of the solution to the differential equation as we may want. Notice that the coefficients just found involve only the variables a0 and a1 . From the recurrence equation, we see that this will be the case for all coefficients of the power series solution. Thus, a0 and a1 are arbitrary constants and these variables will be our arbitrary variables in the general solution. Hence, substituting the values for the coefficients that we found above into the solution y(x) =
n=0 an xn = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + , 512 Exercises 8.3
yields the solution y(x) = a0 + a1 x + = a0 a1  a0 2 (a1 + a0 ) 3 2a1 + a0 4 x + x + x + 2 6 12 x2 x3 x4 x2 x3 x4  + + + a1 x +   + . 1 2 6 12 2 6 6 19. Since x = 0 is an ordinary point for the given equation, we seek for a power series expansion of a general solution of the form y(x) =
n=0 an x n y (x) =
n=1 nan xn1 . Substituting y(x) and y (x) into the given equation, we obtain nan x
n=1 n1  2x
n=0 an x =
n=1 n nan x n1 
n=0 2an xn+1 = 0. We shift the indices of summations so that they sum over the same powers of x. In the first sum, we let k = n  1. Then n = k + 1 and the summation starts from k = 0. In the second sum, let k = n + 1. Then n = k  1 and k = 1 when n = 0. Thus we have (k + 1)ak+1x 
k k=0 k=1 2ak1x = a1 +
k=1 k [(k + 1)ak+1  2ak1 ]xk = 0. In order for this power series to equal zero, each coefficient must be zero. That is, a1 = 0, (k + 1)ak+1  2ak1 = 0, k1 a1 = 0, ak+1 = 2ak1 /(k + 1) , k 1. Since a1 = 0, it follows from this recurrence relation that all odd coefficients are zeros. Indeed, a3 = For even coefficients, we have k = 1 : a2 = 2a0 /2 , k = 3 : a4 = 2a2 /4 = 2[2a0 /2]/4 = 22 a0 /(2 4) , k = 5 : a6 = 2a4 /6 = 2[22 a0 /(2 4)]/6 = 23 a0 /(2 4 6) , . . . 513 2a1 = 0, 3 a5 = 2a3 = 0, 5 etc. Chapter 8
The pattern for the even coefficients is now apparent. Namely, a2k = 2k a0 2k a0 a0 = k = , 2 4 (2k) 2 (1 2 k) k! k = 1, 2, . . . . This formula remains correct for k = 0 as well with 0! := 1. Thus y(x) =
k=0 a0 2k x2k x = a0 , k! k! k=0 where a0 is an arbitrary constant. 21. Since x = 0 is an ordinary point for this differential equation, we will assume that the solution has a power series expansion with a positive radius of convergence about the point x = 0. Thus, we have y(x) =
n=0 an x n y (x) =
n=1 nan x n1 y (x) =
n=2 n(n  1)an xn2 . By plugging these expressions into the differential equation, we obtain y  xy + 4y =
n=2 n(n  1)an x n2 x nan x
n=1 n1 +4
n=0 an xn = 0 n=2 n(n  1)an xn2 
n=1 nan xn +
n=0 4an xn = 0. In order for each power series to sum over the same powers of x, we will shift the index in the first summation by letting k = n  2, and we will let k = n in the other two power series. Thus, we have (k + 2)(k + 1)ak+2x 
k k=0 k=1 kak x +
k=0 k 4ak xk = 0. Next we want all of the summations to start at the same point. Therefore, we will take the first term in the first and last power series out of the summation sign. This yields (2)(1)a2 x0 + 514 (k + 2)(k + 1)ak+2 xk 
k=1 k=1 kak xk + 4a0 x0 + 4ak xk = 0 4ak xk = 0 k=1 2a2 + 4a0 +
k=1 (k + 2)(k + 1)ak+2 xk 
k=1 kak xk +
k=1 Exercises 8.3 2a2 + 4a0 +
k=1 [(k + 2)(k + 1)ak+2 + (k + 4)ak ] xk = 0. By setting each coefficient of the power series equal to zero, we see that 2a2 + 4a0 = 0 a2 = 4a0 = 2a0 , 2 ak+2 = (k + 2)(k + 1)ak+2 + (k + 4)ak = 0 (k  4)ak , (k + 2)(k + 1) k 1, where we have solved the recurrence equation, the last equation above, for ak+2 , the coefficient with the largest subscript. Thus, we have k=1 k=2 k=3 k=4 k=5 k=6 k=7 k=8 k=9 a1 3a1 = , 32 2 2a2 (2)(4)a0 a0 a4 = = = , 43 432 3 (3)(1)a1 a1 a3 = = , a5 = 54 5432 40 a6 = 0, (3)(1)(1)a1 a1 a5 = = , a7 = 76 765432 560 2a6 a8 = = 0, 87 (3)(1)(1)(3)a1 3a7 a9 = = , 98 9! 4a8 a10 = = 0, 10 9 (3)(1)(1)(3)(5)a1 5a9 = . a11 = 11 10 11! a3 = Now we can see a pattern starting to develop. (Note that it is easier to determine sucha pattern if we consider specific coefficients that have not been multiplied out.) We first note that a0 and a1 can be chosen arbitrarily. Next we notice that the coefficients with even subscripts larger than 4 are zero. We also see that the general formula for a coefficient with an odd subscript is given by a2n+1 = (3)(1)(1) (2n  5)a1 . (2n + 1)! 515 Chapter 8
Notice that this formula is also valid for a3 and a5 . Substituting these expressions for the coefficients into the solution y(x) =
n=0 an xn = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + , yields y(x) = a0 + a1 x  2a0 x2  a1 3 a0 4 a1 5 x + x + x + 2 3 40 (3)(1)(1) (2n  5)a1 2n+1 + x + (2n + 1)! (3)(1)(1) (2n  5) 2n+1 x3 x5 + ++ x + + a1 x  2 40 (2n + 1)! = a0 1  2x2 + = a0
2 x4 3 x4 1  2x + + a1 x + 3 k=1 (3)(1)(1) (2k  5) 2k+1 . x (2k + 1)! 29. Since x = 0 is an ordinary point for this differential equation, we can assume that a solution to this problem is given by y(x) =
n=0 an x n y (x) =
n=1 nan x n1 y (x) =
n=2 n(n  1)an xn2 . By substituting the initial conditions, y(0) = 1 and y (0) = 2, into the first two equations above, we see that y(0) = a0 = 1, and y (0) = a1 = 2. Next we will substitute the expressions found above for y(x), y (x), and y (x) into the differential equation to obtain y + y  xy =
n=2 n(n  1)an x n2 +
n=1 nan x n1 x
n=0 an xn = 0 n=2 n(n  1)an xn2 +
n=1 nan xn1 
n=0 an xn+1 = 0. By setting k = n  2 in the first power series above, k = n  1 in the second power series above, and k = n + 1 in the last power series, we can shift the indices so that x is raised to 516 Exercises 8.3
the power k in each power series. Thus, we obtain (k + 2)(k + 1)ak+2x +
k=0 k=0 k (k + 1)ak+1 x 
k k=1 ak1 xk = 0. We can start all of the summations at the same point if we remove the first term from each of the first two power series above. Therefore, we have (2)(1)a2 +
k=1 (k + 2)(k + 1)ak+2xk + (1)a1 +
k=1 (k + 1)ak+1 xk 
k=1 ak1 xk = 0 2a2 + a1 +
k=1 [(k + 2)(k + 1)ak+2 + (k + 1)ak+1  ak1 ] xk = 0. By equating coefficients, we see that all of the coefficients of the terms in the power series above must be zero. Thus, we have 2a2 + a1 = 0 a1 , 2 (k + 2)(k + 1)ak+2 + (k + 1)ak+1  ak1 = 0 ak1  (k + 1)ak+1 , k 1. ak+1 = (k + 2)(k + 1) a2 = Thus, we see that a0 a1 a0  2a2 = + . 32 6 6 Using the fact that a0 = 1 and a1 = 2, which we found from the initial conditions, we k=1 a3 = calculate (2) = 1, 2 1 1 2 = . a3 = + 6 6 6 a2 = By substituting these coefficients, we obtain the cubic polynomial approximation y(x) = 1  2x + x2  x3 . 6 The graphs of the linear, quadratic and cubic polynomial approximations are easily generated by using the software supplied with the text. 517 Chapter 8
31. The point x0 = 0 is an ordinary point for the given equation since p(x) = 2x/(x2 + 2) and q(x) = 3/(x2 + 2) are analytic at zero. Hence we can express a general solution in the form y(x) =
n=0 an xn . Substituting this expansion into the given differential equation yields (x + 2)
n=2 2 n(n  1)an x n2 + 2x
n=1 nan x n1 +3
n=0 an xn = 0 n=2 n(n  1)an xn +
n=2 2n(n  1)an xn2 +
n=1 2nan xn +
n=0 3an xn = 0. To sum over like powers xk , we put k = n  2 into the second summation and k = n into the other summations. This gives k(k  1)ak xk +
k=2 k=0 2(k + 2)(k + 1)ak+2 xk +
k=1 2kak xk +
k=0 3ak xk = 0. Next we separate the terms corresponding to k = 0 and k = 1 and combine the rest under one summation. (4a2 + 3a0 ) + (12a3 + 5a1 )x +
k=2 [k(k  1)ak + 2(k + 2)(k + 1)ak+2 + 2kak + 3ak ] xk = 0. Setting the coefficients equal to zero and simplifying, we get 4a2 + 3a0 = 0, 12a3 + 5a1 = 0, (k 2 + k + 3)ak + 2(k + 2)(k + 1)ak+2 = 0, a2 = 3a0 /4 , a3 = 5a1 /12 , ak+2 = (k 2 + k + 3)ak /[2(k + 2)(k + 1)], From the initial conditions, we have a0 = y(0) = 1 518 and a1 = y (0) = 2. k 2. k2 Exercises 8.3
Therefore, a2 = 3(1)/4 = 3/4 , a3 = 5(2)/12 = 5/6 , and the cubic polynomial approximation for the solution is y(x) = a0 + a1 x + x2 x2 + a3 x3 = 1 + 2x  3x2 5x3  . 4 6 33. In Problem 7, Exercises 8.2 we showed that the radius of convergence of a power series a2n x2n is = 1/ L, where n=0 L = lim In the series (13), a2n = (1)n /n! and so L = lim Therefore, L = 0 and = . 1 (1)n+1 /(n + 1)! = 0. = lim n /n! n n + 1 (1) a2(n+1) . a2n n n 35. With the given values of parameters, we have an initial value problem 0.1q (t) + 1 + Simplifying yields q (t) + (10 + t) q (t) + 5q(t) = 0, q(0) = 10, q (0) = 0. n=0 t 1 q (t) + q(t) = 0, 10 2 q(0) = 10, q (0) = 0. The point t = 0 is an ordinary point for this equation. Let q(t) = equation, we obtain an tn be the power series expansion of q(t) about t = 0. Substituting this series into the above differential n(n  1)an t
n=2 n2 + (10 + t)
n=1 n2 nan t n1 +5
n=0 an tn = 0 n=2 n(n  1)an t +
n=1 10nan t n1 +
n=1 nan t +
n=0 n 5an tn = 0. 519 Chapter 8
Setting k = n  2 in the first summation, k = n  1 in the second summation, and k = n in the last two summations, we obtain (k + 2)(k + 1)ak+2 t +
k=0 k=0 k 10(k + 1)ak+1 t +
k=1 k kak t +
k=0 k 5ak tk = 0. Separating the terms corresponding to k = 0 and combining the rest under one sum yields (2a2 + 10a1 + 5a0 ) +
k=1 [(k + 2)(k + 1)ak+2 + 10(k + 1)ak+1 + (k + 5)ak ] tk = 0. Setting the coefficients equal to zero, we obtain the recurrence relations 2a2 + 10a1 + 5a0 = 0, (k + 2)(k + 1)ak+2 + 10(k + 1)ak+1 + (k + 5)ak = 0, Next we use the initial conditions to find a0 and a1 . a0 = q(0) = 10, From the first equation in (8.7) we have a2 = 10a1  5a0 = 25. 2 a1 = q (0) = 0. k 1. (8.7) Taking k = 1 and k = 2 in the second equation in (8.7), we find a3 and a4 . k = 1 : 6a3 + 20a2 + 6a1 = 0 a3 = (20a2 + 6a1 )/6 = 250/3, a4 = (30a3 + 7a2 )/12 = 775/4. k = 2 : 12a4 + 30a3 + 7a2 = 0 Hence q(t) = 10 + (0)t  25t2 + EXERCISES 8.4: 250t3 775t4 250t3 775t4  + = 10  25t2 +  + . 3 4 3 4 Equations with Analytic Coefficients, page 456 3 . Therefore, singular points will occur 1 + x + x2 3 1 i. x= 2 2 3. For this equation, p(x) = 0 and q(x) = when 1 + x + x2 = 0 520 Exercises 8.4
Thus, x = 1 is an ordinary point for this equation, and we can find a power series solution with a radius of convergence of at least the minimum of the distances between 1 and points (1/2) ( 3/2)i, which, in fact, are equal. Recall that the distance between two complex numbers, z1 = a + bi and z2 = c + di, is given by dist (z1 , z2 ) = (a  c)2 + (b  d)2 . Thus, the distance between (1 + 0 i) and (1/2) + ( 3/2)i is 1 1  2
2 3 + 0 2 2 = 9 3 + = 3. 4 4 Therefore, the radius of convergence for the power series solution of this differential equation about x = 1 will be at least = 3. 9. We see that x = 0 and x = 2 are the only singular points for this differential equation and, thus, x = 1 is an ordinary point. Therefore, according to Theorem 5 on page 451 of the text, there exists a power series solution of this equation about the point x = 1 with a radius of convergence of at least one, the distance from 1 to either 0 or 2. That is, we have a general solution for this differential equation of the form y(x) =
n=0 an (x  1)n , which is convergent for all x at least in the interval (0, 2), the interval on which the inequality x  1 < 1 is satisfied. To find this solution we will proceed as in Example 3 on page 453 of the text. Thus, we make the substitution t = x  1, which implies that x = t + 1. (Note that dx/dt = 1.) We then define a new function Y (t) := y(t + 1) = y(x) dy dY dx dy dy = = 1 = dt dx dt dx dx 2 dY d dY d dy d2 y = = = dt2 dt dt dt dx dx2 dx dt = d2 y . dx2 521 Chapter 8
Hence, with the substitutions t = x  1 and Y (t) = y(t + 1), we transform the differential equation, (x2  2x) y (x) + 2y(x) = 0, into the differential equation (t + 1)2  2(t + 1) y (t + 1) + 2y(t + 1) = 0 (t + 1)2  2(t + 1) Y (t) + 2Y (t) = 0 t2  1 Y (t) + 2Y (t) = 0. (8.8) To find a general solution to (8.8), we first note that zero is an ordinary point of equation (8.8). Thus, we can assume that we have a power series solution of equation (8.8) of the form Y (t) =
n=0 an tn , which converges for all t in (1, 1). (This means that x = t + 1 will be in the interval (0, 2) as desired.) Substituting into equation (8.8) yields t2  1 n=2 n(n  1)an tn2 + 2
n=2 n=0 n an tn = 0 n2 n(n  1)an t 
n=2 n(n  1)an t +
n=0 2an tn = 0. Making the shift in the index, k = n  2, in the second power series above and replacing n with k in the other two power series allows us to take each summation over the same power of t. This gives us k(k  1)ak t 
k k=2 k=0 (k + 2)(k + 1)ak+2 t +
k=0 k 2ak tk = 0. In order to start all of these summations at the same point, we must take the first two terms out of the summation sign in the last two power series. Thus we have, k(k  1)ak tk  (2)(1)a2  (3)(2)a3 t 
k=2 k=2 (k + 2)(k + 1)ak+2 tk +2a0 + 2a1 t +
k=2 2ak tk = 0 522 Exercises 8.4 2a0  2a2 + (2a1  6a3 ) t +
k=2 [k(k  1)ak  (k + 2)(k + 1)ak+2 + 2ak ] tk = 0. For this power series to equal zero, each coefficient must be zero. Thus, we have 2a0  2a2 = 0 a2 = a0 , 2a1  6a3 = 0 a3 = a1 , 3 k(k  1)ak  (k + 2)(k + 1)ak+2 + 2ak = 0, k(k  1)ak + 2ak , (k + 2)(k + 1) k2 (k 2  k + 2)ak , (k + 2)(k + 1) ak+2 = k2 ak+2 = k 2. Therefore, we see that k=2 k=3 a4 = a2 a0 4a2 = = , 43 3 3 2a1 8a3 a5 = = , etc. 54 15 Plugging these values for the coefficients into the power series solution, Y (t) =
n=0 an tn = a0 + a1 t + a2 t2 + a3 t3 + a4 t4 + , yields a1 t3 a0 t4 2a1 t5 + + + Y (t) = a0 + a1 t + a0 t + 3 3 15 t4 t3 2t5 + . Y (t) = a0 1 + t2 + + + a1 t + + 3 3 15
2 Lastly, we want to change back to the independent variable x. To do this, we recall that Y (t) = y(t + 1). Thus, if t = x  1, then Y (t) = Y (x  1) = y ([x  1] + 1) = y(x). Thus, we replace t with x1 in the solution just found, and we obtain a power series expansion for a general solution in the independent variable x. Substituting, we have y(x) = a0 1 + (x  1)2 + 1 2 1 (x  1)4 + + a1 (x  1) + (x  1)3 + (x  1)5 + . 3 3 15 523 Chapter 8
17. Here p(x) = 0 and q(x) =  sin x both of which are analytic everywhere. Thus, x = is an ordinary point for this differential equation, and there are no singular points. Therefore, by Theorem 5 on page 451 of the text, we can assume that this equation has a general power series solution about the point x = with an infinite radius of convergence (i.e., = ). That is, we assume that we have a solution to this differential equation given by y(x) =
n=0 an (x  ) n y (x) =
n=1 nan (x  )n1 , which converges for all x. If we apply the initial conditions, y() = 1 and y () = 0, we see that a0 = 1 and a1 = 0. To find a general solution of this differential equation, we will combine the methods of Example 3 and Example 4 on pages 453455 of the text. Thus, we will first define a new function, Y (t), using the transformation t = x  . Thus, we define Y (t) := y(t + ) = y(x). Hence, by the chain rule (using the fact that x = t + which implies that dx/dt = 1), we have dY /dt = (dy/dx)(dx/dt) = dy/dx, and similarly d2 Y /dt2 = d2 y/dx2 . We now solve the transformed differential equation d2 Y  sin(t + )Y (t) = 0 dt2 d2 Y + (sin t)Y (t) = 0, dt2 (8.9) where we have used the fact that sin(t + ) =  sin t. When we have found the solution Y (t), we will use the fact that y(x) = Y (x  ) to obtain the solution to the original differential equation in terms of the independent variable x. Hence, we seek a power series solution to equation (8.9) of the form Y (t) =
n=0 an t n Y (t) =
n=1 nan t n1 Y (t) =
n=2 n(n  1)an tn2 . Since the initial conditions, y() = 1 and y () = 0, transform into Y (0) = 1 and Y (0) = 0, we must have Y (0) = a0 = 1 524 and Y (0) = a1 = 0. Exercises 8.4
Next we note that q(t) = sin t is an analytic function with a Maclaurin series given by sin t =
n=0 t3 t5 t7 (1)n t2n+1 =t +  + . (2n + 1)! 6 120 5040 By substituting the expressions that we found for Y (t), Y (t), and sin t into equation (8.9), we obtain n(n  1)an tn2 + t 
n=2 t5 t7 t3 +  + 6 120 5040 an tn = 0.
n=0 Therefore, expanding this last equation (and explicitly showing only terms of up to order four), yields 2a2 + 6a3 t + 12a4 t2 + 20a5 t3 + 30a6 t4 + + t a0 + a1 t + a2 t2 + a3 t3 + t3  (a0 + a1 t + ) + = 0 6 2a2 + 6a3 t + 12a4 t2 + 20a5 t3 + 30a6 t4 + + t a0 + a1 t + a2 t2 + a3 t3 + a0 t3 a1 t4 +    + = 0. 6 6 By grouping these terms according to their powers of t, we obtain a0 3 a1 4 t + 30a6 + a3  t + = 0. 2a2 + (6a3 + a0 ) t + (12a4 + a1 ) t2 + 20a5 + a2  6 6 Setting these coefficients to zero and recalling that a0 = 1 and a1 = 0 yields the system of equations 2a2 = 0 6a3 + a0 = 0 a2 = 0, a3 = 1 a0 = , 6 6 a1 = 0, 12 12a4 + a1 = 0 a4 = a0 =0 20a5 + a2  6 a1 =0 30a6 + a3  6 1 a0  a2 1 a5 = 6 = 6 = , 20 20 120 1 a1  a3 0+ 6 = 1 . a6 = 6 = 30 30 180 525 Chapter 8
Plugging these coefficients into the power series solution Y (t) =
n=0 an tn = a0 + a1 t + a2 t2 + , yields the solution to equation (8.9): t5 t6 t3 t5 t6 t3 + + = 1 + + + . Y (t) = 1 + 0 + 0  + 0 + 6 120 180 6 120 180 Lastly we want to find the solution to the original equation with the independent variable x. In order to do this, we recall that t = x  and Y (x  ) = y(x). Therefore, by substituting these values into the equation above, we obtain the solution y(x) = 1  1 1 1 (x  )3 + (x  )5 + (x  )6 + . 6 120 180 21. We assume that this differential equation has a power series solution with a positive radius of convergence about the point x = 0. This is reasonable because all of the coefficients and the forcing function g(x) = sin x are analytic everywhere. Thus, we assume that y(x) =
n=0 an x n y (x) =
n=1 nan xn1 . By substituting these expressions and the Maclaurin expansion for sin x into the differential equation, y (x)  xy(x) = sin x, we obtain nan x
n=1 n1 x
n=0 an x =
n=0 n (1)n x2n+1 . (2n + 1)! In the first power series on the left, we make the shift k = n  1. In the second power series on the left, we make the shift k = n + 1. Thus, we obtain (k + 1)ak+1 x 
k k=0 k=1 ak1 x =
n=0 k (1)n x2n+1 . (2n + 1)! Separating out the first term of the first power series on the left yields a1 +
k=1 (k + 1)ak+1 x 
k k=1 ak1 x =
n=0 k (1)n x2n+1 (2n + 1)! 526 Exercises 8.4 a1 +
k=1 [(k + 1)ak+1  ak1 ] xk =
n=0 (1)n x2n+1 . (2n + 1)! Therefore, by expanding both of the power series, we have a1 + (2a2  a0 ) x + (3a3  a1 ) x2 + (4a4  a2 ) x3 + (5a5  a3 ) x4 x5 x7 x3 +  + . + (6a6  a4 ) x5 + (7a7  a5 ) x6 + = x  6 120 5040 By equating the coefficients of like powers of x, we obtain a1 = 0, 2a2  a0 = 1 3a3  a1 = 0 4a4  a2 = 1 6 a2 = a3 = a0 + 1 , 2 a1 = 0, 3 a2  1/6 a0 1 = + , 4 8 12 a3 = 0, 5 a0 11 a4  1/120 = + . 6 48 720 a4 = a5 = 5a5  a3 = 0 6a6  a4 = 1 120 a6 = Substituting these coefficients into the power series solution and noting that a0 is an arbitrary number, yields y(x) =
n=0 an xn 1 11 a0 1 a0 a0 + + + x2 + 0 + x4 + 0 + x6 + 2 2 8 12 48 720 1 1 1 1 2 1 4 11 6 x + x + x + . 1 + x2 + x4 + x6 + + 2 8 48 2 12 720 = a0 + 0 + = a0 27. Observe that x = 0 is an ordinary point for this differential equation. Therefore, we can assume that this equation has a power series solution about the point x = 0 with a positive 527 Chapter 8
radius of convergence. Thus, we assume that y(x) =
n=0 an x n y (x) =
n=1 nan x n1 y (x) =
n=2 n(n  1)an xn2 . The Maclaurin series for tan x is tan x = x + x3 2x5 + + , 3 15 which is given in the table on the inside front cover of the text. Substituting the expressions for y(x), y (x), y (x), and the Maclaurin series for the function tan x into the differential equation, (1  x2 )y  y + y = tan x, yields 1  x2 n=2 n(n  1)an xn2 
n=2 n=1 n2 nan xn1 +
n=0 an xn = x + n x3 2x5 + + 3 15 n(n  1)an x 
n=2 n(n  1)an x 
n=1 nan x n1 +
n=0 an xn x3 2x5 + + . 3 15 =x+ By shifting the indices of the power series on the lefthand side of this equation, we obtain (k + 2)(k + 1)ak+2x 
k k=0 k=2 k(k  1)ak x 
k k=0 (k + 1)ak+1x +
k=0 k ak xk = x + x3 2x5 + + . 3 15 Removing the first two terms from the summation notation in the first, third and fourth power series above yields (2)(1)a2 + (3)(2)a3 x +
k=2 (k + 2)(k + 1)ak+2 x 
k k=2 k(k  1)ak xk  (1)a1  (2)a2 x 
k=2 (k + 1)ak+1xk + a0 + a1 x +
k=2 ak xk = x + x3 2x5 + + 3 15 (2a2  a1 + a0 ) + (6a3  2a2 + a1 ) x +
k=2 [(k + 2)(k + 1)ak+2  k(k  1)ak  (k + 1)ak+1 + ak ] xk = x + x3 2x5 + + . 3 15 528 Exercises 8.5
By equating the coefficients of the two power series, we see that 2a2  a1 + a0 = 0 6a3  2a2 + a1 = 1 4 3a4  2 1a2  3a3 + a2 = 0 a2 = a1  a0 , 2 1  a0 2a2  a1 + 1 = , 6 6 a1  2a0 + 1 a2 + 3a3 = . 12 24 a3 = a4 = Therefore, noting that a0 and a1 are arbitrary, we can substitute these coefficients into the power series solution y(x) = y(x) = a0 + a1 x + = a0 n n=0 an x = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + to obtain a0 1 a1 a0 2 1 a0 a1   x3 +  + x4 + x + 2 2 6 6 24 12 24 1 1 4 1 4 1 1 x + + a1 x + x2 + x + 1  x2  x3  2 6 12 2 24 1 3 1 4 + x + x + . 6 24 CauchyEuler (Equidimensional) Equations Revisited, page 460 EXERCISES 8.5: 5. Notice that, since x > 0, we can multiply this differential equation by x2 and rewrite it to obtain d2 y dy + 13y = 0.  5x 2 dx dx We see that this is a CauchyEuler equation. Thus, we will assume that a solution has the x2 form y(x) = xr y (x) = rxr1 y (x) = r(r  1)xr2 . Substituting these expressions into the differential equation above yields r(r  1)xr  5rxr + 13xr = 0 r 2  6r + 13 xr = 0 r 2  6r + 13 = 0. We obtained this last equation by using the assumption that x > 0. (We also could arrive at this equation by using equation (4) on page 458 of the text.) Using the quadratic formula, we 529 Chapter 8
see that the roots to this equation are r= 6 36  52 = 3 2i. 2 Therefore, using formulas (5) and (6) on page 458 of the text with complex conjugates roots (and using Euler's formula), we have two linearly independent solutions give by y1 (x) = x3 cos(2 ln x), y2 (x) = x3 sin(2 ln x). Hence the general solution to this equation is given by y(x) = c1 x3 cos(2 ln x) + c2 x3 sin(2 ln x). 7. This equation is a third order CauchyEuler equation, and, thus, we will assume that a solution has the form y(x) = xr . This implies that y (x) = rxr1 y (x) = r(r  1)xr2 y (x) = r(r  1)(r  2)xr3 . By substituting these expressions into the differential equation, we obtain [r(r  1)(r  2) + 4r(r  1) + 10r  10] xr = 0 r 3 + r 2 + 8r  10 xr = 0 r 3 + r 2 + 8r  10 = 0. By inspection we see that r = 1 is a root of this last equation. Thus, one solution to this differential equation will be given by y1 (x) = x and we can factor the indicial equation above as follows: (r  1)(r 2 + 2r + 10) = 0. Therefore, using the quadratic formula, we see that the roots to this equation are r = 1, 13i. Thus, we can find two more linearly independent solutions to this equation by using Euler's formula as was done on page 458 of the text. Thus, three linearly independent solutions to this problem are given by y1 (x) = x, y2 (x) = x1 cos(3 ln x), y3 (x) = x1 sin(3 ln x). Hence, the general solution to this differential equation is y(x) = c1 x + c2 x1 cos(3 ln x) + c3 x1 sin(3 ln x). 530 Exercises 8.5
13. We first must find two linearly independent solutions to the associated homogeneous equation. Since this is a CauchyEuler equation, we assume that there are solutions of the form y(x) = xr y (x) = rxr1 y (x) = r(r  1)xr2 . Substituting these expressions into the associated homogeneous equation yields [r(r  1)  2r + 2] xr = 0 r 2  3r + 2 = 0 (r  1)(r  2) = 0. Thus, the roots to this indicial equation are r = 1, 2. Therefore, a general solution to the associated homogeneous equation is yh (x) = c1 x + c2 x2 . For the variation of parameters method, let y1 (x) = x and y2 (x) = x2 , and then assume that a particular solution has the form yp (x) = v1 (x)y1 (x) + v2 (x)y2 (x) = v1 (x)x + v2 (x)x2 . In order to find v1 (x) and v2 (x), we would like to use formula (10) on page 195 of the text. To use equation (10), we must first find the Wronskian of y1 and y2 . Thus, we compute W [y1 , y2 ] (x) = y1 (x)y2 (x)  y2 (x)y1 (x) = 2x2  x2 = x2 . Next we must write the differential equation given in this problem in standard form. When we do this, we see that g(x) = x5/2 . Therefore, by equation (10), we have v1 (x) = and x5/2 x2 dx = x2 (x5/2 )dx = 2 3/2 x 3 x5/2 x dx = x2 Thus, a particular solution is given by v2 (x) = yp (x) = 2 3/2 x+ x 3 x7/2 dx = 2 5/2 x . 5 2 5/2 4 1/2 x2 = x x . 5 15 4 1/2 x . 15 531 Therefore, a general solution of the nonhomogeneous differential equation is given by y(x) = yh (x) + yp (x) = c1 x + c2 x2 + Chapter 8
19. (a) For this linear differential operator L, we have L [xr ] (x) = x3 r(r  1)(r  2)xr3 + x rxr1  xr = r(r  1)(r  2)xr + rxr  xr = r 3  3r 2 + 3r  1 xr = (r  1)3 xr . (b) From part (a) above, we see that r = 1 is a root of multiplicity three of the indicial equation. Thus, we have one solution given by y1 (x) = x. (8.10) To find two more linearly independent solutions, we use a method similar to that used in the text. By taking the partial derivative of L [xr ] (x) = (r  1)3 xr with respect to r, we have (r  1)3 xr = 3(r  1)2 xr + (r  1)3 xr ln x {L [xr ] (x)} = r r 2 3(r  1)2 xr + (r  1)3 xr ln x {L [xr ] (x)} = 2 r r = 6(r  1)xr + 6(r  1)2 xr ln x + (r  1)3 xr (ln x)2 . Since r  1 is a factor of every term in {L [xr ] (x)} /r and 2 {L [xr ] (x)} /r 2 above, we see that {L [xr ] (x)} r and 2 {L [xr ] (x)} r 2 = 0,
r=1 (8.11) = 0,
r=1 (8.12) We can use these facts to find the two solutions that we seek. In order to find a second solution, we would like an alternative form for {L [xr ] (x)} r Using the fact that L[y](x) = x3 y (x) + xy (x)  y(x) 532 .
r=1 Exercises 8.5
and proceeding as in equation (9) on page 458 of the text with w(r, x) = xr , we have 3w w +x {L [xr ] (x)} = {L[w](x)} = x3 w 3 r r r x x 4w 4w w 2w w 2w  = x3 3 + x  +x = x3 rx3 rx r x r xr r 3 w w w w = x3 3 +x  (x), =L x r x r r r where we are using the fact that mixed partials of w(r, x) are equal. Therefore, combining this with equation (8.11) above yields {L [xr ] (x)} r L
r=1 xr r = L xr ln x
r=1 r=1 = L[x ln x] = 0. Thus, a second linearly independent solution is given by y2 (x) = x ln x. To find a third solution, we will use equation (8.12) above. Hence, we would like to find an alternative form for 2 {L [xr ] (x)} /r 2 . To do this, we use the fact that
4 w 2w r 3 w {L [x ] (x)} = x  , +x 3 r rx rx r which we found above and the fact that mixed partial derivatives of w(r, x) are equal. Thus, we have 2 w 4w 2w {L [xr ] (x)} =  {L [xr ] (x)} = +x x3 2 3 r r r r rx rx r 5 3 2 5 3 w w w w w 2w  2 = x3 2 3 + x 2  2 = x3 3 2 + x r x r x r x r xr 2 r 3 2 2 2 2 w w w w = x3 3 +x  2 =L (x) = 0. 2 2 x r x r r r 2 Therefore, combining this with equation (8.12) above yields 2 {L [xr ] (x)} r 2 =L
r=1 2 (xr ) r 2 = L x(ln x)2 = 0,
r=1 533 Chapter 8
where we have used the fact that 2 xr /r 2 = xr (ln x)2 . Thus we see that another solution is y3 (x) = x(ln x)2 , which, by inspection, is linearly independent from y1 and y2 . Thus, a general solution to the differential equation is y(x) = C1 x + C2 x ln x + C3 x(ln x)2 . EXERCISES 8.6: Method of Frobenius, page 472 5. By putting this equation in standard form, we see that p(x) ==  and 3 3 = . 2  1)2 2 (x + 1)2 (x (x  1) Thus, x = 1, 1 are singular points of this equation. To check if x = 1 is regular, we note q(x) = that 3 1 and (x  1)2 q(x) = . 2 (x + 1) (x + 1)2 These functions are analytic at x = 1. Therefore, x = 1 is a regular singular point for this (x  1)p(x) =  differential equation. Next we check the singular point x = 1. Here (x + 1)p(x) =  1 (x  1)(x + 1) x1 1 x1 = = , (x2  1)2 (x  1)2 (x + 1)2 (x  1)(x + 1)2 is not analytic at x = 1. Therefore, x = 1 is an irregular singular point for this differential equation. 13. By putting this equation in standard form, we see that x2  4 (x  2)(x + 2) x+2 = = , 2  x  2)2 2 (x + 1)2 (x (x  2) (x  2)(x + 1)2 6x . q(x) = (x  2)2 (x + 1)2 ) p(x) = Thus, we have (x  2)p(x) = 534 x+2 (x + 1)2 and (x  2)2 q(x) = 6x . (x + 1)2 Exercises 8.6
Therefore, x = 2 is a regular singular point of this differential equation. We also observe that x+2 4 = = p0 , 2 x2 x2 (x + 1) 9 12 4 6x lim (x  2)2 q(x) =  lim =  =  = q0 . 2 x2 x2 (x + 1) 9 3 lim (x  2)p(x) = lim Thus, we can use equation (16) on page 463 of the text to obtain the indicial equation r(r  1) + 4r 4  =0 9 3 r2  5r 4  = 0. 9 3 By the quadratic formula, we see that the roots to this equation and, therefore, the exponents of the singularity x = 2, are given by 5 + 457 25 + 432 r1 = = , 18 18 5  457 r2 = . 18 5+ 21. Here p(x) = x1 and q(x) = 1. This implies that xp(x) = 1 and x2 q(x) = x2 . Therefore, we see that x = 0 is a regular singular point for this differential equation, and so we can use the method of Frobenius to find a solution to this problem. (Note also that x = 0 is the only singular point for this equation.) Thus, we will assume that this solution has the form w(r, x) = x We also notice that r n=0 an x =
n=0 n an xn+r . p0 = lim xp(x) = lim 1 = 1,
x0 x0 q0 = lim x q(x) = lim x2 = 0.
x0 x0 2 Hence, we see that the indicial equation is given by r(r  1) + r = r 2 = 0. This means that r1 = r2 = 0. Since x = 0 is the only singular point for this differential equation, we observe that the series solution w(0, x) which we will find by the method of 535 Chapter 8
Frobenius converges for all x > 0. To find the solution, we note that w(r, x) =
n=0 an xn+r w (r, x) =
n=0 (n + r)an xn+r1 (n + r)(n + r  1)an xn+r2 .
n=0 w (r, x) = Notice that the power series for w and w start at n = 0. Substituting these expressions into the differential equation and simplifying yields (n + r)(n + r  1)an xn+r +
n=0 n=0 (n + r)an xn+r +
n=0 an xn+r+2 = 0. Next we want each power series to sum over xk+r . Thus, we let k = n in the first and second power series and shift the index in the last power series by letting k = n + 2. Therefore, we have (k + r)(k + r  1)ak xk+r +
k=0 k=0 (k + r)ak xk+r +
k=2 ak2 xk+r = 0. We will separate out the first two terms from the first two power series above so that we can start all of our power series at the same place. Thus, we have (r  1)ra0 xr + r(1 + r)a1 x1+r +
r (k + r)(k + r  1)ak xk+r
k=2 1+r +ra0 x + (1 + r)a1 x r +
k=2 1+r (k + r)ak x k+r +
k=2 ak2 xk+r = 0 [r(r  1) + r] a0 x + [r(r + 1) + (r + 1)] a1 x +
k=2 [(k + r)(k + r  1)ak + (k + r)ak + ak2 ] xk+r = 0. By equating coefficients and assuming that a0 = 0, we obtain r(r  1) + r = 0 (the indicial equation), (r + 1)2 a1 = 0, [r(r + 1) + (r + 1)] a1 = 0 536 Exercises 8.6
and, for k 2, the recurrence relation (k + r)(k + r  1)ak + (k + r)ak + ak2 = 0 ak = ak2 , (k + r)2 k 2. Using the fact (which we found from the indicial equation above) that r1 = 0, we observe that a1 = 0. Next, using the recurrence relation (and the fact that r1 = 0), we see that ak = Hence, k=2 k=3 k=4 k=5 k=6 a0 , 4 a1 = 0, a3 = 9 a2 =  a4 = ak2 , k2 k 2. a2 = 16 a3 a5 = = 0, 25 a0  a4 a0 = 64 =  . a6 = 36 36 2304 a0 4 = a0 , 16 64 Substituting these coefficients into the solution w(0, x) =
n=0 an xn = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 + , we obtain the series solution for x > 0 given by w(0, x) = a0 1  1 4 1 1 2 x + x  x6 + . 4 64 2304 25. For this equation, we see that xp(x) = x/2 and x2 q(x) = (x + 3)/4. Thus, x = 0 is a regular singular point for this equation and we can use the method of Frobenius to find a solution. To this end, we compute lim xp(x) = lim x = 0, x0 2 and lim x2 q(x) = lim 3 (x + 3) = . x0 4 4 537 x0 x0 Chapter 8
Therefore, by equation (16) on page 463 of the text, the indicial equation is r(r  1)  3 =0 4 4r 2  4r  3 = 0 (2r + 1)(2r  3) = 0. This indicial equation has roots r1 = 3/2 and r2 = 1/2. By the method of Frobenius, we can assume that a solution to this differential equation will have the form w(r, x) =
n=0 an xn+r w (r, x) =
n=0 (n + r)an xn+r1 (n + r  1)(n + r)an xn+r2 ,
n=0 w (r, x) = where r = r1 = 3/2. Since x = 0 is the only singular point for this equation, we see that the solution, w(3/2, x), converges for all x > 0. The first step in finding this solution is to plug w(r, x) and its first and second derivatives (which we have found above by term by term differentiation) into the differential equation. Thus, we obtain 4(n + r  1)(n + r)an x
n=0 n+r +
n=0 2(n + r)an x n+r+1 
n=0 an x n+r+1 
n=0 3an xn+r = 0. By shifting indices, we can sum each power series over the same power of x, namely xk+r . Thus, with the substitution k = n in the first and last power series and the substitution k = n + 1 in the two remaining power series, we obtain 4(k + r  1)(k + r)ak xk+r +
k=0 k=1 2(k + r  1)ak1 xk+r 
k=1 ak1 xk+r 
k=0 3ak xk+r = 0. Next removing the first term (the k = 0 term) from the first and last power series above and writing the result as a single power series yields 4(r  1)ra0 xr +
k=1 4(k + r  1)(k + r)ak xk+r +
k=1 2(k + r  1)ak1xk+r 
k=1 ak1 x k+r  3a0 x 
r k=1 3ak xk+r = 0 538 Exercises 8.6 [4(r  1)r  3] a0 xr +
k=1 [4(k + r  1)(k + r)ak + 2(k + r  1)ak1  ak1  3ak ] xk+r = 0. By equating coefficients we see that each coefficient in the power series must be zero. Also we are assuming that a0 = 0. Therefore, we have 4(r  1)r  3 = 0, (the indicial equation), k 1. 4(k + r  1)(k + r)ak + 2(k + r  1)ak1  ak1  3ak = 0, Thus, the recurrence equation is given by ak = (3  2k  2r)ak1 , 4(k + r  1)(k + r)  3 k 1. Therefore, for r = r1 = 3/2, we have ak = 2kak1 , 4(k + 1/2)(k + 3/2)  3 k1 ak = ak1 , 2(k + 2) k 1. Thus, we see that k=1 k=2 k=3 k=4 a1 = a0 23 a1 a2 = 24 a2 a3 = 25 a3 a4 = 26 a0 , 20 3! a0 a0 = = 1 , 2234 2 4! a0 = 2 , 2 5! a0 = 3 . 2 6! = Inspection of this sequence shows that we can write the nth coefficient, an , for n 1 as an = (1)n a0 . 2n1 (n + 2)! Substituting these coefficients into the solution given by w 3 ,x 2 =
n=0 an xn+(3/2) , 539 Chapter 8
yields a power series solution for x > 0 given by w 3 ,x 2 = a0 x
3/2 + a0
n=1 (1)n xn+(3/2) . 2n1 (n + 2)! But since substituting n = 0 into the general coefficient, an , yields (1)0 a0 /(21 2!) = a0 , the solution that we found above can be written as w 3 ,x 2 = a0
n=0 (1)n xn+(3/2) . 2n1 (n + 2)! 27. In this equation, we see that p(x) = 1/x and q(x) = 1. Thus, the only singular point is x = 0. Since xp(x) = 1 and x2 q(x) = x2 , we see that x = 0 is a regular singular point for this equation and so we can use the method of Frobenius to find a solution to this equation. We also note that the solution that we find by this method will converge for all x > 0. To find this solution we observe that p0 = lim xp(x) = lim (1) = 1
x0 x0 and q0 = lim x2 q(x) = lim (x2 ) = 0.
x0 x0 Thus, according to equation (16) on page 463 of the text, the indicial equation for the point x = 0 is r(r  1)  r = 0 r(r  2) = 0. Therefore, the roots to the indicial equation are r1 = 2, r2 = 0. Hence, we will use the method of Frobenius to find the solution w(2, x). If we let w(r, x) =
n=0 an xn+r , then w (r, x) =
n=0 (n + r)an x n+r1 , and w (r, x) =
n=0 (n + r  1)(n + r)an xn+r2 . By substituting these expressions into the differential equation and simplifying, we obtain (n + r  1)(n + r)an x
n=0 n+r1 
n=0 (n + r)an x n+r1 
n=0 an xn+r+1 = 0. 540 Exercises 8.6
Next we shift the indices by letting k = n  1 in the first two power series above and k = n + 1 in the last power series above. Therefore, we have (k + r)(k + r + 1)ak+1 xk+r 
k=1 (k + r + 1)ak+1xk+r 
k=1 k=1 ak1 xk+r = 0. We can start all three of these summations at the same term, the k = 1 term, if we separate out the first two terms (the k = 1 and k = 0 terms) from the first two power series. Thus, we have (r  1)ra0 xr1 + r(r + 1)a1 xr +
k=1 (k + r)(k + r + 1)ak+1 xk+r r ra0 x r1  (r + 1)a1 x 
k=1 (k + r + 1)ak+1 x k+r 
k=1 ak1 xk+r = 0 [(r  1)r  r] a0 xr1 + [r(r + 1)  (r + 1)] a1 xr +
k=1 [(k + r)(k + r + 1)ak+1  (k + r + 1)ak+1  ak1 ] xk+r = 0. By equating coefficients and assuming that a0 = 0, we obtain r(r  1)  r = 0, (r + 1)(r  1)a1 = 0, (k + r)(k + r + 1)ak+1  (k + r + 1)ak+1  ak1 = 0, k 1, (the indicial equation), (8.13) where the last equation above is the recurrence relation. Simplifying this recurrence relation yields ak+1 = ak1 , (k + r + 1)(k + r  1) k 1. (8.14) Next we let r = r1 = 2 in equation (8.13) and in the recurrence relation, equation (8.14), to obtain 3a1 = 0 ak+1 a1 = 0, ak1 , k 1. = (k + 3)(k + 1) 541 Chapter 8
Thus, we have k=1 k=2 k=3 k=4 k=5 a2 = a3 = a4 = a5 = a6 = a0 , 42 a1 = 0, 53 a0 a0 a0 a2 = = 4 = 4 , 64 6442 2 32211 2 3! 2! a3 = 0, 75 a4 a0 a0 = = 6 . 86 8 6 24 3! 2! 2 4! 3! By inspection we can now see that the coefficients of the power series solution w(2, x) are a2n1 = 0 and a0 , (n + 1)!n! for all n 1. Thus, substituting these coefficients into the power series solution yields the a2n = 22n solution w(2, x) = a0
n=0 x2n+2 . 22n (n + 1)!n! 35. In applying the method of Frobenius to this third order linear differential equation, we will seek a solution of the form w(r, x) =
n=0 an xn+r w (r, x) =
n=0 (n + r)an xn+r1 (n + r  1)(n + r)an xn+r2
n=0 w (r, x) = w (r, x) = (n + r  2)(n + r  1)(n + r)an xn+r3 ,
n=0 where we have differentiated term by term. Substituting these expressions into the differential equation and simplifying yields 542 Exercises 8.6 6(n + r  2)(n + r  1)(n + r)an x
n=0 n+r +
n=0 13(n + r  1)(n + r)an xn+r +
n=0 (n + r)an xn+r +
n=0 (n + r)an xn+r+1 +
n=0 an xn+r+1 = 0. By the shift of index k = n + 1 in the last two power series above and the shift k = n in all of the other power series, we obtain 6(k + r  2)(k + r  1)(k + r)ak xk+r +
k=0 13(k + r  1)(k + r)ak xk+r
k=0 +
k=0 (k + r)ak x k+r +
k=1 (k  1 + r)ak1 x k+r +
k=1 ak1 xk+r = 0. Next we remove the first term from each of the first three power series above so that all of these series start at k = 1. Thus, we have 6(r  2)(r  1)ra0 xr +
k=1 r 6(k + r  2)(k + r  1)(k + r)ak xk+r +13(r  1)ra0 x +
k=1 13(k + r  1)(k + r)ak x k+r + ra0 x +
k=1 r (k + r)ak xk+r ak1 xk+r = 0
k=1 + [6(r  2)(r  1)r + 13(r  1)r + r] a0 x k=1 r (k  1 + r)ak1 xk+r + +
k=1 [6(k + r  2)(k + r  1)(k + r)ak + 13(k + r  1)(k + r)ak +(k + r)ak + (k  1 + r)ak1 + ak1 ] xk+r = 0. (8.15) If we assume that a0 = 0 and set the coefficient of xr equal to zero, we find that the indicial equation is 6(r  2)(r  1)r + 13(r  1)r + r = 0 r 2 (6r  5) = 0. Hence, the roots to the indicial equation are 0, 0, and 5/6. We will find the solution associated with the largest of these roots. That is, we will find w(5/6, x). Also, from equation (8.15), we 543 Chapter 8
see that we have the recurrence relation 6(k + r  2)(k + r  1)(k + r)ak + 13(k + r  1)(k + r)ak +(k + r)ak + (k  1 + r)ak1 + ak1 = 0, ak1 ak = , k 1. 6(k + r  2)(k + r  1) + 13(k + r  1) + 1 k1 If we assume that r = 5/6, then this recurrence relation simplifies to ak = Therefore, we have k=1 k=2 k=3 a1 = a0 , 11 a0 a1 a2 = = , 34 374 a2 a0 a3 = = . 69 25, 806 n+(5/6) , n=0 an x ak1 , k(6k + 5) k 1. By substituting these coefficients into the solution w(5/6, x) = w 5 ,x 6 = a0 x
5/6 we obtain x11/6 x17/6 x23/6  +  + . 11 374 25, 806 41. If we let z = 1/x ( dz/dx = 1/x2 ), then we can define a new function Y (z) as Y (z) := y Thus, by the chain rule, we have dY dy = = dx dx x2 dY dz dz dx = dY dz  1 x2 (8.16) (8.17) 1 z = y(x). dY dy = . dx dz Therefore, using the product rule and chain rule, we see that d2 Y d d2 y = = 2 2 dx dx dx 544 dY dx = d dx  1 x2 dY dz (by (8.16) above) Exercises 8.6
1 d  2 dx x dY 2 = 3 x dz = 2 = x3 2 dY + = 3 x dz Hence, we have dY dz +  1 x2 1 x2 d dx 2 dY dz 2 dY dz dz dx (by product rule) (by chain rule) since dz 1 = 2 dx x +  2 dY 1 +  2 dz x 2 1 dY . x4 dz 2 x3 d2 Y dz 2 1 d2 Y d2 Y d2 y dY dY + +z 2 . =2 =2 (8.18) dx2 dz x dz 2 dz dz By using the fact that Y (z) = y(x) and equations (8.17) and (8.18) above, we can now transform the original differential equation into the differential equation 2 d2 Y dY dY +z 2 + Y = 0 dz dz dz zY + 3Y  Y = 0. (8.19) We will now solve this transformed differential equation. To this end, we first note that p(z) = and 3 z zp(z) = 3, 1 z 2 g(z) = z. z Therefore, z = 0 is a regular singular point of this equation and so infinity is a regular singular q(z) point of the original equation. To find a power series solution for equation (8.19), we first compute p0 = lim zp(z) = 3
z0 and q0 = lim z 2 q(z) = 0.
z0 Thus, the indicial equation for equation (8.19) is r(r  1) + 3r = 0 r(r + 2) = 0. Hence, this indicial equation has roots r1 = 0 and r2 = 2. We seek a solution of the form w(r, z) =
n=0 an z n+r . 545 Chapter 8
Substituting this expression into equation (8.19) above yields z
n=0 (n + r  1)(n + r)an z n+r2 + 3
n=0 (n + r)an z n+r1 
n=0 an z n+r = 0. By simplifying, this equation becomes (n + r  1)(n + r)an z
n=0 n+r1 +
n=0 3(n + r)an z n+r1 
n=0 an z n+r = 0. Making the shift of index k = n  1 in the first two power series and k = n in the last power series allows us to sum each power series over the same powers of z, namely z k+r . Thus, we have (k + r)(k + r + 1)ak+1 z
k=1 k+r +
k=1 3(k + r + 1)ak+1 z n+r 
k=0 ak z k+r = 0. By removing the first term from the first two power series above, we can write these three summations as a single power series. Therefore, we have (r  1)ra0 z r1 +
k=0 (k + r)(k + r + 1)ak+1z k+r +3ra0 z r1 +
k=0 3(k + r + 1)ak+1 z n+r 
k=0 ak z k+r = 0 [(r  1)r + 3r] a0 z r1 +
k=0 [(k + r)(k + r + 1)ak+1 + 3(k + r + 1)ak+1  ak ] z k+r = 0. Equating coefficients and assuming that a0 = 0 yields the indicial equation, (r  1)r + 3r = 0, and the recurrence relation (k + r)(k + r + 1)ak+1 + 3(k + r + 1)ak+1  ak = 0, ak , k 0. ak+1 = (k + r + 1)(k + r + 3) Thus, with r = r1 = 0, we obtain the recurrence relation ak+1 = 546 ak , (k + 1)(k + 3) k 3. k0 Exercises 8.7
Since a0 is an arbitrary number, we see from this recurrence equation that the next three coefficients are given by k=0 k=1 k=2 a0 , 3 a1 a2 = = 8 a2 = a3 = 15 a1 = a0 , 24 a0 . 360 Thus, from the method of Frobenius, we obtain a power series solution for equation (8.19) given by Y (z) = w(0, z) =
n=0 an z n = a0 1 + 1 2 1 3 1 z+ z + z + . 3 24 360 In order to find the solution of the original differential equation, we again make the substitution z = 1/x and Y (z) = Y (x1 ) = y(x). Therefore, in the solution found above, we replace the z's with 1/x to obtain the solution given by y(x) = Y x1 = a0 1 + EXERCISES 8.7: 1 2 1 3 1 1 x + x + x + . 3 24 360 Finding a Second Linearly Independent Solution, page 482 3. In Problem 21 of Exercises 8.6, we found one power series solution for this differential equation about the point x = 0 given by y1 (x) = 1  1 2 1 4 1 x + x  x6 + , 4 64 2304 where we let a0 = 1. We also found that the roots to the indicial equation are r1 = r2 = 0. Thus, to find a second linearly independent solution about the regular singular point x = 0, we will use part (b) of Theorem 7 on page 475 of the text. Therefore, we see that this second linearly independent solution will have the form given by y2 (x) = y1 (x) ln x +
n=1 bn xn
1 y2 (x) = y1 (x) ln x + x y1 (x) +
n=1 nbn xn1 547 Chapter 8 y2 (x) = y1 (x) ln x + 2x1 y1 (x)  x2 y1 (x) + n(n  1)bn xn2 .
n=1 Substituting these expressions into the differential equation yields x2 y1 (x) ln x + 2x1 y1 (x)  x2 y1 (x) + +x y1 (x) ln x + x1 y1 (x) + which simplifies to x y1 (x) ln x + 2xy1 (x)  y1 (x) +
n=1 2 n(n  1)bn xn2
n=1 nbn xn1
n=1 + x2 y1 (x) ln x +
n=1 bn xn = 0, n(n  1)bn xn nbn x + x y1 (x) ln x +
n=1 n=1 n 2 +xy1 (x) ln x + y1 (x) + 2 2 bn xn+2 = 0, x y1 (x) + xy1 (x) + x y1 (x) ln x + 2xy1 (x) +
n=1 n(n  1)bn x +
n n=1 nbn x +
n=1 n bn xn+2 = 0. Therefore, since y1 (x) is a solution to the differential equation, the term in braces is zero and the above equation reduces to 2xy1 (x) +
n=1 n(n  1)bn x +
n n=1 nbn x +
n=1 n bn xn+2 = 0. Next we make the substitution k = n + 2 in the last power series above and the substitution k = n in the other two power series so that we can sum all three of the power series over the same power of x, namely xk . Thus, we have 2xy1 (x) +
k=1 k(k  1)bk x +
k k=1 kbk x +
k=3 k bk2 xk = 0. By separating out the first two terms in the first two summations above and simplifying, we obtain 2xy1 (x) + 0 + 2b2 x +
k=3 2 k k(k  1)bk x + b1 x + 2b2 x +
k=3 2 kbk x +
k=3 k bk2 xk = 0 548 Exercises 8.7 2xy1 (x) + b1 x + 4b2 x2 + k 2 bk + bk2 xk = 0. (8.20) k=3 By differentiating the series for y1 (x) term by term, we obtain 1 3 1 1 5 x  x + . y1 (x) =  x + 2 16 384 Thus, substituting this expression for y1 (x) into equation (8.20) above and simplifying yields 1 1 6 x + x4  x + + b1 x + 4b2 x2 + k 2 bk + bk2 xk = 0. 8 192 k=3
2 Therefore, by equating coefficients, we see that b1 = 0; 4b2  1 = 0 9b3 + b1 = 0 1 + 16b4 + b2 = 0 8 25b5 + b3 = 0 1 + 36b6 + b4 = 0 192 Substituting these coefficients into the solution b2 = b3 b4 b5 b6 1 ; 4 = 0; 3 ; = 128 = 0; 11 . = 13, 824 y2 (x) = y1 (x) ln x +
n=1 bn xn , yields 1 2 3 4 11 x  x + x6 + . 4 128 13, 824 Thus, a general solution of this differential equation is given by y2 (x) = y1 (x) ln x + y(x) = c1 y1 (x) + c2 y2 (x), where y1 (x) = 1  1 2 1 4 1 x + x  x6 + , 4 64 2304 1 3 4 11 x + x6 + . y2 (x) = y1 (x) ln x + x2  4 128 13, 824 549 Chapter 8
7. In Problem 25 of Section 8.6, we found a solution to this differential equation about the regular singular point x = 0 given by y1 (x) =
n=0 (1)n xn+(3/2) 1 1 7/2 = x3/2  x5/2 + x + , n1 (n + 2)! 2 6 48 where we let a0 = 1. We also found that the roots to the indicial equation for this problem are r1 = 3/2 and r2 = 1/2, and so r1  r2 = 2. Thus, in order to find a second linearly independent solution about x = 0, we will use part (c) of Theorem 7 on page 475 of the text. Therefore, we will assume that this second solution has the form y2 (x) = Cy1 (x) ln x +
n=0 bn xn(1/2) , b0 = 0 1 1 y2 (x) = Cy1 (x) ln x + C y1 (x) + n bn xn(3/2) x 2 n=0 1 1 3 y2 (x) = Cy1 (x) ln x + 2C y1 (x)  C 2 y1 (x) + n x x 2 n=0 n 1 2 bn xn(5/2) . Substituting these expressions into the differential equation yields 4x
2 3 1 1 n Cy1 (x) ln x + 2C y1 (x)  C 2 y1 (x) + x x 2 n=0 +2x
2 n 1 bn xn(5/2) 2 bn xn(3/2) bn xn(1/2) = 0.
n=0 1 1 n Cy1 (x) ln x + C y1 (x) + x 2 n=0 (x + 3) Cy1 (x) ln x + Multiplying through, we get 4x2 Cy1 (x) ln x + 8Cxy1 (x)  4Cy1 (x) +
2 4 n
n=0 3 2 n 1 bn xn(1/2) 2 1 2 + 2x Cy1 (x) ln x + 2Cxy1 (x) +
n=0 2 n bn xn+(1/2) 3bn xn(1/2) = 0,  Cxy1 (x) ln x +
n=0 bn x n+(1/2) + 3Cy1(x) ln x +
n=0 550 Exercises 8.7
which simplifies to C 4x2 y1 (x) + 2x2 y1 (x)  xy1 (x)  3y1 (x) ln x + 8Cxy1 (x) + 2C(x  2)y1 (x) +
n=0 (2n  3) (2n  1) bn x  n(1/2) +
n=0 (2n  1) bn xn+(1/2) bn xn+(1/2) 
n=0 n=0 3bn xn(1/2) = 0. Since y1 (x) is a solution to the differential equation, the term in brackets is zero. By shifting indices so that each power series is summed over the same power of x, we have 8Cxy1 (x) + 2C(x  2)y1(x) +
k=0 (2k  3) (2k  1) bk xk(1/2) k(1/2) +
k=1 (2k  3) bk1 x 
k=1 bk1 x k(1/2) 
k=0 3bk xk(1/2) = 0. By writing all of these summations as a single power series (noting that the k = 0 term of the first and last summations add to zero), we obtain 8Cxy1 (x) + 2C(x  2)y1(x) +
k=0 (2k  3) (2k  1) bk xk(1/2) +
k=1 [(2k  3) (2k  1) bk + (2k  3) bk1  bk1  3bk ] xk(1/2) = 0. Substituting into this equation the expressions for y1 (x) and y1 (x) given by y1 (x) =
n=0 (1)n xn+(3/2) , 2n1 (n + 2)! y1 (x) =
n=0 (1)n [n + (3/2)]xn+(1/2) , 2n1 (n + 2)! yields n=0 8C(1)n [n + (3/2)]xn+(3/2) 2C(1)n xn+(5/2) + 2n1 (n + 2)! 2n1 (n + 2)! n=0 
n=0 4C(1)n xn+(3/2) + 2n1 (n + 2)! [4k(k  2)bk + 2(k  2)bk1 ] xk(1/2) = 0,
k=1 551 Chapter 8
where we have simplified the expression inside the last summation. Combining the first and third power series yields n=0 8C(1)n (n + 1)xn+(3/2) 2C(1)n xn+(5/2) + 2n1 (n + 2)! 2n1 (n + 2)! n=0 +
k=1 [4k(k  2)bk + 2(k  2)bk1 ] xk(1/2) = 0, (8.21) By writing out the terms up to order x7/2 , we obtain 8C x3/2  1 5/2 3 7/2 1 x + x + + 2C x5/2  x7/2 + 3 16 6 + (4b1  2b0 )x1/2 + (12b3 + 2b2 )x5/2 + (32b4 + 4b3 )x7/2 + = 0. Setting the coefficients equal to zero, yields 4b1  2b0 = 0 8C = 0 (8/3)C + 2C + 12b3 + 2b2 = 0 b1 = b0 /2; C = 0; b3 = b2 /6; b4 = b3 /8 = b2 /48. (2/3)C  (1/3)C + 32b4 + 4b3 = 0 From this we see that b0 and b2 are arbitrary constants and that C = 0. Also, since C = 0, we can use the last power series in (8.21) to obtain the recurrence equation bk = bk1 /(2k). Thus, every coefficient after b4 will depend only on b2 (not on b0 ). Substituting these coefficients into the solution, y2 (x) = Cy1 (x) ln x +
n=0 bn xn(1/2) , yields y2 (x) = b0 x1/2  1 1/2 1 1 7/2 x x + + b2 x3/2  x5/2 + 2 6 48 The expression in the brackets following b2 is just the series expansion for y1 (x). Hence, in order to obtain a second linearly independent solution, we must choose b0 to be nonzero. Taking b0 = 1 and b2 = 0 gives y2 (x) = x1/2  552 1 1/2 x . 2 Exercises 8.7
Therefore, a general solution is y(x) = c1 y1 (x) + c2 y2 (x), where y1 (x) = x3/2  1 5/2 1 7/2 x + x + 6 48 and y2 (x) = x1/2  1 1/2 x . 2 17. In Problem 35 of Section 8.6, we assumed that there exists a series solution to this problem of the form w(r, x) = n=0 an xn+r . This led to the equation (cf. equation (8.15), of the solution to Problem 35, Exercises 8.6) r (6r  5)a0 x +
r k=1 2 (k + r)2 [6(k + r)  5]ak + (k + r)ak1 xk+r = 0. (8.22) From this we found the indicial equation r 2 (6r  5) = 0, which has roots r = 0, 0, 5/6. By using the root 5/6, we found the solution w(5/6, x). Hence one solution is y1 (x) = x5/6  x23/6 x11/6 x17/6 +  + , 11 374 25, 806 where we have chosen a0 = 1 in w(5/6, x). We now seek two more linearly independent solutions to this differential equation. To find a second linearly independent solution, we will use the root r = 0 and set the coefficients in equation (8.22) to zero to obtain the recurrence relation k 2 (6k  5)ak + kak1 = 0, Solving for ak in terms of ak1 gives ak = Thus, we have k=1 k=2 a1 = a0 , a0 a1 = , a2 = 14 14 553 ak1 , k(6k  5) k 1. k 1. Chapter 8
k=3 k=4 k=5 a0 a2 = , 39 546 a0 a3 = , a4 = 76 41, 496 a4 a0 a5 = = . 125 5, 187, 000 a3 = Plugging these coefficients into the solution w(0, x) and setting a0 = 1 yields a second linearly independent solution y2 (x) = 1  x + 1 2 1 3 1 1 x  x + x4  x5 + . 14 546 41, 496 5, 187, 000 To find a third linearly independent solution, we will use the repeated root r = 0 and assume that, as in the case of second order equations with repeated roots, the solution that we seek will have the form y3 (x) = y2 (x) ln x +
n=1 cn xn . Since the first three derivatives of y3 (x) are given by y3 (x) = y2 (x) ln x + x y2 (x) +
n=1 1 ncn xn1 , y3 (x) = y2 (x) ln x + 2x1 y2 (x)  x2 y2 (x) +
1 2 (n  1)ncn xn2 ,
n=1 3 y3 (x) = y2 (x) ln x + 3x y2 (x)  3x y2 (x) + 2x y2 (x) +
n=1 (n  2)(n  1)ncn xn3 , substituting y3 (x) into the differential equation yields 6x3 y (x) + 13x2 y (x) + x + x2 y (x) + xy(x) = 6x3 y2 (x) ln x + 3x1 y2 (x)  3x2 y2 (x) + 2x3 y2 (x) + +13x + x+x 554
2 2 1 2 (n  2)(n  1)ncn xn3
n=1 y2 (x) ln x + 2x y2 (x)  x y2 (x) +
n=1 1 (n  1)ncn xn2 y2 (x) ln x + x y2 (x) +
n=1 ncn x n1 + x y2 (x) ln x +
n=1 cn xn = 0. Exercises 8.7
Since y2 (x) is a solution to the given equation, this simplifies to 18x y2 (x) + 8xy2 (x) + xy2 (x) +
n=1 2 6(n  2)(n  1)ncn xn
n +
n=1 13(n  1)ncn x +
n=1 ncn x +
n=1 n ncn x n+1 +
n=1 cn xn+1 = 0. By shifting indices and then starting all of the resulting power series at the same point, we can combine all of the summations above into a single power series. Thus, we have 18x2 y2 (x) + 8xy2 (x) + xy2 (x) +c1 x +
k=2 [6(k  2)(k  1)kck + 13(k  1)kck + kck + kck1] xk = 0. (8.23) By computing y2 (x) and y2 (x), we obtain 1 1 1 2 x x + x3 + , 7 182 10374 1 1 1 y2 (x) =  x+ x2 + . 7 91 3458 y2 (x) = 1 + By substituting these expressions into equation (8.23), we have 18x2 x x2 x x2 x3 1  + + + 8x 1 +  + + 7 91 3458 7 182 10374 x2 x3 x4 +x 1  x +  + + + c1 x + 14 546 41, 496 Writing out the terms up to orderx3 we find (7 + c1 ) x + (6k 3  5k 2 )ck + kck1 xk = 0. k=2 19 31 + 28c2 + 2c1 x2 +  + 117c3 + 3c2 x3 + = 0. 7 182 By equating coefficients to zero, we obtain 7 + c1 = 0 19/7 + 28c2 + 2c1 = 0 c1 = 7; c2 = 117/196; c3 = 4997/298116. 555 31/182 + 117c3 + 3c2 = 0 Chapter 8
Therefore, plugging these coefficients into the expansion y3 (x) = y2 (x) ln x +
n=1 cn xn , yields a third linearly independent solution is given by y3 (x) = y2 (x) ln x + 7x  Thus, a general solution is y(x) = c1 y1 (x) + c2 y2 (x) + c3 y3 (x), where x11/6 x17/6 x23/6 +  + , 11 374 25, 806 x3 x4 x5 x2  +  + , y2 (x) = 1  x + 14 546 41, 496 5, 187, 000 4997x3 117x2 y3 (x) = y2 (x) ln x + 7x  + + . 196 298, 116 y1 (x) = x5/6  23. We will try to find a solution of the form 117 2 4997 3 x + x + . 196 298116 y(x) =
n=0 an xn+r y (x) =
n=0 (n + r)an xn+r1 (n + r)(n + r  1)an xn+r2 .
n=0 y (x) = Therefore, we substitute these expressions into the differential equation to obtain x2 y + y  2y = x2 (n + r)(n + r  1)an xn+r2 +
n=0 n=0 (n + r)an xn+r1  2
n=0 an xn+r = 0 k=0 (k + r)(k + r  1)ak xk+r + (k + r + 1)ak+1 xk+r 
k=1 k=0 2ak xk+r = 0 556 Exercises 8.7 ra0 xr1 +
k=0 [(k + r)(k + r  1)ak + (k + r + 1)ak+1  2ak ] xk+r = 0, where we have changed all of the indices and the starting point for the second summation so that we could write these three power series as a single power series. By assuming that a0 = 0 and ra0 xr1 = 0, we see that r = 0. Plugging r = 0 into the coefficients in the summation and noting that each of these coefficients must be zero yields the recurrence relation k(k  1)ak + (k + 1)ak+1  2ak = 0, ak+1 = (2  k)ak , k 0. k0 Thus, we see that the coefficients of the solution are given by k=0 k=2 a1 = 2a0 ; k = 1 a3 = 0 ; k=3 a2 = a1 = 2a0 ; a4 = a3 = 0 . Since each coefficient is a multiple of the previous coefficient, we see that an = 0 for n 3. If we take a0 = 1, then one solution is y1 (x) = 1 + 2x + 2x2 . We will now use the reduction of order procedure described in Problem 31, Section 6.1, on page 326 of the text to find a second linearly independent solution. Thus we seek for a solution of the form y(x) = y1 (x)v(x) y (x) = y1 (x)v(x) + y1 (x)v (x) y (x) = y1 (x)v(x) + 2y1(x)v (x) + y1 (x)v (x). Substituting y(x), y (x), and y (x) into the given equation yields x2 y + y  2y = x2 [y1 v + 2y1 v + y1 v ] + [y1 v + y1 v ]  2 [y1 v] = = x2 y1 v + 2x2 y1 + y1 v + x2 y1 + y1  2y1 v x2 y1 v + 2x2 y1 + y1 v = 0 557 Chapter 8
(since y1 is a solution, the coefficient at v equals to zero). With w = v , the last equation becomes a first order separable equation which can be solved by methods of Section 2.2. Namely, x2 y1 (x) w (x) + 2x2 y1 (x) + y1 (x) w(x) = 0 2y (x) dw 2x2 y1 (x) + y1 (x) 1 = dx =  + 2 dx 2 y (x) w x 1 y1 (x) x dx 1 2y (x)dx  = 2 ln y1(x) + ln w =  y1 (x) x2 x 1 e1/x , = w(x) = exp 2 ln y1(x) + x [y1 (x)]2 where we have taken zero integration constant and positive function w. Since (8.24) and [y1 (x)]2 = 4x4 + 8x3 + 8x2 + 4x + 1 2 3 4 x x x + + + e1/x = 1 + x1 + 2 6 24 (we have used the Maclaurin expansion for ez with z = 1/x), performing long division with descending powers of x in each polynomial, we see that e = [y1 (x)]2
1/x 1 + x1 + x2 x3 x4 + + + 1 1 1 2 6 24 = x4  x5 + x6 + . 4 + 8x3 + 8x2 + 4x + 1 4x 4 4 8 Therefore, (8.24) yields 1 1 1 v (x) = w(x) = x4  x5 + x6 + 4 4 8 1 4 1 5 1 6 1 1 4 1 5 x  x + x + dx =  x3 + x  x + v(x) = 4 4 8 12 16 40 and so 1 4 1 5 1 x  x + y(x) = y1 (x)v(x) = 1 + 2x + 2x2  x3 + 12 16 40 1 2 1 3 1 x  x + . =  x1  6 24 120 is a second linearly independent solution. Thus, a general solution to this differential equation is given by y(x) = c1 y1 (x) + c2 y2 (x), where y1 (x) = 1 + 2x + 2x2 558 and 1 1 2 1 3 x  x + . y2 (x) =  x1  6 24 120 Exercises 8.8
EXERCISES 8.8: Special Functions, page 493 1. For this problem, we see that = 1/2, + + 1/4, and = 2. First we note that is not an integer. Next, by solving in the last two equations above simultaneously for and , we see that either = 1 and = 1 or = 2 and = 1. Therefore, by assuming that = 1 and = 2, equations (10) on page 485 and (17) on page 486 of the text yield the two solutions y1 (x) = F 1, 2; 1 ;x 2 and y2 (x) = x1/2 F 3 5 3 , ; ;x . 2 2 2 Therefore, a general solution for this differential equation is given by y(x) = c1 F Notice that 1, 2; 1 ; x + c2 x1/2 F 2 3 5 3 , ; ;x . 2 2 2 F (, ; ; x) = 1 +
n=0 ()n ()n n ()n ()n n x =1+ x = F (, ; ; x). n!()n n!()n n=0 1 ; x + c2 x1/2 F 2 5 3 3 , ; ;x . 2 2 2 Therefore, letting = 2 and = 1 yields an equivalent form of the same solution given by y(x) = c1 F 13. This equation can be written as x2 y + xy + x2  1 y = 0. 4 2, 1; Thus, 2 = 1/4 which implies that = 1/2. Since this is not an integer (even though 2 is an integer), by the discussion on page 487 of the text, two linearly independent solutions to this problem are given by equations (25) and (26) also on page 487, that is y1 (x) = J1/2 (x) =
n=0 (1)n x n! (3/2 + n) 2 (1)n x n! (1/2 + n) 2 2n+1/2 ,
2n1/2 y2 (x) = J1/2 (x) =
n=0 . Therefore, a general solution to this differential equation is given by y(x) = c1 J1/2 (x) + c2 J1/2 (x). 559 Chapter 8
15. In this problem = 1. Thus, one solution to this differential equation is given by y1 (x) = J1 (x) =
n=0 (1)n x n! (2 + n) 2 2n+1 . By the discussion on page 487 of the text, J1 (x) and J1 (x) are linearly dependent. Thus, J1 (x) will not be a second linearly independent solution for this problem. But, a second linearly independent solution will be given by equation (30) on page 488 of the text with m = 1. That is we have y2 (x) = Y1 (x) = lim cos()J (x)  J (x) . sin() 1 Therefore, a general solution to this differential equation is given by y(x) = c1 J1 (x) + c2 Y1 (x). 21. Let y(x) = x J (x). Then, by equation (31) on page 488 of the text, we have y (x) = x J1 (x). Therefore, we see that y (x) = Dx [y (x)] = Dx [x J1 (x)] = Dx x x1 J1 (x) = x1 J1 (x) + xDx x1 J1 (x) = x1 J1 (x) + x J2 (x). Notice that in order to take the last derivative above, we have again used equation (31) on page 488 of the text. By substituting these expressions into the lefthand side of the first differential equation given in the problem, we obtain xy + (1  2)y + xy = x x1 J1 (x) + x J2 (x) + (1  2) [x J1 (x)] + x [x J (x)] = x J1 (x) + x+1 J2 (x) + x J1 (x)  2x J1 (x) + x+1 J (x). Notice that by equation (33) on page 488 of the text, we have J (x) = 560 2(  1) J1 (x)  J2 (x) x (8.25) Exercises 8.8 x+1 J (x) = 2(  1)x J1 (x)  x+1 J2 (x). Replacing x+1 J (x) in equation (8.25) with the above expression and simplifying yields xy + (1  2)y + xy = x J1 (x) + x+1 J2 (x) + x J1 (x) 2x J1 (x) + 2(  1)x J1 (x)  x+1 J2 (x) = 0. Therefore, y(x) = x J (x) is a solution to this type of differential equation. In order to find a solution to the differential equation xy  2y + xy = 0, we observe that this equation is of the same type as the equation given above with 1  2 = 2 Thus, a solution to this equation will be y(x) = x3/2 J3/2 (x) = x3/2 = 3 . 2 n=0 (1)n x n!(5/2 + n) 2 2n+3/2 . 29. In Legendre polynomials, n is a fixed nonnegative integer. Thus, in the first such polynomial, n equals zero. Therefore, we see that [n/2] = [0/2] = 0 and, by equation (43) on page 491 of the text, we have P0 (x) = 20 Similarly, we have n=1 n=2 n=3 n=4 1 2 2 2 3 2 4 2 =0 =1 =1 =2 (1)0 2! 1 x = x, 1!0!1! (1)0 4! 2 (1)1 2! 0 3x2  1 x + x = , P2 (x) = 22 2!0!2! 1!1!0! 2 (1)0 6! 3 (1)1 4! 1 5x3  3x x + x = , P3 (x) = 23 3!0!3! 2!1!1! 2 (1)0 8! 4 (1)1 6! 2 (1)2 4! 0 P4 (x) = 24 x + x + x 4!0!4! 3!1!2! 2!2!0! 35x4  30x2 + 3 . = 8 P1 (x) = 21 561 (1)0 0! 0 x = 1. 0!0!0! Chapter 8
37. Since the Taylor series expansion of an analytic function f (t) about t = 0 is given by f (t) =
n=0 f (n) (0) n t , n!
2 we see that H(x) is just the nth derivative of y(t) = e2txt with respect to t evaluated at the point t = 0 (treating x as a fixed parameter). Therefore, we have y(t) = e2txt
2 2 2 2 H0 (x) = y(0) = e0 = 1, H1 (x) = y (0) = 2xe0 = 2x, H2 (x) = y (0) = [2 + (2x)2 ] e0 = 4x2  2, H3 (x) = y (0) = 8x3  12x. y (t) = (2x  2t)e2txt y (t) = [2 + (2x  2t)2 ] e2txt y (t) = [6(2x  2t) + (2x  2t)3 ] e2txt 39. To find the first four Laguerre polynomials, we need to find the first four derivatives of the function y(x) = xn ex . Therefore, we have y (0) (x) = xn ex , y (x) = nxn1  xn ex , y (x) = n(n  1)xn2  2nxn1 + xn ex , y (x) = n(n  1)(n  2)xn3  3n(n  1)xn2 + 3nxn1  xn ex . Substituting these expressions into Rodrigues's formula and plugging in the appropriate values of n yields L0 (x) = ex 0! ex L1 (x) = 1! ex L2 (x) = 2! ex L3 (x) = 3! x0 ex = 1, 1x11  x1 ex = 1  x, 2(2  1)x22  2 2x21 + x2 ex = 2  4x + x2 , 2 6  18x + 9x2  x3 . 6 3(3  1)(3  2)x33  3 3(3  1)x32 + 3 3x31  x3 ex = 562 Review Problems
REVIEW PROBLEMS: page 497 1. (a) To construct the Taylor polynomials pn (x) = y(0) + y (0) y (n) (0) n y (0) 2 x+ x ++ x 1! 2! n! approximating the solution to the given initial value problem, we need y(0), y (0), etc. y(0) is provided by the initial condition, y(0) = 1. The value of y (0) can be deduced from the differential equation itself. We have y (0) = (0)y(0)  y(0)2 = (0)(1)  (1)2 = 1. Differentiating both sides of the given equation, y = xy  y 2 , and substituting x = 0 into the resulting equation, we get y = y + xy  2yy y (0) = y(0) + (0)y (0)  2y(0)y (0) = (1) + (0)(1)  2(1)(1) = 3. Differentiating once more yields y = y + y + xy  2y y  2yy Thus, p3 (x) = 1 + y (0) = (1) + (1) + (0)(3)  2(1)(1)  2(1)(3) = 10. 3 1 10 3 3x2 5x3 x + x2 + x = 1x+  . 1! 2! 3! 2 3 (b) The values of z(0) and z (0) are given. Namely, z(0) = 1 and z (0) = 1. Substituting x = 0 into the given equation yields z (0)  (0)3 z (0) + (0)z(0)2 = 0 z (0) = 0. We now differentiate the given equation and evaluate the result at x = 0. z  3x2 z  x3 z + z 2 + 2xzz = 0 563 Chapter 8 z (0) = 3(0)2z (0) + (0)3 z (0)  z(0)2  2(0)z(0)z (0) = 1. One more differentiation yields z (4)  6xz  3x2 z  3x2 z  x3 z + 2zz + 2zz + 2xz z + 2xzz = 0 Hence, p4 (x) = 1 + z (4) (0) = 4z(0)z (0) = 4. 1 1 3 4 4 x3 x4 0 x + x2 + x + x = 1 + x  + . 1! 2! 3! 4! 6 6 3. (a) Since both p(x) = x2 and q(x) = 2 are analytic at x = 0, a general solution to the given equation is also analytic at this point. Thus, it has an expansion y=
k=0 ak xk y =
k=1 kak xk1 k(k  1)ak xk2 .
k=2 y = Substituting these expansions for y, y , and y into the original equation yields k(k  1)ak x
k=2 k2 +x 2 kak x
k=1 k1 2
k=0 ak xk = 0 k=2 k(k  1)ak xk2 +
k=1 kak xk+1 
k=0 2ak xk = 0. We now shift the indices of summation so that all three sums contain like powers xn . In the first sum, we let k  2 = n; in the second sum, let k + 1 = n; and let k = n in the third sum. This yields (n + 2)(n + 1)an+2 xn +
n=0 n=2 (n  1)an1 xn 
n=0 2an xn = 0. Separating the terms corresponding to n = 0 and n = 1, and combining the rest under one summation, we obtain (2a2  2a0 ) + (6a3  2a1 ) x +
n=2 [(n + 2)(n + 1)an+2 + (n  1)an1  2an ] xn = 0. 564 Review Problems
Therefore, 2a2  2a0 = 0, 6a3  2a1 = 0, (n + 2)(n + 1)an+2 + (n  1)an1  2an = 0, This yields a2 = a0 , Hence, y(x) = a0 + a1 x + a2 x2 + a3 x3 + = a0 + a1 x + a0 x2 + = a0 1 + x2 + + a1 x + x3 + . 3 a1 3 x + 3 a3 = a1 , 3 and an+2 = 2an  (n  1)an1 , (n + 2)(n + 1) n 2. n 2. 5. Clearly, x = 2 is an ordinary point for the given equation because p(x) = x  2 and q(x) = 1 are analytic everywhere. Thus we seek for a solution of the form w(x) =
k=0 ak (x  2)k . Differentiating this power series yields w (x) =
k=1 kak (x  2) k1 and w (x) =
k=2 k(k  1)ak (x  2)k2 . Therefore, w + (x  2)w  w =
k=2 k(k  1)ak (x  2)k2 + (x  2)
k=1 kak (x  2)k1 
k=0 ak (x  2)k = 0 k=2 k(k  1)ak (x  2)k2 +
k=1 kak (x  2)k 
k=0 ak (x  2)k = 0. Shifting the index of summation in the first sum yields (n + 2)(n + 1)an+2 (x  2) +
n n=0 n=1 nan (x  2) 
n n=0 an (x  2)n = 0 565 Chapter 8 2a2 +
n=1 (n + 2)(n + 1)an+2 (x  2) n +
n=1 nan (x  2)  a0 +
n n=1 an (x  2)n = 0 (2a2  a0 ) +
n=1 [(n + 2)(n + 1)an+2 + nan  an ] (x  2)n = 0, where we have separated the terms corresponding to n = 0 and collected the rest under one summation. In order that the above power series equals zero, it must have all zero coefficients. Thus, 2a2  a0 = 0, (n + 2)(n + 1)an+2 + nan  an = 0, a2 = a0 /2 , an+2 = (1  n)an /[(n + 2)(n + 1)] , n 1. n1 For n = 1 and n = 2, the last equation gives a3 = 0 and a4 = a2 /12 = a0 /24. Therefore, w(x) = a0 + a1 (x  2) + a2 (x  2)2 + a3 (x  2)3 + a4 (x  2)4 + a0 1 a0 (x  2)4 + = a0 + a1 (x  2) + (x  2)2 + (0)(x  2)3  2 24 (x  2)2 (x  2)4 = a0 1 +  + + a1 (x  2). 2 24 7. (a) The point x = 0 is a regular singular point for the given equation because p(x) = and the limits p0 = lim xp(x) = lim (5) = 5,
x0 x0 5 5x = , 2 x x q(x) = 9x , x2 q0 = lim x q(x) = lim (9  x) = 9
x0 x0 2 exist. The indicial equation (3) on page 461 of the text becomes r(r  1) + (5)r + 9 = 0 566 r 2  6r + 9 = 0 (r  3)2 = 0. Review Problems
Hence, r = 3 is the exponent of the singularity x = 0, and a solution to the given differential equation has the form y=x
3 ak x =
k=0 k=0 k ak xk+3 . Substituting this power series into the given equation yields x
2 ak x
k=0 k+3  5x
k=0 ak x
k+3 k+3 + (9  x)
k=0 ak xk+3
k+3 =0 k=0 (k + 3)(k + 2)ak x 
k=0 5(k + 3)ak x + (9  x)
k=0 ak xk+3 = 0 k=0 [(k + 3)(k + 2)  5(k + 3) + 9] ak xk+3 
k=0 ak xk+4 = 0 k=0 k 2 ak xk+3  n2 an xn+3  ak xk+4 = 0
k=0 n=1 an1 xn+3 = 0
n=1 n=1 n2 an  an1 xn+3 = 0 . Thus, n2 an  an1 = 0 This recurrence relation yields n = 1 : a1 = a0 /(1)2 = a0 , n = 2 : a2 = a1 /(2)2 = a0 /4 , n = 3 : a3 = a2 /(3)2 = (a0 /4) /9 = a0 /36 . Therefore, y(x) = x3 a0 + a1 x + a2 x2 + a3 x3 + a0 a0 3 x5 x6 x + = a0 x3 + x4 + + + . = x3 a0 + a0 x + x2 + 4 36 4 36 or an = an1 , n2 n 1. 567 Chapter 8 568 CHAPTER 9: Matrix Methods for Linear Systems
EXERCISES 9.1: Introduction, page 507 3. We start by expressing righthand sides of all equations as dot products. x + y + z = [1, 1, 1] [x, y, z], 2z  x = [1, 0, 2] [x, y, z], 4y = [0, 4, 0] [x, y, z]. Thus, by definition of the product of a matrix and column vector, the matrix form is given by 1 1 1 x x y = 1 0 2 y . 0 4 0 z z 7. First we have to express the second derivative, y , as a first derivative in order to rewrite the equation as a first order system. Denoting y by v we get y = v, mv + bv + ky = 0 y = v, or v = b k y v. m m Expressing the righthand side of each equation as a dot product, we obtain v = [0, 1] [y, v], Thus, the matrix form of the system is y v 11. Introducing the auxiliary variables x1 = x, x2 = x , x3 = y, x4 = y , 569 = 0 1 y v .  b k b k y  v =  , [y, v]. m m m m k/m b/m Chapter 9
we can rewrite the given system in normal form: x1 = x2 x3 = x4 x2 + 3x1 + 2x3 = 0 x4  2x1 = 0 Since x2 = [0, 1, 0, 0] [x1 , x2 , x3 , x4 ], x4 = [0, 0, 0, 1] [x1 , x2 , x3 , x4 ], the matrix is given by 3x1  2x3 = [3, 0, 2, 0] [x1 , x2 , x3 , x4 ], 2x1 = [2, 0, 0, 0] [x1 , x2 , x3 , x4 ], or x1 = x2 x2 = 3x1  2x3 x3 = x4 x4 = 2x1 . x2 = 3 0 2 0 x2 . 0 1 x3 0 0 x3 2 0 0 0 x4 x4 EXERCISES 9.2: Review 1: Linear Algebraic Equations, page 512 3. By subtracting 2 times the first equation from the second, we eliminate x1 from the latter. Similarly, x1 is eliminated from the third equation by subtracting the first equation from it. So we get x1 + 2x2 + x3 = 3, 3x3 = x2  3x3 = 6, 6 or (interchanging last two equations) x1 + 2x2 + x3 = 3, x2  3x3 = 6, x3 = 2. x1 0 1 0 0 x1 The second unknown, x2 , can be eliminated from the first equation by subtracting 2 times the first one from it: x1 + 7x3 = 15, x2  3x3 = 6, x3 = 2. 570 Exercises 9.2
Finally, we eliminate x3 from the first two equations by adding (7) times and 3 times, respectively, the third equation. This gives x1 = 1, x2 = 0, x3 = 2. 7. Subtracting 3 times the first equation from the second equation yields x1 + 3x2 = 0, 0 = 0. The last equation is trivially satisfied, so we ignore it. Thus, just one equation remains: x1 + 3x2 = 0 x1 = 3x2 . Choosing x2 as a free variable, we get x1 = 3s, x2 = s, where s is any number. 9. We eliminate x1 from the first equation by adding (1  i) times the second equation to it: [2  (1 + i)(1  i)]x2 = 0, x1  (1 + i)x2 = 0. Since (1  i)(1 + i) = 12  i2 = 1  (1) = 2, we obtain 0 = 0, x1  (1 + i)x2 = 0 x2 =  1 1 + i x1 = x1 . 1+i 2 Assigning an arbitrary complex value to x1 , say 2s, we see that the system has infinitely many solutions given by x1 = 2s, x2 = (1 + i)s, where s is any complex number. 11. It is slightly more convenient to put the last equation at the top: x1 + x2 + 5x3 = 0, 2x1 + x3 = 1, 3x1 + x2 + 4x3 = 1. 571 Chapter 9
We then eliminate x1 from the second equation by adding 2 times the first one to it; and by subtracting 3 times the first equation from the third, we eliminate x1 in the latter. x1 + x2 + 5x3 = 0, 2x2  11x3 = 1, 2x2 + 11x3 = 1. To make the computations more convenient, we multiply the first equation by 2. 2x1 + 2x2 + 10x3 = 0, 2x2  11x3 = 1, 2x2 + 11x3 = 1. Now we add the second equation to each of the remaining, and obtain 2x1  x3 = 1, 2x2  11x3 = 1, 0=0 or 2x1  x3 = 1, 2x2  11x3 = 1. Choosing x3 as free variable, i.e., x3 = s, yields x1 = (s + 1)/2, x2 = (11s + 1)/2,  < s < . 13. The given system can be written in the equivalent form (2  r)x1  3x2 = 0, x1  (2 + r)x2 = 0. The variable x1 can be eliminated from the first equation by subtracting (2  r) times the second equation: [3 + (2  r)(2 + r)]x2 = 0, x1  (2 + r)x2 = 0 or (1  r 2 ) x2 = 0, x1  (2 + r)x2 = 0. If 1  r 2 = 0, i.e., r = 1, then the first equation implies x2 = 0. Substituting this into the second equation, we get x1 . Thus, the given system has a unique (zero) solution for any r = 1, in particular, for r = 2. 572 Exercises 9.3
If r = 1 or r = 1, then the first equation in the latter system becomes trivial 0 = 0, and the system degenerates to x1  (2 + r)x2 = 0 x1 = (2 + r)x2 . Therefore, there are infinitely many solutions to the given system of the form x1 = (2 + r)s, In particular, for r = 1 we obtain x1 = 3s, x2 = s, s (, ). x2 = s, s (, ), r = 1 . EXERCISES 9.3: Review 2: Matrices and Vectors, page 521 5. (a) AB = 1 2 2 3 1 2 2 3 1 0 1 1 1 1 2 1 = 12 02 23 03 = = 1 2 1 3 = . (b) AC = 1  4 1  2 2  6 2  3 5 1 8 1 . (c) By the Distributive Property of matrix multiplication given on page 515 of the text, we have A (B + C) = AB + AC = 1 2 1 3 + 5 1 8 1 = 6 3 9 4 . 13. Authors note: We will use Ri + cRj Rk to denote the row operation "add row i to c times row j and place the result into row k." We will use cRj Rk to denote the row operation "multiply row j by c and place the result into row k." As in Example 1 on page 517 of the text, we will perform rowreduction on the matrix [AI]. Thus, we have [AI] = 2 1 2 3 1 1 0 1 0 0 0 1 0 0 0 1 573 1 1 Chapter 9 R2 + R1 R2 2R3 + 3R1 R3 R1  R3 R1 R1 /2 R1 R3 R2 R2 R3 R2 + R3 R2 1 0 0 1 0 0 2 1 1 0 2 0 0 1 0 1 1 0 0 0 1 1 0 0 1 1 0 3 0 2 2 0 2 1 1 3 0 1 0 1 0 1 1 0 0 1 1 0 1 0 0 1 0 0 1 0 2 3 0 2 1 1 0 1 0 1 2 1 2 . 1 1 0 Therefore, the inverse matrix is A
1 1 0 1 2 1 2 . = 1 1 0 To check the algebra, it's a good idea to multiply A by A1 to verify that the product is the identity matrix. 19. Authors note: We will use Ri + cRj Rk to denote the row operation "add row i to c times row j and place the result into row k." We will use cRj Rk to denote the row operation "multiply row j by c and place the result into row k." To find the inverse matrix X1 (t), we will again use the method of Example 1 on page 517 of the text. Thus, we start with [X(t)I] = et et 2e2t et et 4e2t 574 et et e2t 1 0 0 0 1 0 0 0 1 Exercises 9.3 R2  R1 R2 R3  R1 R3 R2 /2 R2 R3 /3 R3 R1  R3 R1 R2  R3 /2 R2 0 0 et 0 0 et 0 0 et 0 0 1 0 0 et et 2et et 0 et e
t e2t e2t e2t e2t 0 0 1 1 0 1 0 1 1 1/3 4/3 0 0 1/2 0 et /2 0 1/3 1/2 0 0 1/2 1/2 1 0 0 0 3e2t et e2t /2 0 e2t 0 e
t 1/3 1 1/3 et 0 0
2t R1  R2 R1 et R1 R1 et R2 R2 e2t R3 R3 1/3 1/2 0 e 0 0 1 0 0 1 et /3 et /2 e2t /3 0 0 1/3 1/3 1/6 1/3 1/2 1/6 1/3 et /2 et /6 . 2t e /3 Thus, the inverse matrix X1 (t) is given by the matrix et et /2 et /2 X1 (t) = et /3 et /2 et /6 0 e2t /3 e2t /3 . 23. We will calculate this determinant by first finding its cofactor expansion about row 1. Therefore, we have 1 0 3 1 0 2 = (1) 1 2  0 + 0 = 2  10 = 12. 1 5 2 5 2 37. We first calculate X (t) by differentiating each entry of X(t). Therefore, we have X (t) = 2e2t 3e3t . 575 2e2t 6e3t Chapter 9
Thus, substituting the matrix X(t) into the differential equation and performing matrix multiplication yields 2e2t 3e3t 2e2t 6e3t = 1 1 2 4 e2t e3t e2t 2e3t = e2t + e2t e3t + 2e3t 2e2t  4e2t 2e3t  8e3t . Since this equation is true, we see that X(t) does satisfy the given differential equation. 39. (a) To calculate A(t) dt, we integrate each entry of A(t) to obtain A(t) dt = t dt 1 dt e, dt e, dt = t2 /2 + c1 et + c2 t + c3 et + c4 . (b) Taking the definite integral of each entry of B(t) yields
1 B(t) dt =
0 1 cos t dt 0 1 sin t dt 0  1 sin t dt 0 1 cos t dt 0 = sin t  cos t 1 0 1 0 cos t sin t 1 0 1 0 = sin 1 1  cos 1 cos 1  1 sin 1 . (c) By the product rule on page 521 of the text, we see that d [A(t)B(t)] = A(t)B (t) + A (t)B(t). dt Therefore, we first calculate A (t) and B (t) by differentiating each entry of A(t) and B(t), respectively, to obtain A (t) = 1 et 0 e
t and B (t) =  sin t  cos t cos t  sin t . Hence, by matrix multiplication we have d [A(t)B(t)] = A(t)B (t) + A (t)B(t) dt  sin t  cos t t et = t cos t  sin t 1 e = et cos t  sin t + 1 et 0 e +
t cos t  sin t sin t et sin t cos t et cos t . cos t + et sin t et cos t  sin t et cos t  t sin t t cos t  et sin t  cos t  et sin t = (1 + et ) cos t + (et  t) sin t (et  t) cos t  (et + 1) sin t et cos t + (et  1) sin t (et  1) cos t  et sin t 576 Exercises 9.4
EXERCISES 9.4: Linear Systems in Normal Form, page 530 1. To write this system in matrix form, we will define the vectors x(t) = col[x(t), y(t)] (which means that x (t) = col[x (t), y (t)]) and f(t) = col[t2 , et ], and the matrix A(t) = 3 1 1 2 . Thus, this system becomes the equation in matrix form given by x (t) y (t) = 3 1 1 2 x(t) y(t) + t2 et . We can see that this equation is equivalent to the original system by performing matrix multiplication and addition to obtain the vector equation x (t) y (t) = 3x(t)  y(t) x(t) + 2y(t) + t2 et = 3x(t)  y(t) + t2 x(t) + 2y(t) + et . Since two vectors are equal only when their corresponding components are equal, we see that this vector equation implies that x (t) = 3x(t)  y(t) + t2 , y (t) = x(t) + 2y(t) + et , which is the original system. 5. This equation can be written as a first order system in normal form by using the substitutions x1 (t) = y(t) and x2 (t) = y (t). With these substitutions this differential equation becomes the system x1 (t) = 0 x1 (t) + x2 (t), x2 (t) = 10x1 (t) + 3x2 (t) + sin t. We can then write this system as a matrix differential equation by defining the vectors x(t) = col[x1 (t), x2 (t)] (which means that x (t) = col[x1 (t), x2 (t)]), f(t) = col[0, sin t], and the matrix A= 0 1 . 577 10 3 Chapter 9
Hence, the system above in normal form becomes the differential equation given in matrix form by x1 (t) x2 (t) = 0 1 x1 (t) x2 (t) 10 3 + 0 sin t . (As in Problem 1 above, we can see that this equation in matrix form is equivalent to the system by performing matrix multiplication and addition and then noting that corresponding components of equal vectors are equal.) 7. This equation can be written as a first order system in normal form by using the substitutions x1 (t) = w(t), x2 (t) = w (t), x3 (t) = w (t), and x4 (t) = w (t). With these substitutions this differential equation becomes the system x1 (t) = 0 x1 (t) + x2 (t) + 0 x3 (t) + 0 x4 (t), x2 (t) = 0 x1 (t) + 0 x2 (t) + x3 (t) + 0 x4 (t), x3 (t) = 0 x1 (t) + 0 x2 (t) + 0 x3 (t) + x4 (t), x4 (t) = x1 (t) + 0 x2 (t) + 0 x3 (t) + 0 x4 (t) + t2 . We can then write this system as a matrix differential equation x = Ax by defining the vectors x(t) = col[x1 (t), x2 (t), x3 (t), x4 (t)] (which means that x (t) = col[x1 (t), x2 (t), x3 (t), x4 (t)]), f(t) = col[0, 0, 0, t2], and the matrix 0 0 1 0 A= 0 0 0 1 1 0 0 0 0 1 0 0 . That is, the given fourth order differential equation is equivalent to the matrix system x2 (t) 0 0 1 0 = x3 (t) 0 0 0 1 1 0 0 0 x4 (t) 578 x1 (t) 0 1 0 0 x2 (t) 0 . + x3 (t) 0 t2 x4 (t) x1 (t) 0 Exercises 9.4
17. Notice that by scalar multiplication these vector functions can e2t e2t 0 0 , e2t , e3t e2t 0 5e2t independent by showing that the only way that we can have e2t e2t 0 c1 0 + c2 e2t + c3 e3t 2t 2t e 0 5e it must be true for t = 0. Thus, c1 , c2 , and 1 0 + c2 c1 5 which is equivalent to the system c1 + c2 = 0, c2 + c3 = 0, 5c1  c2 = 0. By solving the first and last of these equations simultaneously, we see that c1 = c2 = 0. Substituting these values into the second equation above yields c3 = 0. Therefore, the original set of vectors must be linearly independent on the interval (, ). 21. Since it is given that these vectors are solutions to the system x (t) = Ax(t), in order to determine whether they are linearly independent, we need only calculate their Wronskian. If their Wronskian is never zero, then these vectors are linearly independent and so form a fundamental solution set. If the Wronskian is identically zero, then the vectors are linearly dependent, and they do not form a fundamental solution set. Thus, we observe et et W [x1 , x2 , x3 ] (t) = 2et et et e3t 2e3t 579 0 e3t be written as . Thus, as in Example 2 on page 526 of the text, we will prove that these vectors are linearly =0 for all t in (, ) is for c1 = c2 = c3 = 0. Since the equation above must be true for all t, c3 must satisfy 1 0 + c3 1 = 0, 1 1 0 Chapter 9
= et 0 e3t e
t 2e 3t  et 2et e3t e
t 2e 3t + e3t 2et e
t 0 et = et 0 + e4t  et 4e2t + e2t + e3t (2  0) = 2e3t = 0, where we have used cofactors to calculate the determinant. Therefore, this set of vectors is linearly independent and so forms a fundamental solution set for the system. Thus, a fundamental matrix is given by e e e 2et 0 e3t , X(t) = t t 3t e e 2e and a general solution of the system will be et x(t) = X(t)c = c1 2et et
t t 3t + c2 0 + c3 e3t . et 2e3t et e3t 27. In order to show that X(t) is a fundamental matrix for the system, we must first show that each of its column vectors is a solution. Thus, we substitute each of the vectors 6et 3e2t 2e3t x1 (t) = et , x2 (t) = e2t , x3 (t) = e3t t 2t 3t 5e e e into the given system to obtain 0 1 Ax1 (t) = 1 0 Ax2 (t) = 1 1 580 0 1 6 0 1 6 0 6e 6e et = et = x1 (t), 1 t t 0 5e 5e 6e2t 0 3e2t 1 e2t = 2e2t = x2 (t), e2t 2e2t 0 t t Exercises 9.4 0 6 0 Ax3 (t) = 1 0 1 e3t = 3e3t = x3 (t). 1 1 0 e3t 3e3t Therefore, each column vector of X(t) is a solution to the system on (, ). Next we must show that these vectors are linearly independent. Since they are solutions to a differential equation in matrix form, it is enough to show that their Wronskian is never zero. Thus, we find 6et 3e2t 2e3t W (t) = et 5et = 6et e2t e2t e3t e3t + 3e2t et e3t 5et e3t + 2e3t et e2t 5et e2t 2e3t 6e3t e2t e3t e2t e3t = 6et et  et + 3e2t e2t + 5e2t + 2e3t e3t + 5e3t = 20 = 0, where we have used cofactors to calculate the determinant. Hence, these three vectors are linearly independent. Therefore, X(t) is a fundamental matrix for this system. We will now find the inverse of the matrix X(t) by performing rowreduction on the matrix [X(t)I]  [IX1 (t)]. Thus, we have 1 0 0 [X(t)I] = et e2t e3t 0 1 0 t 2t 3t 0 0 1 5e e e t 2t 3t e e e 0 1 0 R2 R1 6et 3e2t 2e3t 1 0 0 R1 R2 t 2t 3t 5e e e 0 0 1 et e2t e3t 0 1 0 R2  6R1 R2 0 6 0 3e2t 8e3t 1 R3 + 5R1 R3 0 4e2t 4e3t 0 5 1 581 6et 3e2t 2e3t Chapter 9 R3 /4 R2 R2 R3 R1 + R2 R1 R3  3R2 R3 1 R3 R3 5 R2  R3 R2 et R1 R1 e2t R2 R2 e3t R3 R3 Therefore, we see that X1 (t) = e2t /5 0 2e2t /5 . e3t /5 9e3t /20 3e3t /20 4e2t /5 et /4 et /4 et e2t e3t 0 0 et 0 0 et 0 0 et 0 1 0 0 e2t e3t 0 5/4 1/4 6 0 3e2t 8e3t 1 0 0 0 1/4 1/4 e2t e3t 0 5/4 1/4 3t 0 5e 1 9/4 3/4 0 1/4 1/4 0 0 2t 3t 0 5/4 1/4 e e 3t 0 e 1/5 9/20 3/20 0 0 0 0 1/4 1/4 1/5 0 1 0 0 e2t 0 e3t 0 e2t /5 4/5 2/5 1/5 9/20 3/20 et /4 et /4 4e2t /5 0 1 0 0 0 1 2e2t /5 . 3t 3t 3t e /5 9e /20 3e /20 We now can use Problem 26 to find the solution to this differential equation for any initial value. For the initial value given here we note that t0 = 0. Thus, substituting t0 = 0 into the matrix X1 (t) above yields X1 (0) = 1/5 1/5 0 1/4 4/5 2/5 . 9/20 3/20 1/4 Hence, we see that the solution to this problem is given by x(t) = X(t)X1 (0)x(0) 582 Exercises 9.4 = et e2t e3t 5et e2t e3t 6et 3e2t 2e3t = et e2t e3t 5et e2t e3t 6et 3e2t 2e3t 1/5 1/5 1/4 1/5 1/20 0 1/4 4/5 2/5 0 1 9/20 3/20 (3/2)et + (3/5)e2t  (1/10)e3t = (1/4)et  (1/5)e2t  (1/20)e3t (5/4)et  (1/5)e2t  (1/20)e3t 1/4 1 . There are two short cuts that can be taken to solve the given problem. First, since we only need X1 (0), it suffices to compute the inverse of X(0), not X(t). Second, by producing X1 (t) we automatically know that det X(0) = 0 and hence X(t) is a fundamental matrix. Thus, it was not really necessary to compute the Wronskian. 33. Let (t) be an arbitrary solution to the system x (t) = A(t)x(t) on the interval I. We want to find c = col(c1 , c2 , . . . , cn ) so that (t) = c1 x1 (t) + c2 x2 (t) + + cn xn (t), where x1 , x2 , . . . , xn are n linearly independent solutions for this system. Since c1 x1 (t) + c2 x2 (t) + + cn xn (t) = X(t)c , where X(t) is the fundamental matrix whose columns are the vectors x1 , x2 , . . . , xn , this equation can be written as (t) = X(t)c (9.1) Since x1 , x2 , . . . , xn are linearly independent solutions of the system x (t) = A(t)x(t), their Wronskian is never zero. Therefore, as was discussed on page 528 of the text, X(t) has an inverse at each point in I. Thus, at t0 , a point in I, X1 (t0 ) exists and equation (9.1) becomes (t0 ) = X (t0 ) c X1 (t0 ) (t0 ) = X1 (t0 ) X (t0 ) c = c . Hence, if we define c0 to be the vector c0 = X1 (t0 ) (t0 ), then equation (9.1) is true at the point t0 (i. e. (t0 ) = X(t0 )X1 (t0 )(t0 )). To see that, for this definition of c0 , equation (9.1) is true for all t in I (and so this is the vector that we seek), notice that (t) and X(t)c0 583 Chapter 9
are both solutions to same initial value problem (with the initial value given at the point t0 ). Therefore, by the uniqueness of solutions, Theorem 2 on page 525 of the text, these solutions must be equal on I, which means that (t) = X(t)c0 for all t in I. EXERCISES 9.5: Homogeneous Linear Systems with Constant Coefficients, page 541 5. The characteristic equation for this matrix is given by 1r A  rI = 0 0 0 r 2 0 2 r = (1  r) r 2 2 r = (1  r) r 2  4 = (1  r)(r  2)(r + 2) = 0. Thus, the eigenvalues of this matrix are r = 1, 2, 2. Substituting the eigenvalue r = 1, into equation (A  rI)u = 0 yields 0 0 which is equivalent to the system u2 + 2u3 = 0, 2u2  u3 = 0. This system reduces to the system u2 = 0, u3 = 0, which does not assign any value to u1 . Thus, we can let u1 be any value, say u1 = s, and u2 = 0, u3 = 0 and the system given by (9.2) will be satisfied. From this we see that the eigenvectors associated with the eigenvalue r = 1 are given by u1 = col (u1 , u2 , u3) = col(s, 0, 0) = scol(1, 0, 0). For r = 2 we observe that the equation (A  rI)u = 0 becomes 0 1 0 0 u1 (A  2I)u = 0 2 2 u2 = 0 , u3 0 0 2 2 584 0 0 u1 0 (9.2) (A  I)u = 0 1 2 u2 = 0 , 0 u3 2 1 Exercises 9.5
whose corresponding system of equations reduces to u1 = 0, u2 = u3 . Therefore, we can pick u2 to be any value, say u2 = s (which means that u3 = s), and we find that the eigenvectors for this matrix associated with the eigenvalue r = 2 are given by u2 = col (u1 , u2 , u3 ) = col(0, s, s) = scol(0, 1, 1). For r = 2, we solve the equation 3 0 0 u1 0 (A + 2I)u = 0 2 2 u2 = 0 , u3 0 2 2 0 which reduces to the system u1 = 0, u2 = u3 . Hence, u3 is arbitrary, and so we will let u3 = s (which means that u2 = s). Thus, solutions to this system and, therefore, eigenvectors for this matrix associated with the eigenvalue r = 2 are given by the vectors u3 = col (u1 , u2 , u3) = col(0, s, s) = scol(0, 1, 1). 13. We must first find the eigenvalues and eigenvectors associated with the given matrix A. Thus, we note that the characteristic equation for this matrix is given by 1r A  rI = 2 2 (1  r) r 3 2 r 3 3 r 2 3 r 2 2 3 2 r +2 2 r 2 3 =0 =0 (1  r) r 2  9  2(2r  6) + 2(6 + 2r) = (1  r)(r  2)(r + 2) = 0 (r + 3)[(1  r)(r  3) + 8] = 0 (r + 3)(r  5)(r + 1) = 0. Therefore, the eigenvalues are r = 3, 1, 5. To find an eigenvector associated with the eigenvalue r = 3, we must find a vector u = col(u1 , u2 , u3 ) which satisfies the equation (A + 3I)u = 0. Thus, we have 0 4 2 2 u1 (A + 3I)u = 2 3 3 u2 = 0 u3 0 2 3 3 2 0 0 u1 0 0 1 1 u2 = 0 , u3 0 0 0 0 585 Chapter 9
where we have obtained the last equation above by using elementary row operations. This equation is equivalent to the system u1 = 0, u2 = u3 . Hence, if we let u3 have the arbitrary value s1 , then we see that, for the matrix A, the eigenvectors associated with the eigenvalue r = 3 are given by u = col (u1 , u2, u3 ) = col (0, s1 , s1 ) = s1 col(0, 1, 1). Thus, if we choose s1 = 1, then vector u1 = col(0, 1, 1) is one eigenvector associated with this eigenvalue. For the eigenvalue r = 1, we must find a vector u which satisfies the equation (A + I)u = 0. Thus, we see that 2 2 2 u1 0 1 2 0 u1 0 (A + I)u = 2 1 3 u2 = 0 u3 2 3 1 0 0 1 1 u2 = 0 , u3 0 0 0 0 which is equivalent to the system u1 = 2u2 , u3 = u2 . Therefore, if we let u2 = s2 , then we see that vectors which satisfy the equation (A + I)u = 0 and, hence, eigenvectors for the matrix A associated with the eigenvalue r = 1 are given by u = col (u1 , u2 , u3) = col (2s2 , s2 , s2 ) = s2 col(2, 1, 1). By letting s2 = 1, we find that one such vector will be the vector u2 = col(2, 1, 1). In order to find an eigenvector associated with the eigenvalue r = 5, we will solve the equation (A  5I)u = 0. Thus, we have (A  5I)u = 4 2 2 2 u1 0 3 u2 = 0 3 5 u3 0 0 1 1 u2 = 0 , u3 0 0 0 0 1 0 1 u1 0 2 5 which is equivalent to the system u1 = u3 , u2 = u3 . Thus, if we let u3 = s3 , then, for the matrix A, the eigenvectors associated with the eigenvalue r = 5 are given by u = col (u1 , u2 , u3) = col (s3 , s3 , s3 ) = s3 col(1, 1, 1). 586 Exercises 9.5
Hence, by letting s3 = 1, we see that one such vector will be the vector u3 = col(1, 1, 1). Therefore, by Corollary 1 on page 538 of the text, we see that a fundamental solution set for this equation is given by e3t u1 , et u2 , e5t u3 . Thus, a general solution for this system is 0 1 x(t) = c1 e3t u1 + c2 et u2 + c3 e5t u3 = c1 e3t 1 + c2 et 1 2 1 + c3 e5t 1 . 1 1 21. A fundamental matrix for this system has three columns which are linearly independent solutions. Therefore, we will first find three such solutions. To this end, we will first find the eigenvalues for the matrix A by solving the characteristic equation given by r A  rI = 0 8 r r 1 r 1 0 1 =0 14 7  r  0 1 8 7r =0 (r  1)(r  2)(r  4) = 0. 14 7  r r 3  7r 2 + 14r  8 = 0 Hence, this matrix has three distinct eigenvalues, r = 1, 2, 4, and, according to Theorem 6 on page 538 of the text, the eigenvectors associated with these eigenvalues will be linearly independent. Thus, these eigenvectors will be used in finding the three linearly independent solutions which we seek. To find an eigenvector, u = col(u1 , u2 , u3), associated with the eigenvalue r = 1, we will 1 1 0 1 (A  I)u = 8 14 solve the 0 u1 u2 1 6 u3 equation 0 = 0 0 (A  I)u = 0. Therefore, 1 0 0 1 0 0 we have 1 u1 u2 1 0 u3 0 = 0 , 0 which is equivalent to the system u1 = u3 , u2 = u3 . Thus, by letting u3 = 1 (which implies that u1 = u2 = 1), we find that one eigenvector associated with the eigenvalue r = 1 is given 587 Chapter 9
by the vector u1 = col(1, 1, 1). To find an eigenvector associated with the eigenvalue r = 2, we solve the equation 2 1 0 u1 0 0 2 1 u2 = 0 (A  2I)u = u3 8 14 5 0 4 0 1 u1 0 0 2 1 u2 = 0 , u3 0 0 0 0 which is equivalent to the system 4u1 = u3 , 2u2 = u3 . Hence, letting u3 = 4 implies that u1 = 1 and u2 = 2. Therefore, one eigenvector associated with the eigenvalue r = 2 is the vector u2 = col(1, 2, 4). In order to find an eigenvector associated with the eigenvalue r = 4, we will solve the equation 4 1 0 (A4I)u = 0 4 1 8 14 3 u1 0 u1 0 u2 = 0 0 u3 0 4 1 u2 = 0 , 0 0 0 0 u3 16 0 1 which is equivalent to the system 16u1 = u3 , 4u2 = u3. Therefore, letting u3 = 16 implies that u1 = 1 and u2 = 4. Thus, one eigenvector associated with the eigenvalue r = 4 is the vector u3 = col(1, 4, 16). Therefore, by Theorem 5 on page 536 of the text (or Corollary 1), we see that three linearly independent solutions of this system are given by et u1 , e2t u2 , and e4t u3 . Thus, a fundamental matrix for this system will be the matrix e4t et e2t et 2e2t 4e4t . t 2t 4t e 4e 16e 33. Since the coefficient matrix for this system is a 3 3 real symmetric matrix, by the discussion on page 540 of the text, we know that we can find three linearly independent eigenvectors for this matrix. Therefore, to find the solution to this initial value problem, we must first find three such eigenvectors. To do this we first find eigenvalues for this matrix. Therefore, we solve the characteristic equation given by 1r A  rI = 2 2 588 2 1r 2 2 2 1r =0 Exercises 9.5 (1  r) 1r 2 2 1r +2 2 2 2 1r +2 2 1  r 2 2 =0 (1  r) (1  r)2  4 + 2 [2(1  r) + 4] + 2 [4  2(1  r)] = 0 (1  r)(r  3)(r + 1) + 8(r + 1) = (r + 1)(r  5)(r + 1) = 0. Thus, the eigenvalues are r = 1 and r = 5, with r = 1 an eigenvalue of multiplicity two. In order to find an 4 2 (A  5I)u = 2 eigenvector associated with the eigenvalue r = 5, we solve the equation 0 0 2 2 1 0 1 u1 u1 u2 = 0 u2 = 0 . 0 1 4 2 1 u3 u3 0 0 2 4 0 0 0 This equation is equivalent to the system u1 = u3 , u2 = u3 . Thus, if we let u3 = 1, we see that for this coefficient matrix an eigenvector associated with the eigenvalue r = 5 is given by the vector u1 = col(u1 , u2, u3 ) = col(1, 1, 1). We must now find two more linearly independent eigenvectors for this coefficient matrix. By the discussion above, these eigenvectors will be associated with the eigenvalue r = 1. Thus, we solve the equation 1 1 1 u1 2 2 2 0 (A + I)u = 2 0 0 2 2 u2 = 0 0 0 0 0 2 2 2 0 u3 u1 0 u2 = 0 , (9.3) 0 u3 which is equivalent to the equation u1  u2 + u3 = 0. Therefore, if we arbitrarily assign the value s to u2 and v to u3 , we see that u1 = s  v, and solutions to equation (9.3) above will be given by u= sv s v 1 1 = s 1 + v 0 . 1 0 By taking s = 1 and v = 0, we see that one solution to equation (9.3) will be the vector u2 = col(1, 1, 0). Hence, this is one eigenvector for the coefficient matrix. Similarly, by letting s = 0 and v = 1, we find a second eigenvector will be the vector u3 = col(1, 0, 1). Since the eigenvectors u1 , u2 , and u3 are linearly independent, by Theorem 5 on page 536 of the text, 589 Chapter 9
we see that a general solution for this system will be 1 1 5t t x(t) = c1 e 1 + c2 e 1 1 0 given by + c3 et 1 0 . 1 To find a solution which satisfies the initial condition, 1 1 1 x(0) = c1 1 + c2 1 + c3 0 1 0 1 we must solve the equation c1 1 1 1 2 1 1 0 c2 = 3 . 1 0 1 2 c3 This equation can be solved by either using elementary row operations on the augmented matrix associated with this equation or by solving the system c1 + c2  c3 = 2, c1 + c2 = 3, c1 + c3 = 2. By either method we find that c1 = 1, c2 = 2, and c3 = 1. Therefore, the solution to this initial value problem is given by 1 1 1 1  2et 1 + et 0 x(t) = e 1 0 1 e5t  2et  et 3et + e5t = e5t  2et + 0 = 2et  e5t . e5t + 0 + et et + e5t
5t 37. (a) In order to find the eigenvalues for the matrix A, we will solve the characteristic equation 2r A  rI = 0 0 590 1 2r 0 6 5 2r =0 (2  r)3 = 0. Exercises 9.5
Thus, r = 2 is an eigenvalue of multiplicity three. To find the eigenvectors for the matrix A associated with this eigenvalue, we solve 0 1 (A  2I)u = 0 0 0 0 the equation 6 u1 0 5 u2 = 0 . 0 u3 0 This equation is equivalent to the system u2 = 0, u3 = 0. Therefore, we can assign u1 to be any arbitrary value, say u1 = s, and we find that the vector u = col(u1 , u2, u3 ) = col(s, 0, 0) = scol(1, 0, 0) will solve this equation and will, thus, be an eigenvector for the matrix A. We also notice that the vectors u = scol(1, 0, 0) are the only vectors that will solve this equation, and, hence, they will be the only eigenvectors for the matrix A. (b) By taking s = 1, we find that, for the matrix A, one eigenvector associated with the eigenvalue r = 2 will be the vector u1 = col(1, 0, 0). Therefore, by the way eigenvalues and eigenvectors were defined (as was discussed in the text on page 533), we see that one solution to the system x = Ax will be given by the vector 1 x1 (t) = e2t u1 = e2t 0 . 0 (c) We know that u1 = col(1, 0, 0) is an eigenvector for the matrix A associated with the eigenvalue r = 2. Thus, u1 satisfies the equation (A  2I)u1 = 0 Au1 = 2u1 . (9.4) We want to find a constant vector u2 = col(v1 , v2 , v3 ) such that x2 (t) = te2t u1 + e2t u2 will be a second solution to the system x = Ax. To do this, we will first show that x2 will satisfy the equation x = Ax if and only if the vector u2 satisfies the equation 591 Chapter 9
(A  2I)u2 = u1 . To this end, we find that x2 (t) = 2te2t u1 + e2t u1 + 2e2t u2 = 2te2t u1 + e2t (u1 + 2u2 ) , where we have used the fact that u1 and u2 are constant vectors. We also have Ax2 (t) = A (te2t u1 + e2t u2 ) = A (te2t u1 ) + A (e2t u2 ) , = te2t (Au1 ) + e2t (Au2 ) , = 2te2t u1 + e2t Au2 , distributive property of matrix multiplication (page 515 of the text) associative property of matrix multiplication (page 515 of the text) by equation (9.4) above. Thus, if x2 (t) is to be a solution to the given system we, must have x2 (t) = Ax2 (t) 2te2t u1 + e2t (u1 + 2u2 ) = 2te2t u1 + e2t Au2 e2t (u1 + 2u2 ) = e2t Au2 . By dividing both sides of this equation by the nonzero term e2t , we obtain u1 + 2u2 = Au2 (A  2I)u2 = u1 . Since all of these steps are reversible, if a vector u2 satisfies this last equation, then x2 (t) = te2t u1 + e2t u2 will be a solution to the system x = Ax. Now we can use the formula (A  2I)u2 = u1 to find the vector u2 = col(v1 , v2 , v3 ). Hence, we solve the equation 0 1 6 1 v1 0 0 5 v2 = 0 . (A  2I)u2 = v3 0 0 0 0 This equation is equivalent to the system v2 +6v3 = 1, 5v3 = 0, which implies that v2 = 1, v3 = 0. Therefore, the vector u2 = col(0, 1, 0) will satisfy the equation (A  2I)u2 = u1 and, thus, 1 0 x2 (t) = te2t 0 + e2t 1 0 0 592 Exercises 9.5
will be a second solution to the given system. We can see by inspection x2 (t) and x1 (t) = e2t u1 are linearly independent. (d) To find a third linearly independent solution to this system we will try to find a solution t2 of the form x3 (t) = e2t u1 + te2t u2 + e2t u3 , where u3 is a constant vector that we must 2 find, and u1 and u2 are the vectors that we found in parts (b) and (c), respectively. To find the vector u3 , we will proceed as we did in part (c) above. We will first show that x3 (t) will be a solution to the given system if and only if the vector u3 satisfies the equation (A  2I)u3 = u2 . To do this we observe that x3 (t) = te2t u1 + t2 e2t u1 + e2t u2 + 2te2t u2 + 2e2t u3 . Also, using the facts that (A  2I)u1 = 0 and (A  2I)u2 = u1 we have Ax3 (t) = A t2 2t e u1 + te2t u2 + e2t u3 2 t2 2t e u1 2 + A (te2t u2 ) + +A (e2t u3 ) , distributive property Au2 = u1 + 2u2 , (9.6) Au1 = 2u1 (9.5) = A = t2 2t e (Au1 ) + te2t (Au2 ) + e2t (Au3 ) , 2 t2 2t e (2u1 ) + te2t (u1 + 2u2 ) + e2t Au3 , 2 associative property = equations (9.5) and (9.6) = t2 e2t u1 + te2t u1 + 2te2t u2 + e2t Au3 . Therefore, for x3 (t) to satisfy the given system, we must have x3 (t) = Ax3 (t) 593 Chapter 9 te2t u1 + t2 e2t u1 + e2t u2 + 2te2t u2 + 2e2t u3 = t2 e2t u1 + te2t u1 + 2te2t u2 + e2t Au3 e2t u2 + 2e2t u3 = e2t Au3 u2 + 2u3 = Au3 (A  2I)u3 = u2 . Again since these steps are reversible, we see that, if a vector u3 satisfies the equation (A  2I)u3 = u2 , then the vector t2 2t x3 (t) = e u1 + te2t u2 + e2t u3 2 will be a third linearly independent solution to the given system. Thus, we can use this equation to find the vector u3 = col(v1 , v2 , v3 ). Hence, we solve 0 1 6 0 v1 0 0 5 v2 = 1 . (A  2I)u3 = v3 0 0 0 0 This equation is equivalent to the system v2 + 6v3 = 0, 5v3 = 1, which implies that v3 = 1/5, v2 = 6/5. Therefore, if we let u3 = col(0, 6/5, 1/5), then 1 0 0 t2 2t 2t 2t x3 (t) = e 0 + te 1 + e 6/5 2 0 0 1/5 will be a third solution to the given system and we see by inspection that this solution is linearly independent from the solutions x1 (t) and x2 (t). (e) Notice that (A  2I)3 u3 = (A  2I)2 [(A  2I)u3 ] = (A  2I)2 u2 = (A  2I) [(A  2I)u2 ] = (A  2I)u1 = 0. 43. According to Problem 42, we will look for solutions of the form x(t) = tr u, where r is an eigenvalue for the coefficient matrix and u is an associated eigenvector. To find the eigenvalues 594 Exercises 9.5
for this matrix, we solve the equation A  rI = 1r 1 3 5r =0 (1  r)(5  r) + 3 = 0 r 2  6r + 8 = 0 (r  2)(r  4) = 0. Therefore, the coefficient matrix has the eigenvalues r = 2, 4. Since these are distinct eigenvalues, Theorem 6 on page 538 of the text assures us that their associated eigenvectors will be linearly independent. To find an eigenvector u = col(u1, u2 ) associated with the eigenvalue r = 2, we solve the system (A  2I)u = 1 3 1 3 u1 u2 = 0 0 , which is equivalent to the equation u1 + 3u2 = 0. Thus, if we let u2 = 1 then, in order to satisfy this equation, we must have u1 = 3. Hence, we see that the vector u1 = col(3, 1) will be an eigenvector for the coefficient matrix of the given system associated with the eigenvalue r = 2. Therefore, according to Problem 42, one solution to this system will be given by x1 (t) = t2 u1 = t2 3 1 . To find an eigenvector associated with the eigenvalue r = 4, we solve the equation (A  4I)u = 3 3 1 1 u1 u2 = 0 0 , which is equivalent to the equation u1 = u2 . Thus, if we let u2 = 1, then we must have u1 = 1 and so an eigenvector associated with the eigenvalue r = 4 will be given by the vector u2 = col(1, 1). Therefore, another solution to the given system will be x2 (t) = t4 u2 = t4 1 1 . 595 Chapter 9
Clearly the solutions x1 (t) and x2 (t) are linearly independent. So the general solution to the given system with t > 0 will be x(t) = c1 t2 3 1 + c2 t4 1 1 = c1 3t2 t2 + c2 t4 t4 . EXERCISES 9.6: Complex Eigenvalues, page 549 3. To find the eigenvalues for the matrix A, we solve the characteristic equation given by 1r A  rI = 0 0 (1  r) 1 2 1r 1 1r 1 1r 1 1 1r 0+0=0 =0 (1  r) (1  r)2 + 1 = (1  r) r 2  2r + 2 = 0. By this equation and the quadratic formula, we see that the roots to the characteristic equation and, therefore, the eigenvalues for the matrix A are r = 1, and r = 1i. To find an eigenvector u = col(u1, u2 , u3 ) associated with the real eigenvalue r = 1, we solve the system 0 2 1 0 u1 (A  I)u = 0 0 1 u2 = 0 , u3 0 1 0 0 which implies that u2 = 0, u3 = 0. Therefore, we can set u1 arbitrarily to any value, say u1 = s. Then the vectors u = col(u1 , u2, u3 ) = col(s, 0, 0) = scol(1, 0, 0) will satisfy the above equation and, therefore, be eigenvectors for the matrix A. Hence, if we set s = 1, we see that one eigenvector associated with the eigenvalue r = 1 will be the vector u1 = col(1, 0, 0). Therefore, one solution to the given system will be 1 t t x1 (t) = e u1 = e 0 . 0 596 Exercises 9.6
In order to find an eigenvector z = col(z1 , z2 , z3 ) associated with the complex eigenvalue r = 1 + i, we solve the equation z1 0 [A  (1 + i)I]z = 0 This equation is equivalent to the system i 0 2 i 1 1 1 z2 = 0 . i 0 z3 iz1 + 2z2  z3 = 0 and  iz2 + z3 = 0. Thus, if we let z2 = s, then we see that z3 = is and iz1 = 2z2 + z3 = 2s + is z1 = 2is  s = s  2is , (i)(iz1 ) = (i)(2s + is) where we have used the fact that i2 = 1. Hence, eigenvectors associated with the eigenvalue r = 1+i will be z = scol(12i, 1, i). By taking s = 1, we see that one eigenvector associated with this eigenvalue will be the vector 1  2i z1 = 1 i = 1 + i 0 . 1 0 1 2 Thus, by the notation on page 545 of the text, we have = 1, = 1, a = col(1, 1, 0), and b = col(2, 0, 1). Therefore, according to formulas (6) and (7) on page 546 of the text, two more linearly independent solutions to the given system will be given by x2 (t) = (et cos t)a  (et sin t)b and x3 (t) = (et sin t)a + (et cos t)b. Hence, the general solution to the system given in this problem will be x(t) = c1 x2 (t) + c2 x3 (t) + c3 x1 (t) 1 2 t 1  c1 et sin t 0 = c1 e cos t 0 1 1 2 + c2 et sin t 1 + c2 et cos t 0 0 1 1 + c3 et 0 . 0 597 Chapter 9
7. In order to find a fundamental matrix for this system, we must first find three linearly independent solutions. Thus, we seek the eigenvalues for the matrix A by solving the characteristic equation given by r A  rI = 0 0 r 0 1 r 1 =0 r 0+0 =0 r r 2 + 1 = 0. r 1 r 1 1 Hence, the eigenvalues for the matrix A will be r = 0 and r = i. To find an eigenvector u = col(u1, u2 , u3 ) associated with the real eigenvalue 0 0 1 (A  0I)u = 0 0 1 0 1 0 r = 0, we solve the equation u1 0 u2 = 0 , 0 u3 which is equivalent to the system u3 = 0, u2 = 0. Thus, if we let u1 have the arbitrary value u1 = s, then the vectors u = col(u1 , u2, u3 ) = col(s, 0, 0) = scol(1, 0, 0) will satisfy this equation and will, therefore, be eigenvectors for the matrix A associated with the eigenvalue r = 0. Hence, by letting s = 1, we find that one of these eigenvectors will be the vector u = col(1, 0, 0). Thus, one solution to the given system will be 1 0 x1 (t) = e u = 0 . 0 To find two more linearly independent solutions for this system, we will first look for an eigenvector associated with the complex eigenvalue r = i. That is, we seek a vector, say, z = col(z1 , z2 , z3 ) which satisfies the equation 0 i 0 1 z1 (A  iI)z = 0 i 1 z2 = 0 z3 0 0 1 i 598 , Exercises 9.6
which is equivalent to the system iz1 = z3 and iz2 = z3 . Thus, if we let z3 be any arbitrary value, say z3 = is, (which means that we must have z1 = s and z2 = s), then we see that the vectors, given by z = col(z1 , z2 , z3 ) = col(s, s, is) = scol(1, 1, i), will be eigenvectors for the matrix A associated with the eigenvalue r = i. Therefore, by letting s = 1, we find that one of these eigenvectors will be the vector 1 1 0 z = 1 = 1 + i 0 . i 0 1 From this, by the notation given on page 546 of the text, we see that = 0, = 1, a = col(1, 1, 0), and b = col(0, 0, 1). Therefore, by formulas (6) and (7) on page 546 of the text, two more linearly independent solutions for this system will be cos t 0 cos t  cos t  0 =  cos t x2 (t) = (cos t)a  (sin t)b = sin t sin t 0 and sin t 0 sin t x3 (t) = (sin t)a + (cos t)b =  sin t + 0 =  sin t . cos t cos t 0 Finally, since a fundamental matrix for the system given in this problem must have three columns which are linearly independent solutions of the system, we see that such a fundamental matrix will be given by the matrix 1 cos t sin t X(t) = 0  cos t  sin t 0  sin t cos t . 599 Chapter 9
17. We will assume that t > 0. According to Problem 42 in Exercises 9.5, a solution to this CauchyEuler system will have the form x(t) = tr u, where r is an eigenvalue for the coefficient matrix of the system and u is an eigenvector associated with this eigenvalue. Therefore, we first must find the eigenvalues for this matrix by solving the characteristic equation given by 1  r A  rI = 2 0 (1  r) 1 1  r 1 1  r 1 1 1  r 0 1 1  r + 2 1 0 1  r =0 =0 (1  r) (1  r)2  1 + 2(1  r) = (1 + r) r 2 + 2r + 2 = 0. From this equation and by using the quadratic formula, we see that the eigenvalues for this coefficient matrix will be r = 1, 1 i. The eigenvectors associated with the real eigenvalue r = 1 will be the vectors u = col(u1 , u2 , u3 ) which satisfy the equation u1 0 0 1 0 (A + I)u = 2 0 1 u2 = 0 , 0 0 1 0 u3 which is equivalent to the system u2 = 0, 2u1 + u3 = 0. Thus, by letting u1 = 1 (which means that u3 = 2), we see that the vector u = col(u1 , u2 , u3 ) = col(1, 0, 2) satisfies this equation and is, therefore, an eigenvector of the coefficient matrix associated with the eigenvalue r = 1. Hence, according to Problem 42 of Exercises 9.5, we see that a solution to this CauchyEuler system will be given by 1 x1 (t) = t1 u = t1 0 = 0 . 2 2t1 600 t1 Exercises 9.6
To find the eigenvectors z = col(z1 , z2 , z3 ) associated with the complex eigenvalue r = 1 + i, we solve the equation (A  (1 + i)I)z = 2 i 1 i 0 z1 0 1 z2 = 0 0 1 i z3 z1 0 = 0 , z2 z3 0 i 1 0 0 i 1 0 0 0 0 which is equivalent to the system iz1  z2 = 0, iz2 + z3 = 0. Thus, if we let z1 = 1, we must let z2 = i and z3 = 1 in order to satisfy this system. Therefore, one eigenvector for the coefficient matrix associated with the eigenvalue r = 1+i will be the vector z = col(1, i, 1) and another solution to this system will be x(t) = t1+i z. We would like to find real solutions to this problem. Therefore, we note that by Euler's formula we have t1+i = t1 ti = t1 ei ln t = t1 [cos(ln t) + i sin(ln t)], where we have made use of our assumption that t > 0. Hence, the solution that we have just found becomes x(t) = t1+i z = t1 [cos(ln t) + i sin(ln t)]z 1 t1 cos(ln t) = t1 [cos(ln t) + i sin(ln t)] i = t1 sin(ln t) t1 cos(ln t) 1 + i t1 cos(ln t) . t1 sin(ln t) t1 sin(ln t) Thus, by Lemma 2 (adapted to systems) on page 172 of the text we see that two more linearly independent solutions to this CauchyEuler system will be x2 (t) = t1 sin(ln t) t1 cos(ln t) t1 cos(ln t) and x3 (t) = t1 cos(ln t) , t1 sin(ln t) 601 t1 sin(ln t) Chapter 9
and, hence, a general solution will be given by t1 t1 cos(ln t) x(t) = c1 0 + c2 t1 sin(ln t) 1 2t t1 cos(ln t) + c3 t1 cos(ln t) . 1 t sin(ln t) t1 sin(ln t) EXERCISES 9.7: Nonhomogeneous Linear Systems, page 555 3. We must first find the general solution to the corresponding homogeneous system. Therefore, we first find the eigenvalues for the coefficient matrix A by solving the characteristic equation given by 1r A  rI = 2 2 (1  r) 1r 2 2 1r 2 2 1r 2 2 1r +2 2 2 2 1r +2 2 1  r 2 2 =0 =0 (1  r) (1  r)2  4 + 2 [2(1  r)  4] + 2 [4  2(1  r)] = 0 (1  r)(r 2  2r  3) + 4(2r  6) = 0 (1  r)(r + 1)(r  3) + 8(r  3) = (r  3)(r 2  9) = (r  3)(r  3)(r + 3) = 0. Thus, the eigenvalues for the matrix A are r = 3, 3, where r = 3 is an eigenvalue of multiplicity two. Notice that, even though the matrix A has only two distinct eigenvalues, we are still guaranteed three linearly independent eigenvectors because A is a 33 real symmetric matrix. To find an eigenvector associated with the eigenvalue r = 3, we must find a vector u = col(u1, u2 , u3 ) which satisfies the system u1 0 4 2 2 2 4 2 u2 = 0 (A + 3I)u = 0 2 2 4 u3 u1 1 0 1 0 0 1 1 u2 = 0 , 0 0 0 0 u3 which is equivalent to the system u1 + u3 = 0, u2 + u3 = 0. Hence, by letting u3 = 1, we must have u1 = u2 = 1, and so the vector u1 = col(1, 1, 1) will then satisfy the above system. 602 Exercises 9.7
Therefore, this vector is an eigenvector for the matrix A associated with the eigenvalue r = 3. Thus, one solution to the corresponding homogeneous system is given by 1 3t 3t x1 (t) = e u1 = e 1 . 1 To find eigenvectors u = col(u1 , u2 , u3) associated with the eigenvalue r = 3, we solve the equation given by 2 2 2 2 2 u1 0 (A  3I)u = 2 2 2 u2 = 0 , 0 2 u3 which is equivalent to the equation u1 + u2  u3 = 0. Thus, if we let u3 = s and u2 = v, then we must have u1 = s  v. Hence, solutions to the above equation and, therefore, eigenvectors for A associated with the eigenvalue r = 3 will be 1 sv u = v = s 0 1 s the vectors 1 + v 1 , 0 where s and v are arbitrary scalars. Therefore, letting s = 1 and v = 0 yields the eigenvector u2 = col(1, 0, 1). Similarly, by letting s = 0 and v = 1, we obtain the eigenvector u3 = col(1, 1, 0), which we can see by inspection is linearly independent from u2 . Hence, two more solutions to the corresponding homogeneous system which are linearly independent from each other and from x1 (t) are given by 3t 3t and 1 0 x2 (t) = e u2 = e 1 1 1 . x3 (t) = e u3 = e 0
3t 3t Thus, the general solution to the corresponding homogeneous system will be 1 1 1 xh (t) = c1 e3t 1 + c2 e3t 0 + c3 e3t 1 . 1 1 0 603 Chapter 9
To find a particular solution to the nonhomogeneous system, we note that 2 2et f(t) = 4et = et 4 = et g , 2 2et where g = col(2, 4, 2). Therefore, we will assume that a particular solution to the nonhomogeneous system will have the form xp (t) = et a, where a = col(a1 , a2 , a3 ) is a constant vector which must be determined. Hence, we see that xp (t) = et a. By substituting xp (t) into the given system, we obtain et a = Axp (t) + f(t) = Aet a + et g = et Aa + et g . Therefore, we have et a = et Aa + et g a = Aa + g a1 2 0 2 2 2 0 2 a2 = 4 2 2 2 0 a3 augmented matrix or by solving the system 2a2  2a3 = 2a1  2a3 = 2, 4, . (I  A)a = g The last equation above can be solved by either performing elementary row operations on the 2a1  2a2 = 2. Either way, we obtain a1 = 1, a2 = 0, and a3 = 1. Thus, a particular solution to the nonhomogeneous system will be given by 1 xp (t) = et a = et 0 , 1 and so the general solution to the nonhomogeneous system will be 1 1 1 1 3t + c2 e3t 0 + c3 e3t 1 + et 0 . x(t) = xh (t) + xp (t) = c1 e 1 1 1 0 1 604 Exercises 9.7
13. We must first find a fundamental matrix for the corresponding homogeneous system x = Ax. To this end, we first find the eigenvalues of the matrix A by solving the characteristic equation given by A  rI = 2r 3 1 2  r =0 (2  r)(2  r) + 3 = 0 r2  1 = 0 . Thus, the eigenvalues of the coefficient matrix A are r = 1. The eigenvectors associated with the eigenvalue r = 1 are the vectors u = col(u1 , u2) which satisfy the equation (A  I)u = 1 1 u1 u2 = 0 0 . 3 3 This equation is equivalent to the equation u1 + u2 = 0. Therefore, if we let u1 = 1, then we have u2 = 1, so one eigenvector of the matrix A associated with the eigenvalue r = 1 is the vector u1 = col(1, 1). Hence, one solution of the corresponding homogeneous system is given by x1 (t) = et u1 = et 1 1 = et et . To find an eigenvector associated with the eigenvalue r = 1, we solve the equation (A + I)u = 3 1 u1 u2 = 0 0 , 3 1 which is equivalent to the equation 3u1 + u2 = 0. Since u1 = 1 and u2 = 3 satisfy this equation, one eigenvector for the matrix A associated with the eigenvalue r = 1 is the vector u2 = col(1, 3). Thus, another linearly independent solution of the corresponding homogeneous system is x2 (t) = et u2 = et 1 3 = et 3et . Hence, the general solution of the homogeneous system is given by xh (t) = c1 et et + c2 et 3et , 605 Chapter 9
and a fundamental matrix is X(t) = et et et 3et . To find the inverse matrix X1 (t), we will perform rowreduction on the matrix [X(t)I]. Thus, we have [X(t)I] =  et e et
t et 3e
t 1 0 0 1 1/2   et et
t 1 0 1 1 (3/2)et (1/2)et . 0 2e 1 0 0 1 (1/2)et 0 3/2 0 et 1/2 1/2 (1/2)et (1/2)et Therefore, we see that X1 (t) = Hence, we have X1 (t)f(t) = and so we have X1 (t)f(t) dt = (5)dt 3 e dt
2t (3/2)et (1/2)et (1/2)et . (3/2)et (1/2)e
t (1/2)et (1/2)e
t 2et 4e
t = 5 3e2t , = 5t (3/2)e2t , where we have taken the constants of integration to be zero. Thus, by equation (8) on page 553 of the text, we see that xp (t) = et et 5t (3/2)e2t = 5tet  (3/2)et 5tet + (9/2)et . et 3et Therefore, by adding xh (t) and xp (t) we obtain x(t) = c1 et et + c2 et 3et + 5tet  (3/2)et 5tet + (9/2)et . We remark that this answer is the same as the answer given in the text as can be seen by replacing c1 by c1 + 9/4. 606 Exercises 9.7
15. We must first find a fundamental matrix for the associated homogeneous system. We will do this by finding the solutions derived from the eigenvalues and the associated eigenvectors for the coefficient matrix A. Therefore, we find these eigenvalues by solving the characteristic equation given by A  rI = 4  r 2 2 1  r =0 r 2 + 5r = 0 . (4  r)(1  r)  4 = 0 Thus, the eigenvalues for the matrix A are r = 5, 0. An eigenvector for this matrix associated with the eigenvalue r = 0 is the vector u = col(u1 , u2) which satisfies the equation Au = 4 2 2 1 u1 u2 = 0 0 . This equation is equivalent to the equation 2u1 = u2 . Therefore, if we let u1 = 1 and u2 = 2, then the vector u1 = col(1, 2) satisfies this equation and is, therefore, an eigenvector for the matrix A associated with the eigenvalue r = 0. Hence, one solution to the homogeneous system is given by x1 (t) = e(0)t u1 = 1 2 . To find an eigenvector associated with the eigenvalue r = 5, we solve the equation (A + 5I)u = 1 2 2 4 u1 u2 = 0 0 , which is equivalent to the equation u1 + 2u2 = 0. Thus, by letting u2 = 1 and u1 = 2, the vector u2 = col(u1 , u2 ) = col(2, 1) satisfies this equation and is, therefore, an eigenvector for A associated with the eigenvalue r = 5. Hence, since the two eigenvalues of A are distinct, we see that another linearly independent solution to the corresponding homogeneous system is given by x2 (t) = e5t u2 = e5t 2 1 = 2e5t e5t . 607 Chapter 9
By combining these two solutions, we see that a general solution to the homogeneous system is xh (t) = c1 1 2 + c2 2e5t e5t and a fundamental matrix for this system is the matrix X(t) = 1 2e5t 2 e5t . We will use equation (10) on page 553 of the text to find a particular solution to the nonhomogeneous system. Thus, we need to find the inverse matrix X1 (t). This can be done, for example, by performing rowreduction on the matrix [X(t)I] to obtain the matrix [IX1 (t)]. In this way, we find that the required inverse matrix is given by X1 (t) = Therefore, we have X1 (t)f(t) = From this we see that X1 (t)f(t) dt = [t1 + (8/5)] dt (4/5)e5t dt = ln t + (8/5)t (4/25)e5t , 1/5 2/5 t1 4 + 2t1 (2/5)e5t (1/5)e5t = t1 + (8/5) (4/5)e5t . 1/5 2/5 (2/5)e5t (1/5)e5t . where we have taken the constants of integration to be zero. Hence, by equation (10) on page 553 of the text, we obtain xp (t) = X(t) = X1 (t)f(t) dt ln t + (8/5)t (4/25)e5t = ln t + (8/5)t  (8/25) 2 ln t + (16/5)t + (4/25) . 1 2e5t 2 e5t Adding xh (t) and xp (t) yields the general solution to the nonhomogeneous system given by x(t) = c1 1 2 + c2 2e5t e5t + ln t + (8/5)t  (8/25) 2 ln t + (16/5)t + (4/25) . 608 Exercises 9.7
21. We will find the solution to this initial value problem by using equation (13) on page 554 of the text. Therefore, we must first find a fundamental matrix for the associated homogeneous system. This means that we must find the eigenvalues and corresponding eigenvectors for the coefficient matrix of this system by solving the characteristic equation A  rI = r 2 r 2  3r + 2 = 0 (r  2)(r  1) = 0. 1 3  r r(3  r) + 2 = 0 Hence, r = 1, 2 are the eigenvalues for this matrix. To find an eigenvector u = col(u1 , u2 ) for this coefficient matrix associated with the eigenvalue r = 1, we solve the system (A  I)u = 1 2 1 2 u1 u2 = 0 0 . This system is equivalent to the equation u1 = 2u2. Thus, u1 = 2 and u2 = 1 is a set of values which satisfies this equation and, therefore, the vector u1 = col(2, 1) is an eigenvector for the coefficient matrix corresponding to the eigenvalue r = 1. Hence, one solution to the homogeneous system is given by x1 (t) = et u1 = et Similarly, by solving the equation (A  2I)u = 2 2 1 1 u1 u2 = 0 0 , 2 1 = 2et et . we find that one eigenvector for the coefficient matrix associated with the eigenvalue r = 2 is u2 = col(u1 , u2) = col(1, 1). Thus, another linearly independent solution to the associated homogeneous problem is given by x2 (t) = e2t u2 = e2t 1 1 = e2t e2t . 609 Chapter 9
By combining these two solutions, we obtain a general solution to the homogeneous system xh (t) = c1 and the fundamental matrix X(t) = 2et et + c2 e2t e2t , 2et e2t et e2t . In order to use equation (13) on page 554 of the text, we must also find the inverse of the fundamental matrix. One way of doing this is to perform rowreduction on the matrix [X(t)I] to obtain the matrix [IX1(t)]. Thus, we find that X1 (t) = From this we see that X1 (s)f(s) = es es es es e2s 2e2s = 2 3es . et et . e2t 2e2t (a) Using the initial condition x(0) = col(5, 4), and t0 = 0, we have X1 (0) = Therefore
t t 1 1 1 2 . X (s)f(s) ds =
t0 0 1 X (s)f(s) ds = 1 t (2)ds 0 t (3es ) ds 0 = 2t 3et  3 , from which it follows that
t X(t)
t0 X1 (s)f(s) ds = 2et e2t et e2t 2t 3et  3 = 4tet + 3et  3e2t 2tet + 3et  3e2t . We also find that X(t)X1 (t0 )x0 = 610 2et e2t et e2t 1 1 1 2 5 4 = 2et e2t et e2t 1 3 = 2et + 3e2t et + 3e2t . Exercises 9.7
Hence, by substituting these expressions into equation (13) on page 554 of the text, we obtain the solution to this initial value problem given by
t x(t) = X(t)X1(t0 )x0 + X(t)
t0 X1 (s)f(s) ds 4tet + 5et 2tet + 4et = 2et + 3e2t et + 3e2t + 4tet + 3et  3e2t 2tet + 3et  3e2t = . (b) Using the initial condition x(1) = col(0, 1), and t0 = 1, we have X1 (1) = Therefore
t t e1 e2 e1 2e2 . X (s)f(s) ds =
t0 1 1 X1 (s)f(s) ds
t (2)ds 1 t (3es ) ds 1 = from which it follows that
t = 2t  2 3e
t  3e1 , X(t)
t0 X1 (s)f(s) ds = 2et e2t et e2t 2t  2 3et  3e1 = 4tet  et  3e2t1 2tet + et  3e2t1 . = We also find that X(t)X (t0 )x0 = =
1 4tet  4et + 3et  3e2t1 2tet  2et + 3et  3e2t1 2et e2t et et e2t e2t 2et e2t e1 e2 e1 2e2 e1 2e2 = 0 1 2et1 + 2e2t2 et1 + 2e2t2 . 611 Chapter 9
Hence, by substituting these expressions into equation (13) on page 554 of the text, we obtain the solution to this initial value problem given by
t x(t) = X(t)X1 (t0 )x0 + X(t)
t0 X1 (s)f(s) ds 4tet  et  3e2t1 2tet + et  3e2t1 . = 2et1 + 2e2t2 et1 + 2e2t2 + = 2et1 + 2e2t2 + 4tet  et  3e2t1 et1 + 2e2t2 + 2tet + et  3e2t1 (c) Using the initial condition x(5) = col(1, 0), and t0 = 5, we have X1 (5) = Therefore
t t e5 e5 e10 2e10 . X (s)f(s) ds =
t0 5 1 X (s)f(s) ds = 1 t (2)ds 5 t (3es ) ds 5 = 2t  10 3et  3e5 , from which it follows that
t X(t)
t0 X1 (s)f(s) ds = 2et e2t et e2t 2t  10 3et  3e5 = 4tet  17et  3e2t5 2tet  7et  3e2t5 . = We also find that X(t)X (t0 )x0 = = 612
1 4tet  20et + 3et  3e2t5 2tet  10et + 3et  3e2t5 2et e2t et et e2t e2t 2et e2t e5 e5 e10 e5 1 0 2et5  e2t10 et5  e2t10 . e10 2e10 = Exercises 9.7
Hence, by substituting these expressions into equation (13) on page 554 of the text, we obtain the solution to this initial value problem given by
t x(t) = X(t)X1 (t0 )x0 + X(t)
t0 X1 (s)f(s) ds 4tet  17et  3e2t5 2tet  7et  3e2t5 . = 2et5  e2t10 et5  e2t10 + = 2et5  e2t10 + 4tet  17et  3e2t5 et5  e2t10 + 2tet  7et  3e2t5 25. (a) We will find a fundamental solutions set for the corresponding homogeneous system by deriving solutions using the eigenvalues and associated eigenvectors for the coefficient matrix. Therefore, we first solve the characteristic equation A  rI = r 1 =0 r 2  3r + 2 = 0 (r  2)(r  1) = 0. 2 3  r r(3  r) + 2 = 0 Therefore, we see that the eigenvalues for the coefficient matrix of this problem are r = 1, 2. Since these eigenvalues are distinct, the associated eigenvectors will be linearly independent, and so the solutions derived from these eigenvectors will also be linearly independent. We find an eigenvector for this matrix associated with the eigenvalue r = 1 by solving the equation (A  I)u = 1 1 2 2 u1 u2 = 0. Since the vector u1 = col(u1, u2 ) = col(1, 1) satisfies this equation, we see that this vector is one such eigenvector and so one solution to the homogeneous problem is given by x1 (t) = et u1 = et 1 1 . 613 Chapter 9
To find an eigenvector associated with the eigenvalue r = 2, we solve the equation (A  2I)u = 2 1 2 1 u1 u2 = 0. The vector u2 = col(u1, u2 ) = col(1, 2) is one vector which satisfies this equation and so it is one eigenvector of the coefficient matrix associated with the eigenvalue r = 2. Thus, another linearly independent solution to the corresponding homogeneous problem is given by x2 (t) = e2t u2 = e2t 1 2 , and a fundamental solution set for this homogeneous system is the set et u1 , e2t u2 , where u1 = col(1, 1) and u2 = col(1, 2). (b) If we assume that xp (t) = tet a for some constant vector a = col(a1 , a2 ), then we have xp (t) = tet a + et a = We also have 0 1 2 3 xp (t) + f(t) = 0 1 tet a1 tet a2 2 3 + et 0 = tet a2 + et 2tet a1 + 3tet a2 . tet a1 tet a2 + et a1 et a2 = tet a1 + et a1 tet a2 + et a2 . Thus, if xp (t) = tet a is to satisfy this system, we must have tet a1 + et a1 tet a2 + et a2 which means that tet a1 + et a1 = tet a2 + et , tet a2 + et a2 = 2tet a1 + 3tet a2 . By dividing out the term et and equating coefficients, this system becomes the system a1 = a2 , a1 = 1, = tet a2 + et 2tet a1 + 3tet a2 , a2 = 2a1 + 3a2 , a2 = 0. 614 Exercises 9.7
Since this set of equations implies that 1 = a1 = a2 = 0, which is of course impossible, we see that this system has no solutions. Therefore, we cannot find a vector a for which xp (t) = tet a is a particular solution to this problem. (c) Assuming that xp (t) = tet a + et b = tet a1 tet a2 + et b1 et b2 = tet a1 + et b1 tet a2 + et b2 , where a = col(a1 , a2 ) and b = col(b1 , b2 ) are two constant vectors, implies that xp (t) = tet a + et a + et b = We also see that 0 1 2 3 xp (t) + f(t) = = 0 1 tet a1 + et b1 tet a2 + et b2 tet a2 + et b2 + et 2tet a1  2et b1 + 3tet a2 + 3et b2 2 3 + et 0 . tet a1 + et a1 + et b1 tet a2 + et a2 + et a2 . Thus, if xp (t) is to satisfy this system, we must have tet a1 + et a1 + et b1 tet a2 + et a2 + et b2 = tet a2 + et b2 + et 2tet a1  2et b1 + 3tet a2 + 3et b2 , (9.7) which implies the system of equations given by tet a1 + et a1 + et b1 = tet a2 + et b2 + et , tet a2 + et a2 + et b2 = 2tet a1  2et b1 + 3tet a2 + 3et b2 . Dividing each equation by et and equating the coefficients in the resulting equations yields the system a1 = a2 , a1 + b1 = b2 + 1 , a2 = 2a1 + 3a2 , a2 + b2 = 2b1 + 3b2 . Taking the pair of equations on the right and simplifying yields the system b1  b2 = 1  a1 , (9.9) 615 (9.8) 2b1  2b2 = a2 . Chapter 9
By multiplying the first of these equations by 2, we obtain the system 2b1  2b2 = 2  2a1 , 2b1  2b2 = a2 , which when subtracted yields 2  2a1 + a2 = 0. Applying the first equation in (9.8) (the equation a1 = a2 ) to this equation yields a1 = a2 = 2. By substituting these values for a1 and a2 into equation (9.9) above we see that both equations reduce to the equation b2 = b1 + 1. (Note also that the remaining equation in (9.8) reduces to the first equation in that set.) Thus, b1 is free to be any value, say b1 = s, and the set of values a1 = a2 = 2, b1 = s, b2 = s + 1, satisfies all of the equations given in (9.8) and, hence, the system given in (9.7). Therefore, particular solutions to the nonhomogeneous equation given in this problem are xp (t) = tet 2 2 + et s s+1 = tet 2 2 + et 0 1 + set 1 1 . But, since the vector u = et col(1, 1) is a solution to the corresponding homogeneous system, the last term can be incorporated into the solution xh (t) and we obtain one particular solution to this problem given by xp (t) = tet 2 2 + et 0 1 . (d) To find the general solution to the nonhomogeneous system given in this problem, we first form the solution to the corresponding homogeneous system using the fundamental solution set found in part (a). Thus, we have xh (t) = c1 et 1 1 + c2 e2t 1 2 . By adding the solution found in part (c) to this solution, we obtain the general solution given by x(t) = c1 et 1 1 + c2 e2t 1 2 + tet 2 2 + et 0 1 . 616 Exercises 9.8
EXERCISES 9.8: The Matrix Exponential Function, page 566 3. (a) From the characteristic equation, A  rI = 0, we obtain 2r A  rI = 3 9 (2  r) 1 1  r 3 1  r 3 1 4  r 1 1 4  r  3 9 1 4  r + (1) 3 1  r 9 3 =0 =0 (2  r)[(1  r)(4  r)  3]  [3(4  r)  9]  [9  9(1  r)] = 0 r3 + 3r 2 + 3r + 1 = (r + 1)3 = 0. Therefore, for the matrix A, r = 1 is an eigenvalue of multiplicity three. Thus, by the CayleyHamilton theorem as stated on page 561 of the text, we have (A + I)3 = 0 (so that r = 1 and k = 3). (b) In order to find eAt , we first notice (as was done in the text on page 560) that eAt = e[I+(A+I)]t , commutative and associative properties of matrix addition = eIt e(A+I)t , = et e(A+I)t . Therefore, to find eAt we need only to find e(A+I)t then multiply the resulting expression by et . By formula (2) on page 558 of the text and using the fact that (A + I)3 = 0 (which implies that (A + I)n = 0 for n 3), we have e(A+I)t = I + (A + I)t + (A + I)2 = I + (A + I)t + (A + I)2 t2 2 t2 2 + + (A + I)n . tn n! + (9.10) 617 property (d) on page 559 of the text [since (A + I)I = I(A + I)] = et Ie(A+I)t , property (e) on page 559 of the text Chapter 9
Since 3 1 1 3 1 1 3 0 1 (A + I)2 = 3 0 1 3 0 1 = 0 0 0 , 9 3 3 9 3 3 9 0 3 equation (9.10) becomes 1 0 0 (A+I)t 0 1 0 + = e 0 0 1 1 + 3t  3t2 /2 = 3t 9t  9t2 /2 Hence, we have eAt = et 1 + 3t  3t2 /2 3t 9t  9t2 /2 t 1 t + t2 /2 t . 3t 3t 9t t 1 t 0 t t 3t 3t + 3(t /2) 0 0 0 2 t /2 0 2 9(t2 /2) 0 3(t2 /2) t + t2 /2 t 3t 1  3t + 3t2 /2 . 3t 1  3t + 3t2 /2 9. By equation (6) on page 562 of the text, we see that eAt = X(t)X1 (0), where X(t) is a fundamental matrix for the system x = Ax. We will construct this fundamental matrix from three linearly independent solutions derived from the eigenvalues and associated eigenvectors for the matrix A. Thus, we solve the characteristic equation r A  rI = 0 1 (r) r 1 r 1 0 1 =0 1 1  r  0 1 1 1r =0 1 1  r r[r(1  r) + 1] + 1 = r 3 + r 2  r + 1 = (r  1)(r 2 + 1) = 0. Therefore, the eigenvalues of the matrix A are r = 1 and r = i. To find an eigenvector 618 Exercises 9.8
u = col(u1, u2 , u3 ) associated with the eigenvalue r = 1, we 1 1 0 u1 0 (A  I)u = 0 0 1 1 u2 = 0 u3 1 1 0 0 solve the system 1 0 1 u1 0 0 1 1 u2 = 0 . u3 0 0 0 0 This system is equivalent to the system u1 = u3 , u2 = u3 . Hence, u3 is free to be any arbitrary value, say u3 = 1. Then u1 = u2 = 1, and so the vector u = col(1, 1, 1) is an eigenvector associated with r = 1. Hence, one solution to the system 1 t t x1 (t) = e u = e 1 = 1 x = Ax is given by et et . et Since the eigenvalue r = i is complex, we want to find two more linearly independent solutions for the system x = Ax derived from the eigenvectors associated with this eigenvalue. These eigenvectors, z = col(z1 , z2 , z3 ), must satisfy the equation z1 i 1 0 (A  iI)z = 0 i 1 z2 1 1 1  i z3 0 = 0 , 0 which is equivalent to the system z1 = z3 , z2 = iz3 . Thus, one solution to this system is z3 = 1, z1 = 1, and z2 = i and so one eigenvector for A associated with the eigenvalue r = i is given by z1 z = z2 = i = 0 + i 1 . 0 1 1 z3 By the notation on page 546 of the text, this means that = 0, = 1, a = col(1, 0, 1) and b = col(0, 1, 0). Therefore, by equations (6) and (7) on page 546 of the text we see that two more linearly independent solutions to the system x = Ax are given  cos t 0 (0)t (0)t   sin t x2 (t) = e (cos t)a  e (sin t)b = 0 cos t 0 by 1 1 0 = sin t , cos t 619  cos t Chapter 9 x3 (t) = e(0)t (sin t)a + e(0)t (cos t)b =  sin t 0 0 +  cos t =  cos t . sin t 0 sin t 1 . 0 0  sin t Thus, a fundamental matrix for this system is et  cos t  sin t X(t) = et sin t  cos t sin t et cos t X(0) = 1 1 1 0 1 1 To find the inverse of the matrix X(0) we can, for example, perform rowreduction on the matrix [X(0)I] to obtain the matrix [IX1 (0)]. Thus, 1/2 0 1 X (0) = 1/2 0 1/2 1 Hence, we obtain eAt = X(t)X1 (0) = et et et  cos t  sin t sin t cos t
t we see that 1/2 1/2 . 1/2 1/2 0 1/2  cos t 1/2 0 1/2 sin t 1/2 1 1/2
t e + cos t  sin t 2 sin t e  cos t  sin t 1 t e  cos t  sin t 2 cos t et  cos t + sin t . = 2 t t e  cos t + sin t 2 sin t e + cos t + sin t 11. The first step in finding eAt using a fundamental matrix for the system x = Ax is to find the eigenvalues for the matrix A. Thus, we solve the characteristic equation 5  r 4 A  rI = 1 0 620 (5  r) r 2 r 2 2 5r 0 2 5r +4 1 2 0 5r =0 =0 Exercises 9.8 (5  r)[r(5  r)  4] + 4(5  r) = r(r  5)2 = 0. Therefore, the eigenvalues of A are r = 0, 5, with r = 5 an eigenvalue of multiplicity two. Next we must find the eigenvectors and generalized eigenvectors for the matrix A and from these vectors derive three linearly independent solutions of the system x = Ax. To find the eigenvector associated 5 4 Au = 1 0 0 2 with the eigenvalue r = 0, we solve the equation 1 0 2 u1 0 0 u1 0 2 5 u2 2 u2 = 0 0 0 0 0 5 u3 u3 0 = 0 . 0 This equation is equivalent to the system u1 = 2u3 , 2u2 = 5u3 and one solution to this system is u3 = 2, u1 = 4, u2 = 5. Therefore, one eigenvector of the matrix A associated with the eigenvalue r = 0 is given by the vector u1 = col (u1 u2 u3 ) = col(4, 5, 2), and so one solution to the system x = Ax is
0 4 5 . x1 (t) = e u1 = 2 To find an eigenvector associated with the eigenvalue 0 4 0 (A  5I)u = 1 5 2 0 2 0 r = 5, we solve the equation u1 0 u2 = 0 , 0 u3 which is equivalent to the system u2 = 0, u1 = 2u3 . One solution to this system is u3 = 1, u1 = 2, u2 = 0. Thus, one eigenvector of the matrix A associated with the eigenvalue r = 5 is the vector u2 = col (u1 u2 u3 ) = col(2, 0, 1), and so another linearly independent solution to the system x = Ax is given by 2 2e5t x2 (t) = e5t u2 = e5t 0 = 0 . 1 e5t 621 Chapter 9
Since r = 5 is an eigenvalue of multiplicity two, we can find a generalized eigenvector (with k = 2) associated with the eigenvalue r = 5 which will be linearly independent from the vector u2 found above. Thus, we solve the equation (A  5I)2 u = 0. Because (9.11) (A  5I)2 = 1 5 2 1 5 2 = 5 25 10 , 2 10 4 0 2 0 0 2 0 we see that equation (9.11) becomes u1 4 20 8 0 5 25 10 u2 = 0 2 10 4 0 u3 This equation is equivalent to the equation u1 + 5u2  2u3 = 0 and is, therefore, satisfied if we let u2 = s, u3 = v, and u1 = 5s  2v for any values of s and v. Hence, solutions to equation (9.11) are given by 2 5 5s  2v u1 = s 1 +v 0 u = u2 = s 1 0 v u3 . u1 0 0 4 0 0 4 0 4 20 8 0 0 1 5 2 0 0 0 u2 = 0 . 0 0 u3 Notice that the vectors vcol(2, 0, 1) are the eigenvectors that we found above associated with the eigenvalue r = 5. Since we are looking for a vector which satisfies equation (9.11) and is linearly independent from this eigenvector we will choose s = 1 and v = 0. Thus, a generalized eigenvector for the matrix A associated with the eigenvalue r = 5 and linearly independent of the eigenvector u2 is given by u3 = col(5, 1, 0). 622 Exercises 9.8
Hence, by formula (8) on page 563 of the text, we see that another linearly independent solution to the system x = Ax is given by x3 (t) = eAt u3 = e5t [u3 + t(A  5I)u3 ] 5 5 0 4 0 = e5t 1 + te5t 1 5 2 1 0 0 0 2 0 5 4 5e5t  4te5t = e5t 1 + te5t 0 = e5t 2te5t 0 2 , where we have used the fact that, by our choice of u3 , (A 5I)2 u3 = 0 and so (A 5I)n u3 = 0 for n 2. (This is the reason why we used the generalized eigenvector to calculate x3 (t). The CayleyHamilton theorem, as given on page 561 of the text, states that A satisfies its characteristic equation, which in this case means that A(A  5I)2 = 0. However, we cannot assume from this fact that (A  5I)2 = 0 because in matrix multiplication it is possible for two nonzero matrices to have a zero product.) Our last step is to find a fundamental matrix for the system x = Ax using the linearly independent solutions found above and then to use this fundamental matrix to calculate eAt . Thus, from these three solutions we obtain the fundamental matrix given by 4 2e 5 X(t) = 0 2 e5t 5t 5e  4te e
5t 5t 5t 2te5t 4 2 5 5 0 1 . X(0) = 2 1 0 We can find the inverse matrix X1 (0) by (for example) performing rowreduction on the matrix [X(0)I] to obtain the matrix [IX1 (0)]. Thus, we find 1 X1 (0) = 1 2 10 21 . 25 5 0 10 623 5 2 Chapter 9
Therefore, by formula (6) on page 562 of the text, we see that 1 5 4 2e5t 5e5t  4te5t 1 2 10 5 eAt = X(t)X1 (0) = 0 e5t 25 5t 5t 2 e 2te 5 0 4 + 29e5t  20te5t 20  20e5t 8 + 8e5t  40te5t 1 = 5 + 5e5t 25 10 + 10e5t 25 2  2e5t + 10te5t 10 + 10e5t 4 + 21e5t + 20te5t 21 10 . 2 17. We first calculate the eigenvalues for the matrix A by solving the characteristic equation r A  rI = 0 r 1 r 1 0 1 =0 2 5 4  r (r) 5 4  r  0 1 2 4  r =0 r[r(4  r) + 5]  2 =  r 3 + 4r 2 + 5r + 2 = (r + 1)2 (r + 2) = 0. Thus, the eigenvalues for A are r = 1, 2, with r = 1 an eigenvalue of multiplicity two. To find an eigenvector u = col(u1 , u2 , u3 ) associated with the eigenvalue r = 1, we solve the equation (A + I)u = 0 1 1 1 0 u1 0 1 u2 = 0 , u3 2 5 3 0 which is equivalent to the system u1 = u3 , u2 = u3 . Therefore, by letting u3 = 1 (so that u1 = 1 and u2 = 1), we see that one eigenvector for the matrix A associated with the eigenvalue r = 1 is the vector u1 = col(u1 , u2 , u3) = col(1, 1, 1). Hence, one solution to the system x = Ax is given by 1 x1 (t) = et u1 = et 1 . 1 624 Exercises 9.8
Since r = 1 is an eigenvalue of multiplicity two, we can find a generalized eigenvector associated with this eigenvalue (with k = 2) which will be linearly independent from the vector u1 . To do this, we solve the equation (A + I)2 u = 0 1 1 0 1 1 0 0 1 1 0 1 1 2 5 3 2 5 3 1 2 1 0 u1 2 4 2 u2 = 0 4 8 4 u3 0 u1 0 u2 = 0 0 u3 0 1 2 1 u1 0 0 0 u2 = 0 , u3 0 0 0 0 which is equivalent to the equation u1 + 2u2 + u3 = 0. This equation will be satisfied if we let u3 = s, u2 = v, and u1 = 2v  s for any values of s and v. Thus, generalized eigenvectors associated with the eigenvalue r = 1 are given by u1 u = u2 = u3 2v  s v s = s 0 + v 1 . 1 0 1 2 Hence, by letting s = 2 and v = 1, we find one such generalized eigenvector to be the vector u2 = col(0, 1, 2), which we see by inspection is linearly independent from u1 . Therefore, by equation (8) on page 563 of the text, we obtain a second linearly independent solution of the system x = Ax given by x2 (t) = eAt u2 = et [u2 + t(A + I)u2 ] 0 1 1 0 t t = e 1 + te 0 1 1 2 2 5 3 0 1 2 625 Chapter 9 0 = et 1 + tet 1 = et 1 + t . 2 1 2t In order to obtain a third linearly independent solution to this system, we will find an eigenvector associated with the eigenvalue r = 2 by solving the equation u1 2 1 0 0 (A + 2I)u = 0 2 1 u2 = 0 . 2 5 2 0 u3 This equation is equivalent to the system 2u1 + u2 = 0, 2u2 + u3 = 0. One solution to this system is given by u1 = 1, u2 = 2, and u3 = 4. Thus, one eigenvector associated with the eigenvalue r = 2 is the vector u3 = col(u1 , u2 , u3) = col(1, 2, 4), and another linearly independent solution to this system is given by 1 2t 2t x3 (t) = e u3 = e 2 . 4 Hence, by combining the three linearly independent solutions that we have just found, we see that a general solution to this system is 1 t x(t) = c1 et 1 + c2 et 1 + t 1 2t 23. In Problem 3, we found that eAt = et 1 + 3t  3t2 /2 3t 9t  9t2 /2 t 1 t + t2 /2 t . 1 1 t + c3 e2t 2 . 4 3t 1  3t + 3t2 /2 626 Exercises 9.8
In order to use the variation of parameters formula (equation (13) on page 565 of the text), we need to find expressions for eAt x0 and t0 = 0. Thus, we first notice that
t t t t 0 eA(ts) f(s) ds, where we have used the fact that e
0 A(ts) f(s) ds =
0 e AtAs f(s) ds = e At 0 eAs f(s) ds . Since f(s) = col(0, s, 0), we observe that 1  3s  3s2 /2 s s + s2 /2 eAs f(s) = es 3s 1  9s  9s2 /2 3s 1 + 3s + 3s2 /2 s2 s2 es = es s = ses . 3s2 es 3s2 Therefore, we have
t t 0 s 0 eA(ts) f(s) ds = eAt
0 0 eAs f(s) ds
t (s2 es )ds 0 t (ses )ds 0 t (3s2 es )ds 0 t 2 , = eAt = e
At 2  e (t  2t + 2) 1 + et (t  1) 6  3et (t2  2t + 2) where we have used integration by parts to evaluate the three integrals above. Next, since x0 = col(0, 3, 0), we see that 1 + 3t  3t2 /2 t t + t2 /2 eAt x0 = et 3t 1 t 2 3t 1  3t + 3t2 /2 9t  9t /2 0 3t 3 = et 3 . 0 9t 627 Chapter 9
Finally, substituting these expressions into the variation of parameters formula (13), page 565 of the text, yields
t x(t) = e x0 + = e
t 0 At eA(ts) f(s) ds t 2 3t 2  e (t  2t + 2) , 3 + eAt 1 + et (t  1) t 2 6  3e (t  2t + 2) 9t where eAt is given above. 628 CHAPTER 10: Partial Differential Equations
EXERCISES 10.2: Method of Separation of Variables, page 587 5. To find a general solution to this equation, we first observe that the auxiliary equation associated with the corresponding homogeneous equation is given by r 2  1 = 0. This equation has roots r = 1. Thus, the solution to the corresponding homogeneous equation is given by yh (x) = C1 ex + C2 ex . By the method of undetermined coefficients, we see that the form of a particular solution to the nonhomogeneous equation is yp (x) = A + Bx, where we have used the fact that neither y = 1 nor y = x is a solution to the corresponding homogeneous equation. To find A and B, we note that yp (x) = B and yp (x) = 0. By substituting these expressions into the original differential equation, we obtain yp (x)  yp (x) = A  Bx = 1  2x. By equating coefficients, we see that A = 1 and B = 2. Substituting these values into the equation for yp (x) yields yp (x) = 1 + 2x. Thus, we see that y(x) = yh (x) + yp (x) = C1 ex + C2 ex  1 + 2x. 629 Chapter 10
Next we try to find C1 and C2 so that the solution y(x) will satisfy the boundary conditions. That is, we want to find C1 and C2 satisfying y(0) = C1 + C2  1 = 0 and y(1) = C1 e + C2 e1 + 1 = 1 + e. From the first equation we see that C2 = 1  C1 . Substituting this expression for C2 into the second equation and simplifying yields e  e1 = C1 e  e1 . Thus, C1 = 1 and C2 = 0. Therefore, y(x) = ex  1 + 2x is the only solution to the boundary value problem. 13. First note that the auxiliary equation for this problem is r 2 + = 0. To find eigenvalues which yield nontrivial solutions we will consider the three cases: < 0, = 0, and > 0. Case 1, < 0: In this case the roots to the auxiliary equation are r =  (where we note that  is a positive number). Therefore, a general solution to the differential equation y + y = 0 is given by y(x) = C1 e x + C2 e x . In order to apply the boundary conditions we need to find y (x). Thus, we have y (x) = C1 e x  C2 e x . By applying the boundary conditions we obtain y(0)  y (0) = C1 + C2  C1 + C2 = 0 1   C1 + 1 +  C2 = 0, and y() = C1 e 630  + C2 e  =0 C2 = C1 e2  . Exercises 10.2
By combining these expressions, we observe that 1  C1  1 +  C1 e2  = 0 C1 1    1 +  e2  = 0. (10.1) This last expression will be true if C1 = 0 or if e But since  > 0, we see that e2 2   1   . = 1 +  > 1 while (1  )/(1 + ) < 1. Therefore, the only way that equation (10.1) can be true is for C1 = 0. This means that C2 must also equal zero and so in this case we have only the trivial solution. Case 2, = 0: In this case we are solving the differential equation y = 0. This equation has a general solution given by y(x) = C1 + C2 x y (x) = C2 . By applying the boundary conditions we obtain y(0)  y (0) = C1  C2 = 0 and y() = C1 + C2 = 0. Solving these equations simultaneously yields C1 = C2 = 0. Thus, we again find only the trivial solution. Case 3, > 0: In this case the roots to the associated auxiliary equation are r = i. Therefore, the general solution is given by y(x) = C1 cos x + C2 sin x x + C2 cos x . y (x) =  C1 sin By applying the boundary conditions, we obtain y(0)  y (0) = C1  C2 = 0 C1 = C2 , 631 Chapter 10
y
0 1 2 3 4 5 6 7 2 4 6 x Figure 10A: The intersection of the graphs y = x and y = tan(x), x > 0. and y() = C1 cos By combining these results, we obtain C2 cos + C2 sin = 0. + sin =0 Therefore, in order to obtain a solution other than the trivial solution, we must solve the equation cos + sin = 0. By simplifying this equation becomes tan =  . To see that there exist values for > 0 which satisfy this equation, we examine the graphs of the equations y = x and y = tan(x). For any values of x > 0 where these two graphs intersect, we set = x2 . These values for will be the eigenvalues that we seek. From the graph in Figure 10A, we see that there are (countably) infinitely many such eigenvalues. These values satisfy the equations tan 632 n + n = 0. Exercises 10.2
As n becomes large, we can also see from the graph that these eigenvalues approach the square of odd multiples of 1/2. That is, (2n  1)2 , 4 when n is large. Corresponding to the eigenvalue n we obtain the solutions n yn (x) = C1n cos (since C1n = n x + C2n sin n x = n C2n cos n x + C2n sin n x n C2n ). Thus yn (x) = Cn n cos n x + sin n x , where Cn is arbitrary. 17. We are solving the problem u(x, t) 2 u(x, t) , =3 t t2 u(0, t) = u(, t) = 0, 0 < x < , t > 0, t > 0, u(x, 0) = sin x  7 sin 3x + sin 5x. A solution to this partial differential equation satisfying the first boundary condition is given in equation (11) on page 582 of the text. By letting = 3 and L = in this equation we obtain the series u(x, t) =
n=1 cn e3n t sin nx . 2 (10.2) To satisfy the initial condition, we let t = 0 in this equation and set the result equal to sin x  7 sin 3x + sin 5x. This yields u(x, t) =
n=1 cn sin nx = sin x  7 sin 3x + sin 5x. By equating the coefficients of the like terms, we see that c1 = 1, c3 = 7, c5 = 1, and all other cn 's are zero. Plugging these values into equation (10.2) gives the solution u(x, t) = e3(1) t sin x  7e3(3) t sin 3x + e3(5) t sin 5x = e3t sin x  7e27t sin 3x + e75t sin 5x . 633
2 2 2 Chapter 10
21. By letting = 3 and L = in formula (24) on page 585 of the text, we see that the solution we want will have the form u(x, t) =
n=1 [an cos 3nt + bn sin 3nt] sin nx . (10.3) Therefore, we see that u = t [3nan sin 3nt + 3nbn cos 3nt] sin nx .
n=1 In order for the solution to satisfy the initial conditions, we must find an and bn such that u(x, 0) =
n=1 an sin nx = 6 sin 2x + 2 sin 6x, and u(x, 0) = t 3nbn sin nx = 11 sin 9x  14 sin 15x.
n=1 From the first condition, we observe that we must have a term for n = 2, 6 and for these terms we want a2 = 6 and a6 = 2. All of the other an 's must be zero. By comparing coefficients in the second condition, we see that we require 3(9)b9 = 11 or b9 = 11 27 and 3(15)b15 = 14 or b15 =  14 . 45 We also see that all other values for bn must be zero. Therefore, by substituting these values into equation (10.3) above, we obtain the solution of the vibrating string problem with = 3, L = and f (x) and g(x) as given. This solution is given by u(x, t) = 6 cos(3 2 t) sin 2x+ 2 cos(3 6 t) sin 6x+ Or by simplifying, we obtain u(x, t) = 6 cos 6t sin 2x + 2 cos 18t sin 6x + 14 11 sin 27t sin 9x  sin 45t sin 15x. 27 45 14 11 sin(3 9 t) sin 9x sin(3 15 t) sin 15x. 27 45 23. We know from equation (11) on page 582 of the text that a formal solution to the heat flow problem is given by u(x, t) =
n=1 cn e2(n) t sin nx , 2 (10.4) 634 Exercises 10.3
where we have made the substitutions = 2 and L = 1. For this function to be a solution to the problem it must satisfy the initial condition u(x, 0) = f (x), 0 < x < 1. Therefore, we let t = 0 in equation (10.4) above and set the result equal to f (x) to obtain u(x, 0) =
n=1 cn sin nx =
n=1 1 sin nx . n2 By equating coefficients, we see that cn = n2 . Substituting these values of cn into equation (10.4) yields the solution u(x, t) =
n=1 n2 e2(n) t sin nx . 2 EXERCISES 10.3: Fourier Series, page 603 5. Note that f (x) = ex cos(3x) = ex cos 3x. Since f (x) = ex cos 3x = ex cos 3x = f (x) unless x = 0 we see that this function is not even. Similarly since f (x) = ex cos 3x = ex cos 3x = f (x), this function is also not odd. 13. For this problem T = 1. Thus, by Definition 1 on page 594 of the text, the Fourier series for this function will be given by a0 + (an cos nx + bn sin nx) . 2 n=1 (10.5) To compute a0 , we use equation (9) given in Definition 1 in the text noting that cos(0x) = 1. Thus, we have
1 a0 =
1 x3 x dx = 3
2 1 =
1 2 1 1  = . 3 3 3 635 Chapter 10
To find an for n = 1, 2, 3, . . ., we again use equation (9) on page 594 of the text. This yields
1 1 an =
1 x cos nx dx = 2
0 2 x2 cos nx dx, where we have used the fact that x2 cos nx is an even function. Thus, using integration by parts twice, we obtain
1 x2 cos nx dx = 2 x2 sin nx n 1 1 x sin nx dx an = 2 = 2
0 
0 2 n + 0 1 cos nx dx 2 cos nx sin n 0  x n n n 1 0 1 n 0 1 0 2 2 = 2 0 + 2 2 (cos n  0)  2 2 n n 1 sin nx n 4 4 4 (1)n  3 3 (sin n  0) = 2 2 (1)n . = 22 n n n To calculate the bn 's, note that since x2 is even and sin nx is odd, their product is odd (see Problem 7 in this section of the text). Since x2 sin nx is also continuous, by Theorem 1 on page 590 of the text, we have
1 bn =
1 x2 sin nx dx = 0 . By plugging these coefficients into equation (10.5) above, we see that the Fourier series associated with x2 is given by 1 4 + (1)n cos nx . 22 3 n=1 n 21. We use Theorem 2 on page 600 of the text. Notice that f (x) = x2 and f (x) = 2x are continuous on [1, 1]. Thus, the Fourier series for f converges to f (x) for 1 < x < 1. Furthermore, f 1+ = lim + x2 = 1
x1 and f 1 = lim x2 = 1. 
x1 636 Exercises 10.3
Hence, 1 1 f 1+ + f 1 = (1 + 1) = 1, 2 2 and so, by Theorem 2, the sum of the Fourier series equals 1 when x = 1. Therefore, the Fourier series converges to f (x) = x2 for  1 x 1. Since the sum function must be periodic with period 2, the sum function is the 2periodic extension of f (x) which we can write as g(x) = (x  2n)2 , 2n  1 x < 2n + 1, n = 0, 1, 2, . . . . 29. To calculate the coefficients of this expansion we use formula (20) on page 599 of the text. Thus we have c0 = 0 = 0, P0 P0 2 where we have used the fact that f (x) is an odd function. To find c1 we first calculate the
2 1 1 f (x) dx = denominator to be P1 Therefore, we obtain
1 2 1 1 2 P1 (x) dx =
1 =
1 x3 x dx = 3
2 1 =
1 2 . 3 1 3 c1 = 2 1 3 f (x)P1 (x) dx = 2 2 0 x2 x dx = 3 2 1 =
0 3 . 2 Notice that in order to calculate the above integral, we used the fact that the product of the two odd functions f (x) and P1 (x) is even. To find c2 , we first observe that, since f (x) is odd and P2 (x) is even, their product is odd and so we have
1 f (x)P2 (x) dx = 0.
1 Hence c2 = 1 1 f (x)P2 (x) dx P2
2 = 0 P2 2 = 0. 637 Chapter 10
31. We need to show that Hm (x)Hn (x)ex dx = 0,
 2 for m = n, where m, n = 0, 1, 2. Therefore, we need to calculate several integrals. Let's begin with m = 0, n = 2. Here we see that H0 (x)H2 (x)e
 x2 dx =
 4x2  2 ex dx
N 0 2 = N lim 4x2  2 e
0 x2 dx + lim M M 4x2  2 ex dx . 2 We will first calculate the indefinite integral using integration by parts with the substitution u = x, dv = 2xex dx
2 2 du = dx, v = ex . That is we find 4x2  2 ex dx = 2
2 2x2 ex dx  2
2 2 2 ex dx ex dx = 2xex + C.
2 2 2 = 2 xex + ex dx  2 Substituting this result in for the integrals we are calculating and using L'Hospital's rule to find the limits, yields H0 (x)H2 (x)ex dx =
 2 N lim 2xex 2 N 0 + lim M 2xex 2 0 M 2N 2M + 0 + lim 0  M 2 N2 N M e e 2N 2M =  lim N 2  lim M 2 = 0  0 = 0. N e M e = lim When m = 0, n = 1 and m = 1, n = 2, the integrals are, respectively, H0 (x)H1 (x)e
 x2 dx =
 2xex dx 2 638 Exercises 10.4
y
1 3 5/2 2 3/2  /2 /2 3/2 2 5/2 3 x Figure 10B: The graph of the periodic extension of f . and H1 (x)H2 (x)e
 x2 dx =
 2x(4x2  2)ex dx . 2 In each case the integrands are odd functions and hence their integrals over symmetric intervals of the form (N, N) are zero. Since it is easy to show that the above improper integrals are convergent, we get N = lim
 N N = lim 0 = 0.
N Since we have shown that the 3 integrals above are all equal to zero, the first three Hermite polynomials are orthogonal. EXERCISES 10.4: Fourier Cosine and Sine Series, page 611 3. (a) The periodic extension f(x) on the interval (, ) is 0, 1, f (x) = 0, 1,  < x < /2, /2 < x < 0, 0 < x < /2, /2 < x < , with f(x + 2) = f(x). The graph of this function is given in Figure 10B. (b) Using the formula on page 607 of the text, the odd 2periodic extension fo on the 639 Chapter 10
y
1 3 5/2 2 3/2  /2 /2 3/2 2 5/2 3 x Figure 10C: The graph of the odd 2periodic extension of f . interval (, ) is 1, 0, f (x),  < x < 0, = 0, f (x), 0<x< 1,  < x < /2, /2 < x < 0, 0 < x < /2, /2 < x < , fo (x) = with fo (x + 2) = fo (x). The graph of fo (x) is given in Figure 10C. (c) Using the formula on page 608 of the text, the even 2periodic extension fe on the interval (, ) is 1, 0, f (x),  < x < 0, = 0, f (x), 0<x< 1,  < x < /2, /2 < x < 0, 0 < x < /2, /2 < x < , fe (x) = with fe (x + 2) = fe (x). The graph of fe (x) is given in Figure 10D. 7. Since f is piecewise continuous on the interval [0, ], we can use equation (6) in Definition 2 on page 609 of the text to calculate its Fourier sine series. In this problem T = and f (x) = x2 . Thus we have f (x) =
n=1 bn sin nx , with 2 bn = x2 sin nx dx .
0 640 Exercises 10.4
y
1 3 5/2 2 3/2  /2 /2 3/2 2 5/2 3 7/2 x Figure 10D: The graph of the even 2periodic extension of f . To calculate the coefficients, we use integration by parts twice to obtain 2 2 bn = 2 cos nx 2 x sin nx dx = x + x cos nx dx n 0 n 0 0 2 cos n 2 sin nx 1 =  + 0 + x  sin nx dx n n n 0 n
0 2 0 cos n 2 1 cos nx + 0  n n n n 2 2 cos n + 3 (cos n  cos 0) , =  n n =  where n = 1, 2, 3, . . . . Since cos n = 1 if n is even and cos n = 1 if n is odd for all n = 1, 2, 3, . . ., we see that 2 (1)n 2[(1)n  1] . bn =  + 2 n n3 Therefore, for n = 1, 2, 3, . . ., we have bn = 2(1)n+1 4[(1)n  1] . + n n3 Substituting these coefficients into the Fourier sine series for f (x) = x2 , yields n=1 2(1)n+1 4[(1)n  1] + n n3 sin nx . Since f (x) = x2 and f (x) = 2x are piecewise continuous on the interval [0, ], Theorem 2 on page 600 of the text implies that this Fourier series converges pointwise to f (x) on the 641 Chapter 10
interval (0, ). Hence, we can write f (x) = x =
n=1 2 2(1)n+1 4[(1)n  1] + n n3 sin nx , for x in the interval (0, ). But since the odd 2periodic extension of f (x) is discontinuous at odd multiples of , the Gibbs' phenomenon (see Problem 39 on page 606 of the text) occurs around these points, and so the convergence of this Fourier sine series is not uniform on (0, ). 13. Since f (x) = ex is piecewise continuous on the interval [0, 1], we can use Definition 2 on page 609 of the text to find its Fourier cosine series. Therefore, we have a0 + an cos nx, 2 n=1 1 where an = 2
0 ex cos nx dx . Using the fact that cos 0 = 1, we find the coefficient a0 to be
1 a0 = 2
0 ex dx = 2(e  1). We will use integration by parts twice (or the table of integrals on the inside cover of the text) to calculate the integrals in the remaining coefficients. This yields ex cos nx dx = ex (cos nx + n sin nx) , 1 + n2 2 where n = 1, 2, 3, . . .. Thus, the remaining coefficients are given by
1 an = 2
0 ex cos nx dx = = 2ex (cos nx + n sin nx) 1 + n2 2 2 0 2e(cos n) 2e(1) 2 [(1)n e  1]  = , 1 + n2 2 1 + n2 2 1 + n2 2 n = 1, 2, 3, . . . , where we have used the fact that cos n = 1 if n is even and cos n = 1 if n is odd. By substituting the above coefficients into the Fourier cosine series for f given above, we obtain e =e1+2
x n=1 (1)n e  1 cos nx , 1 + n2 2 642 Exercises 10.4
for 0 < x < 1. Note that we can say that ex for 0 < x < 1 equals its Fourier cosine series because this series converges uniformly. To see this, first notice that the even 2periodic extension of f (x) = ex , 0 < x < 1, is given by fe (x) = ex , 1 < x < 0, ex , 0 < x < 1, with fe (x + 2) = fe (x). Since this extension is continuous on (, ) and fe (x) is piecewise continuous on [1, 1], Theorem 3 on page 601 of the text states that its Fourier series (which is the one we found above) converges uniformly to fe (x) on [1, 1] and so it converges uniformly to f (x) = ex on (0, 1). 17. This problem is the same as the heat flow problem on page 580 of the text with = 5, L = and f (x) = 1  cos 2x. Therefore, the formal solution to this problem is given in equations (11) and (12) on pages 582 and 583 of the text. Thus, the formal solution is u(x, t) =
n=1 cn e5n t sin nx 2 0 < x < , t > 0, (10.6) where f (x) = 1  cos 2x = cn sin nx .
n=1 Therefore, we must find the Fourier sine series for 1  cos 2x. To do this, we can use equations (6) and (7) of Definition 2 on page 609 of the text. Hence, the coefficients are given by cn 2 = 2 = (1  cos 2x) sin nx dx
0 0 2 sin nx dx  cos 2x sin nx dx ,
0 n = 1, 2, 3, . . . . Calculating the first integral above yields 2 sin nx dx = 
0 2 2 (cos n  1) = [1  (1)n ] , n n 643 Chapter 10
where we have used the fact that cos n = 1 if n is even and cos n = 1 if n is odd. To calculate the second integral, we use the fact that 2 cos sin = sin(  ) + sin( + ), to obtain  2 0 1 cos 2x sin nx dx =  sin[(n  2)x] dx +
0 0 sin[(n + 2)x] dx 1 1 {cos[(n  2)]  1} + {cos[(n + 2)]  1} = (n  2) (n + 2) 1 1 = [(1)n  1] + [(1)n  1] . (n  2) (n + 2) Combining these two integrals yields cn = = 2 1 1 [1  (1)n ] + [(1)n  1] + [(1)n  1] n (n  2) (n + 2) 0, if n is even, 4/(n)  2/[(n  2)]  2/[(n + 2)], if n is odd, for n = 1, 2, 3, . . .. Hence, we obtain the formal solution to this problem by substituting these coefficients into equation (10.6) above and setting n = 2k  1. Therefore, we have u(x, t) = 2 k=1 2 1 1 2   e5(2k1) t sin(2k  1)x . 2k  1 2k  3 2k + 1 EXERCISES 10.5: The Heat Equation, page 624 3. If we let = 3, L = , and f (x) = x, we see that this problem has the same form as the problem in Example 1 on page 613 of the text. Therefore, we can find the formal solution to this problem by substituting these values into equation (14) on page 615 of the text. Hence, we have u(x, t) =
n=0 cn cos e 3n2 t cos nx , where f (x) =
n=0 cn cos nx . (10.7) Thus, we must find the Fourier cosine series coefficients for f (x) = x, 0 < x < . (Note that the even 2extension for f (x) = x, 0 < x < , which is given by fe (x) = 644 x, for  < x < 0, x, for 0 < x < , Exercises 10.5
with fe (x + 2) = fe (x), is continuous. Also note that its derivative is piecewise continuous on [.]. Therefore, the Fourier series for this extension converges uniformly to fe . This means that the equality sign in the second equation given in formula (10.7) above is justified for 0 < x < .) To find the required Fourier series coefficients, we use equations (4) and (5) given in Definition 2 on page 609 of the text. Hence, we have a0 x= + an cos nx , 2 n=1 (so that c0 = a0 /2 and cn = an for n = 1, 2, 3, . . .) where 2 a0 = 0 2 x2 x dx = 2 =
0 2 and an = x cos nx dx,
0 for n = 1, 2, 3, . . .. To calculate the second integral above we use integration by parts to obtain 2 2 x 1 x cos nx dx = sin nx  sin nx dx an = n n 0
0 0 = 2 1 cos nx 0  n n 0 = 2 2 (cos n  1) = [(1)n  1] . 2 n n2 Combining these results yields , if n = 0, an = 4/(n2 ), if n is odd, 0, if n is even and n = 0, where n = 0, 1, 2, . . .. The formal solution for this problem is, therefore, found by substituting these coefficients into the first equation given in formula (10.7) above. (Recall that c0 = a0 /2 and cn = an for n = 1, 2, 3, . . ..) Thus, we have 0 4 2 u(x, t) = e cos 0  e3(2k+1) t cos(2k + 1)x 2 2 (2k + 1) k=0 4 2  e3(2k+1) t cos(2k + 1)x = 2 2 k=0 (2k + 1) 645 Chapter 10
7. This problem has nonhomogeneous boundary conditions and so has the same form as the problem in Example 2 on page 616 of the text. By comparing these two problems, we see that for this problem = 2, L = , U1 = 5, U2 = 10, and f (x) = sin 3x  sin 5x. To solve this problem, we assume that the solution consists of a steady state solution v(x) and a transient solution w(x, t). The steady state solution is given in equation (24) on page 617 of the text and is v(x) = 5 + (10  5)x 5 = 5+ x. The formal transient solution is given by equations (39) and (40) on page 619 of the text. By using these equations and making appropriate substitutions, we obtain w(x, t) =
n=1 cn e2n t sin nx , 2 (10.8) where the coefficients (the cn 's) are given by f (x)  v(x) = sin 3x  sin 5x  5  5 x= cn sin nx ,
n=1 0 < x < . Therefore, we must find the Fourier sine series coefficients for the function f (x)  v(x) for 0 < x < . Since the function f (x) = sin 3x  sin 5x is already in the form of a sine series, we only need to find the Fourier sine series for v(x) = 5  5x/ and then add sin 3x  sin 5x to this series. The resulting coefficients are the ones that we need. (Note that the Fourier sine series for 5  5x/ will converge pointwise but not uniformly to 5  5x/ for 0 < x < .) To find the desired Fourier series we use equations (6) and (7) in Definition 2 on page 609 of the text. Thus, with the appropriate substitutions, we have 5x 5  = bn sin nx ,
n=1 where 2 bn = 5 
0 5x sin nx dx . To find the bn 's, we will use integration by parts to obtain bn 10 =  0 10 sin nx dx  2 x sin nx dx
0 646 Exercises 10.5 = 10 x 10 (cos n  1)  2  cos nx n n 0 cos nx dx + 1 n 0 10 10 (cos n  1)  2  cos n + 0 = n n 10 10 (2 cos n  1) = [2(1)n  1] , = n n n = 1, 2, 3, . . . . Thus, the Fourier sine series for sin 3x  sin 5x  5  5x/ is given by 10 5x = sin 3x  sin 5x + [2(1)n  1] sin nx sin 3x  sin 5x  5  n n=1 = sin 3x  sin 5x  10 30 10 30 sin x + sin 2x  sin 3x + sin 4x 2 3 4 30 10  sin 5x + [2(1)n  1] sin nx 5 n n=6 =  5 10 30 5 sin x + sin 2x + 1  sin 4x sin 3x + 2 6  1+ sin 5x +
n=6 10 [2(1)n  1] sin nx . n We therefore obtain the formal transient solution by taking the coefficients from this Fourier series and substituting them in for the cn coefficients in equation (10.8) above. Thus, we find w(x, t) =  5 10 2(3)2 t 5 2(4)2 t 30 2(1)2 t 2 sin x + e2(2) t sin 2x + 1  sin 3x + sin 4x e e e 2 6  1+ e
2(5)2 t sin 5x +
n=6 10 2 [2(1)n  1] e2n t sin nx , n and so the formal solution to the original problem is given by u(x, t) = v(x) + w(x, t) 5 10 5x 30 2t  e sin x + e8t sin 2x + 1  = 5+ 6  1+ e
50t e18t sin 3x + 5 32t e sin 4x 2 sin 5x +
n=6 10 2 [2(1)n  1] e2n t sin nx . n 647 Chapter 10
9. Notice that this problem is a nonhomogeneous partial differential equation and has the same form as the problem given in Example 3 on page 618 of the text. By comparing these problems, we see that here = 1,P (x) = ex , L = , U1 = U2 = 0, and f (x) = sin 2x. As in Example 3, we will assume that the solution is the sum of a steady state solution v(x) and a transient solution w(x, t). The steady state solution is the solution to the boundary value problem v (x) = ex , 0 < x < , v(0) = v() = 0. Thus the steady state solution can be found either by solving this ODE or by substituting the appropriate values into equation (35) given on page 618 of the text. By either method we find e  1 v(x) = x  ex + 1. The formal transient solution is then given by equations (39) and (40) on page 619 of the text. By making the appropriate substitutions into this equation, we obtain w(x, t) =
n=1 cn en t sin nx , 2 (10.9) where the cn 's are given by f (x)  v(x) = sin 2x  e  1 x + ex  1 = cn sin nx .
n=1 Hence, the problem is to find the Fourier sine coefficients for f (x)  v(x). The first term, f (x) = sin 2x, is already in the desired form. Therefore, the Fourier sine series for f (x)  v(x) is sin 2x +
n=1 bn sin nx = b1 sin x + (b2 + 1) sin 2x +
n=3 bn sin nx , where the bn 's are the Fourier sine coefficients for v(x). This implies that if n = 2, then cn = bn and if n = 2, then cn = bn + 1. The bn coefficients are given by equation (7) on page 609 of the text. Thus, we have bn 2 = 
0 e  1 x + ex  1 sin nx dx 648 Exercises 10.5
2 = e  1  0 2 x sin nx dx + e
0 x 2 sin nx dx  sin nx dx .
0 We will calculate each integral separately. The first integral is found by using integration by parts. This yields 2  e
 1 x sin nx dx =
0 2(e 2(e   1) x  cos nx n  1) 0 cos nx dx 2
 + 1 n 0 = 2 2(e  1)  cos n + 0 + 0 = (1)n . n n To find the second integral we use the table of integrals on the inside front cover of the text (or use integration by parts twice) to obtain 2 ex sin nx dx =
0 2 e n cos n + n 2n e (1)n+1 + 1 . = 2 1+n (1 + n2 ) The last integral is found to be 2  sin nx dx =
0 2 2 [cos n  1] = [(1)n  1] . n n By combining all of these results, we find that the Fourier coefficients for v(x) are given by bn = 2(e  1) 2n 2 (1)n + e (1)n+1 + 1 + [(1)n  1] . n (1 + n2 ) n Therefore, the coefficients for the formal transient solution are  2n 2 2(e  1) (1)n + e (1)n+1 + 1 + [(1)n  1] , if n = 2, 2 ) n (1 + n n cn =  e 1 4 + 1  e + 1, if n = 2. 5 Since the formal solution to the PDE given in this problem is the sum of its steady state solution and its transient solution, we find this final solution to be e  1 2 x  ex + 1 + u(x, t) = v(x) + w(x, t) = cn en t sin nx , n=1 where the cn 's are given above. 649 Chapter 10
11. Let u(x, t) = X(x)T (t). Substituting u(x, t) = X(x)T (t) into the PDE yields T (t)X(x) = 4X (x)T (t) X (x) T (t) = = K, 4T (t) X(x) where K is a constant. Substituting the solution u(x, t) = X(x)T (t) into the boundary conditions, we obtain X (0)T (t) = 0, X()T (t) = 0, t > 0. Thus, we assume that X (0) = 0 and X() = 0 since this allows the expressions above to be true for all t > 0 without implying that u(x, t) 0. Therefore, we have the two ODE's X (x) = KX(x), X (0) = X() = 0, and T (t) = 4KT (t), t > 0. (10.11) 0 < x < , (10.10) To solve boundary value problem (10.10), we will examine three cases. Case 1: Assume K = 0. Now equation (10.10) becomes X = 0. The solution is X(x) = ax+b, where a and b are arbitrary constants. To find these constants we use the boundary conditions in (10.10). Thus, we have X (0) = a = 0 and so X() = b = 0 b = 0. a=0 X(x) = b, Therefore, in this case we have only the trivial solution. Case 2: Assume K > 0. In this case the auxiliary equation for equation (10.10) is r 2  K = 0. The roots to this equation are r = K. Thus, the solution is X(x) = C1 e 650 Kx + C2 e Kx , Exercises 10.5
where C1 and C2 are arbitrary constants. To find these constants we again use the boundary conditions in (10.10). We first note that X (x) = C1 Ke Kx  C2 Ke Kx . Therefore, X (0) = C1 K  C2 K = 0 C1 = C2 X(x) = C1 e Kx + e Kx . The other boundary condition implies that X() = C1 e K + e K =0 C1 e2 K + 1 = 0. The only way that the final equation above can be zero is for C1 to be zero. Therefore, we again obtain only the trivial solution. Case 3: Assume K < 0, so K > 0. Then the auxiliary equation for equation (10.10) has the roots r = K = i K. Therefore, the solution is X(x) = C1 sin Kx + C2 cos Kx Kx  C2 K sin Kx . X (x) = C1 K cos Using the boundary condition X (0) = 0, we obtain 0 = X (0) = C1 K cos 0  C2 K sin 0 = C1 K Hence, X(x) = C2 cos C1 = 0. Kx . Applying the other boundary condition yields K 2 K= (2n + 1)2 , 4 n = 0, 1, 2, . . . . 0 = X() = C2 cos K = (2n + 1) Therefore, nontrivial solutions to problem (10.10) above are given by Xn (x) = cn cos 2n + 1 x , 2 n = 0, 1, 2, . . . . 651 Chapter 10
By substituting the values of K into equation (10.11), we obtain T (t) = (2n + 1)2 T (t), This is a separable differential equation, and we find dT = (2n + 1)2 dt T ln T  = (2n + 1)2 t + A dT = (2n + 1)2 dt T 2 Tn (t) = bn e(2n+1) t , t > 0. n = 0, 1, 2, . . . , (where bn = eA ). Hence, by the superposition principle (and since un (x, t) = Xn (x)Tn (t)), we see that the formal solution to the original PDE is u(x, t) =
n=0 bn e (2n+1)2 t 2n + 1 x cn cos 2 =
n=0 an e(2n+1) t cos 2 n+ 1 x , 2 (10.12) where an = bn cn . To find the an 's, we use the initial condition to obtain u(x, 0) = f (x) =
n=0 an cos n+ 1 2 x . (10.13) Therefore, the formal solution to this PDE is given by equation (10.12), where the an 's are given by equation (10.13). 17. This problem is similar to the problem given in Example 4 on page 619 of the text with = 1, L = W = , and f (x, y) = y. The formal solution to this problem is given in equation (52) on page 621 of the text with its coefficients given on pages 621 and 622 in equations (54) and (55). By making appropriate substitutions in the first of these equations, we see that the formal solution to this problem is u(x, y, t) =
m=0 n=1 amn e(m 2 +n2 )t cos mx sin ny . (10.14) We can find the coefficients, a0 n, n = 1, 2, 3, . . ., by using equation (54) on page 621 of the text with the appropriate substitutions. This yields 0 dx dy a0n = 2 2 y sin ny dx dy =
0 0 2 2 y sin ny 0 652 Exercises 10.6 2 = = = y sin ny dy
0 0 (use integration by parts) cos ny dy 2 y  cos ny n + 1 n 0 2  cos n + n 1 sin ny n2 =
0 2 2  cos n = (1)n+1 . n n We will use equation (55) on page 622 of the text to find the other coefficients. Thus for m 1 and n 1, we have amn 4 = 2 4 = 2 y cos mx sin ny dx dy
0 0 0 4 cos mx dx dy = 2 0 y sin ny 0 y sin ny(0) dy = 0. The formal solution to this problem is found by substituting these coefficients into equation (10.14). To do this we first note that the coefficients for any terms containing m = 0 are zero. Hence, only terms containing m = 0 will appears in the summation. Therefore, the formal solution is given by u(x, y, t) =
n=1 2 (1)n+1 n2 t 2 (1)n+1 en t sin ny = 2 e sin ny . n n n=1 EXERCISES 10.6: The Wave Equation, page 636 1. This problem has the form of the problem given in equations (1)(4) on page 625 of the text. Here, however, = 1, L = 1, f (x) = x(1  x), and g(x) = sin 7x. This problem is consistent because f (0) = 0 = f (1), and g(0) = sin 0 = 0 = sin 7 = g(1). The solution to this problem was derived in Section 10.2 of the text and given again in equation (5) on page 625 of the text. Making appropriate substitutions in equation (5) yields a formal 653 Chapter 10
solution given by u(x, t) =
n=1 [an cos nt + bn sin nt] sin nx . (10.15) To find the an 's we note that they are the Fourier sine coefficients for x(1  x) and so are given by equation (7) on page 609 of the text. Thus, for n = 1, 2, 3, . . ., we have 1 1 1 an = 2
0 x(1  x) sin nx dx = 2 0 x sin nx dx 
0 x2 sin nx dx . We will use integration by parts to calculate these two integrals. This yields
1 x sin nx dx = 
0 1 1 cos n =  (1)n n n and
1 x2 sin nx dx = 
0 1 2 1 2 1 1 cos n  2 2  cos n + =  (1)n + 3 3 [(1)n  1]. n n n n n n Therefore, for n = 1, 2, 3, . . ., we see that an = 2  1 1 2 (1)n + (1)n  3 3 [(1)n  1] n n n = 4 [(1)n  1] . n3 3 This can also be expressed as an = 0,
3 3 if n is even, 8/(n ), if n is odd. The bn 's were found in equation (7) on page 626. By making appropriate substitutions in this equations we have sin 7x =
n=1 nbn sin nx. From this we see that for n = 7 7b7 = 1 654 b7 = 1 , 7 Exercises 10.6
and for all other n's, bn = 0. By substituting these coefficients into the formal solution given in equation (10.15) above, we obtain 8 1 sin 7t sin 7x + u(x, t) = cos[(2k + 1)t] sin[(2k + 1)x]. 7 [(2k + 1)]3 k=0 5. First we note that this problem is consistent because g(0) = 0 = g(L) and f (0) = 0 = f (L). The formal solution to this problem is given in equation (5) on page 625 of the text with the coefficients given in equations (6) and (7) on page 626. By equation (7), we see that g(x) = 0 =
n=1 bn nx n sin . L L Thus, each term in this infinite series must be zero and so bn = 0 for all n's. Therefore, the formal solution given in equation (5) on page 625 of the text becomes u(x, t) =
n=1 an cos nt L sin nx . L (10.16) To find the an 's we note that by equation (6) on page 626 of the text these coefficients are the Fourier sine coefficients for f (x). Therefore, by using equation (7) on page 609 of the text, for n = 1, 2, 3, . . . we have nx Lx sin dx La L 0 0 a a L L L 1 2h0 1 nx nx nx dx + dx  dx . = x sin sin x sin L a L La L La L 2 L f (x) sin nx 2 h0 dx = L L a x sin nx dx + h0 L
0 a a L a L an = By using integration by parts, we find x sin nx nx nx xL L2 cos dx =  + 2 2 sin . L n L n L 655 Chapter 10
Therefore, for n = 1, 2, 3, . . ., the coefficients become 2h0 an = L na 1 aL na L2 L2 na  cos + 2 2 sin  cos n  cos a n L n L n(L  a) L 2 2 1 na L aL na L  + 2 2 sin n  sin  cos n + cos La n n L n L na 2h0 L2 sin , 2 2 a(L  a) n L . After simplifying, this becomes an = n = 1, 2, 3, . . . . By substituting this result into equation (10.16) above, we obtain the formal solution to this problem given by u(x, t) = 2h0 L2 2 a(L  a) n=1 na 1 sin cos 2 n L nt L sin nx . L 7. If we let = 1, h(x, t) = tx, L = , f (x) = sin x, and g(x) = 5 sin 2x  3 sin 5x, then we see that this problem has the same form as the problem given in Example 1 on page 627 of the text. The formal solution to the problem in Example 1 is given in equation (16) on page 628 of the text. Therefore, with the appropriate substitutions, the formal solution to this problem is an cos nt + bn sin nt + t u(x, t) =
n=1 1 n hn (s) sin[n(t  s)] ds
0 sin nx . (10.17) The an 's are shown in equation (14) on page 628 of the text to satisfy sin x =
n=1 an sin nx . Thus, the only nonzero term in this infinite series is the term for n = 1. Therefore, we see that a1 = 1 and an = 0 for n = 1. The bn 's are given in equation (15) on page 628 of the text and so must satisfy 5 sin 2x  3 sin 5x =
n=1 nbn sin nx , which implies that 2b2 = 5 656 b2 = 5 2 and 5b5 = 3 3 b5 =  , 5 Exercises 10.6
and bn = 0 for all other values of n. To calculate the integral given in the formal solution we must first find the functions hn (t). To do this, we note that in Example 1, the functions hn (t), n = 1, 2, . . ., are the Fourier sine coefficients for h(x, t) = tx with t fixed. These functions are given below equation (13) on page 628 of the text. (We will assume proper convergence of this series.) Thus, we have 2 hn (t) = = 0 2t tx sin nx dx = x sin nx dx
0 2t 2t 1 2t  cos n + 0 + 2 sin n  sin 0 =  cos n = (1)n+1 , n n n = 1, 2, 3, . . ., where we have used integration by parts to calculate this integral. Substituting this result into the integral in equation (10.17) above yields
t t hn (s) sin[n(t  s)] ds =
0 0 2s (1)n+1 sin[n(t  s)] ds = sin nt 2(1)n+1 2(1)n+1 t  (nt  sin nt) , = n n n2 n3 where n = 1, 2, 3, . . .. By plugging the an 's, the bn 's, and the result we just found into equation (10.17), we obtain the formal solution to this problem given by 3 5 1 2(1)n+1 u(x, t) = cos t sin x + sin 2t sin 2x  sin 5t sin 5x + (nt  sin nt) sin nx 2 5 n n3 n=1 = cos t sin x + 3 5 (1)n+1 sin 2t sin 2x  sin 5t sin 5x + 2 2 5 n3 n=1 t sin nt n sin nx . 11. We will assume that a solution to this problem has the form u(x, t) = X(x)T (t). Substituting this expression into the partial differential equations yields X(x)T (t) + X(x)T (t) + X(x)T (t) = 2 X (x)T (t). Dividing this equation by 2 X(x)T (t) yields X (x) T (t) + T (t) + T (t) = . 2 T (t) X(x) 657 Chapter 10
Since these two expressions must be equal for all x in (0, L) and all t > 0, they can not vary. Therefore, they must both equal a constant, say K. This gives us the two ordinary differential equations T (t) + T (t) + T (t) =K 2 T (t) and X (x) =K X (x)  KX(x) = 0. (10.19) X(x) Substituting u(x, t) = X(x)T (t) into the boundary conditions, u(0, t) = u(L, t) = 0, t > 0, we obtain X(0)T (t) = 0 = X(L)T (t), t > 0. T (t) + T (t) + 1  2 K T (t) = 0 (10.18) Since we are seeking a nontrivial solution to the partial differential equation, we do not want T (t) 0. Therefore, for the above equation to be zero, we must have X(0) = X(L) = 0. Combining this fact with equation (10.19) above yields the boundary value problem given by X (x)  KX(x) = 0, with X(0) = X(L) = 0. This problem was solved in Section 10.2 of the text. There we found that for K = (n/L)2 , n = 1, 2, 3, . . ., we obtain nonzero solutions of the form Xn (x) = An sin nx , L n = 1, 2, 3, . . . . (10.20) Plugging these values of K into equation (10.18) above yields the family of linear ordinary differential equations with constant coefficients given by T (t) + T (t) + 1 + 2 n2 2 L2 T (t) = 0, n = 1, 2, 3, . . . . (10.21) The auxiliary equations associated with these ODE's are r2 + r + 1 + 2 n2 2 L2 = 0. By using the quadratic formula, we obtain the roots to these auxiliary equations. Thus, we have 1 r= 658 14 1+ 2 2 n2 2 L2 1 =  2 L2  4L2  42n2 2 2L Exercises 10.6
1 =  2 3L2 + 42 n2 2 i, 2L n = 1, 2, 3, . . . . Hence, the solutions to the linear equations given in equation (10.21) above are Tn (t) = et/2 Bn cos for n = 1, 2, 3, . . .. By letting n = 3L2 + 42 n2 2 t 2L + Cn sin 3L2 + 42 n2 2 t 2L , 3L2 + 42 n2 2 , 2L (10.22) for n = 1, 2, 3, . . ., this family of solutions can be more easily written as Tn (t) = et/2 [Bn cos n t + Cn sin n t] . Substituting the solutions we have just found and the solutions given in equation (10.20) above into u(x, t) = X(x)T (t), yields solutions to the original partial differential equation given by un (x, t) = Xn (x)Tn (t) = An et/2 [Bn cos n t + Cn sin n t] sin nx , L n = 1, 2, 3, . . . . By the superposition principle, we see that solutions to the PDE will have the form u(x, t) =
n=1 et/2 [an cos n t + bn sin n t] sin nx , L where n is given in equation (10.22) above, an = An Bn , and bn = An Cn . To find the coefficients an and bn , we use the initial conditions u(x, 0) = f (x) and u(x, 0)/t = 0. Therefore, since u(x, t) = t (1/2) et/2 [an cos n t + bn sin n t] +et/2 [an n sin n t + bn n cos n t] sin nx , L n=1 we have u(x, 0) =0= t 
n=1 an nx + bn n sin . 2 L 659 Chapter 10
Hence, each term in this infinite series must be zero which implies that  an + bn n = 0 2 bn = an , 2n n = 1, 2, 3, . . . . Thus, we can write u(x, t) =
n=1 an et/2 cos n t + 1 nx sin n t sin , 2n L (10.23) where n is given above in equation (10.22). To find the an 's, we use the remaining initial condition to obtain u(x, 0) = f (x) =
n=1 an sin nx . L Therefore, the an 's are the Fourier sine coefficients of f (x) and so satisfy
L 2 an = L f (x) sin
0 nx dx . L (10.24) Combining all of these results, we see that a formal solution to the telegraph problem is given by equation (10.23) where n and an are given in equation (10.22) and (10.24), respectively. 15. This problem has the form of the problem solved in Example 2 on page 631 of the text with f (x) = g(x) = x. There it was found that d'Alembert's formula given in equation (32) on page 631 of the text is a solution to this problem. By making the appropriate substitutions in this equation (and noting that f (x + t) = x + t and f (x  t) = x  t), we obtain the solution
x+t 1 1 u(x, t) = [(x + t) + (x  t)] + 2 2
xt s ds = x + 1 2 s2 2 x+t xt = x+ EXERCISES 10.7: 1 1 (x + t)2  (x  t)2 = x + [4tx] = x + tx . 4 4 Laplace's Equation, page 649 3. To solve this problem using separation of variables, we will assume that a solution has the form u(x, y) = X(x)Y (y). Making this substitution into the partial differential equation yields X (x)Y (y) + X(x)Y (y) = 0. 660 Exercises 10.7
By dividing the above equation by X(x)Y (y), we obtain X (x) Y (y) + = 0. X(x) Y (y) Since this equation must be true for 0 < x < and 0 < y < , there must be a constant K such that Y (y) X (x) = = K, 0 < x < , and 0 < y < . X(x) Y (y) This leads to the two ordinary differential equations given by X (x)  KX(x) = 0 and Y (y) + KY (y) = 0 (10.26) (10.25) By making the substitution u(x, y) = X(x)Y (y) into the first boundary conditions, that is, u(0, y) = u(, y) = 0, we obtain X(0)Y (y) = X()Y (y) = 0. Since we do not want the trivial solution which would be obtained if we let Y (y) 0, these boundary conditions imply that X(0) = X() = 0. Combining these boundary conditions with equation (10.25) above yields the boundary value problem X (x)  KX(x) = 0, with X(0) = X() = 0. To solve this problem, we will consider three cases. Case 1: K = 0. For this case, the differential equation becomes X (x) = 0, which has solutions X(x) = A + Bx. By applying the first of the boundary conditions, we obtain X(0) = A = 0 The second boundary condition yields X() = B = 0 B = 0. 661 X(x) = Bx. Chapter 10
Thus, in this case we obtain only the trivial solution. Case 2: K > 0. In this case, the auxiliary equation associated with this differential equation is r 2  K = 0, which has the real roots r = K. Thus, solutions to this problem are given by X(x) = Ae Applying the boundary conditions yields X(0) = A + B = 0 and X() = Be K Kx + Be Kx . A = B X(x) = Be Kx + Be Kx + Be K =0 B e2 K  1 = 0. This last expression is true only if K = 0 or if B = 0. Since we are assuming that K > 0, we must have B = 0 which means that A = B = 0. Therefore, in this case we again find only the trivial solution. Case 3: K < 0. The auxiliary equation associated with the differential equation in this case has the complex valued roots r = Ki, (where K > 0). Therefore, solutions to the ODE for this case are given by X(x) = A cos Kx + B sin Kx . By applying the boundary conditions, we obtain X(0) = A = 0 and X() = B sin K = 0 K = n K = n2 , n = 1, 2, 3, . . . , X(x) = B sin Kx where we have assumed that B = 0 since this would lead to the trivial solution. Therefore, nontrivial solutions Xn (x) = Bn sin nx are obtained when K = n2 , n = 1, 2, 3 . . . . 662 Exercises 10.7
To solve the differential equation given in equation (10.26) above, we use these values for K. This yields the family of linear ordinary differential equations given by Y (y)  n2 Y (y) = 0, n = 1, 2, 3, . . . . The auxiliary equations associated with these ODE's are r 2  n2 = 0, which have the real roots r = n, n = 1, 2, 3, . . .. Hence, the solutions to this family of differential equations are given by Yn (y) = Cn eny + Dn eny , n = 1, 2, 3, . . . . With the substitutions K1n = Cn + Dn and K2n = Cn  Dn , so that Cn = K1n + K2n , 2 and Dn = K1n  K2n , 2 we see that these solutions can be written as Yn (y) = K1n + K2n ny K1n  K2n ny e + e 2 2 eny + eny eny  eny = K1n + K2n = K1n cosh ny + K2n sinh ny . 2 2 This last expression can in turn be written as Yn (y) = An sinh (ny + En ) ,
2 2 where An = K2n  K1n and En = tanh1 (K1n /K2n ). (See Problem 18.) The last boundary condition u(x, ) = X(x)Y () = 0 implies that Y () = 0(since we do not want the trivial solution). Therefore, by substituting into the solutions just found, we obtain Yn () = An sinh (n + En ) . Since we do not want An = 0, this implies that sinh (n + En ) = 0. This will be true only if n + En = 0 or in other words if En = n. Substituting these expressions for En into the family of solutions we found for Y (y), yields Yn (y) = An sinh (ny  n) . 663 Chapter 10
Therefore, substituting the solutions just found for X(x) and Y (y) into un (x, y) = Xn (x)Yn (y) we see that un (x, y) = an sin nx sinh(ny  n), where an = An Bn . By the superposition principle, a formal solution to the original partial differential equation is given by u(x, y) =
n=1 un (x, y) =
n=1 an sin nx sinh(ny  n). (10.27) In order to find an expression for the coefficients an , we will apply the remaining boundary condition, u(x, 0) = f (x). From this condition, we obtain u(x, 0) = f (x) =
n=1 an sin nx sinh(n), which implies that an sinh(n) are the coefficients of the Fourier sine series of f (x). Therefore, by equation (7) on page 609 of the text, we see that (with T = ) 2 an sinh(n) = f (x) sin nx dx
0 2 an = sinh(n) f (x) sin nx dx .
0 Thus, a formal solution to this ODE is given in equation (10.27) with the an 's given by the equation above. 5. This problem has two nonhomogeneous boundary conditions, and, therefore, we will solve two PDE problems, one for each of these boundary conditions. These problems are 2u 2u + = 0, 0 < x < , 0 < y < 1; x2 y 2 u(, y) u(0, y) = = 0, 0 y 1, x x u(x, 0) = cos x  cos 3x, u(x, 1) = 0, 0 x , and 2u 2u + = 0, x2 y 2 664 0 < x < , 0 < y < 1; Exercises 10.7
u(, y) u(0, y) = = 0, 0 y 1, x x u(x, 0) = 0, u(x, 1) = cos 2x, 0 x . If u1 and u2 are solutions to the first and second problems, respectively, then u = u1 + u2 will be a solution to the original problem. To see this notice that 2 u1 2 u2 2 u1 2 u2 + + + x2 x2 y 2 y 2 2 u1 2 u1 2 u2 2 u2 + + = + = 0 + 0 = 0, x2 y 2 x2 y 2 u1 (0, y) u2 (0, y) u(0, y) = + = 0 + 0 = 0, x x x u(, y) u1 (, y) u2 (, y) = + = 0 + 0 = 0, x x x u(x, 0) = u1 (x, 0) + u2 (x, 0) = cos x  cos 3x + 0 = cos x  cos 3x , u(x, 1) = u1 (x, 1) + u2 (x, 1) = 0 + cos 2x = cos 2x . This is an application of the superposition principle. The first of these two problems has the form of the problem given in Example 1 on page 639 of the text with a = , b = 1, and f (x) = cos x  cos 3x. A formal solution to this problem is given in equation (10) on page 641 of the text. Thus, by making the appropriate substitutions, we find that a formal solution to the first problem is 2u 2u + = x2 y 2 u1 (x, y) = E0 (y  1) +
n=1 En cos nx sinh(ny  n). To find the coefficients En , we use the nonhomogeneous boundary condition u(x, 0) = cos x  cos 3x . Thus, we have u1 (x, 0) = cos x  cos 3x = E0 +
n=1 En cos nx sinh(n). 665 Chapter 10
From this we see that for n = 1, E1 sinh(1) = 1 and for n = 3, E3 sinh(3) = 1 E1 = E1 = 1 sinh(1) 1 . sinh(3) For all other values of n, En = 0. By substituting these values into the expression found above for u1 , we obtain the formal solution to the first of our two problems given by u1 (x, y) = cos x sinh(y  1) cos 3x sinh(3y  3)  . sinh(1) sinh(3) (10.28) To solve the second of our problems, we note that, except for the last two boundary conditions, it is similar to the problem solved in Example 1 on page 639 of the text. As in that example, using the separation of variables technique, we find that the ODE X (x)  KX(x) = 0, X (0) = X () = 0, has solutions Xn (x) = an cos nx, when K = n2 , n = 1, 2, 3, . . . . By substituting these values for K into the ODE Y (y) + KY (y) = 0, we again find that a family solutions to this differential equation is given by Y0 (y) = A0 + B0 y, Yn (y) = Cn sinh [n (y + Dn )] , n = 1, 2, 3 . . . . At this point, the problem we are solving differs from the example. The boundary condition u(x, 0) = X(x)Y (0) = 0, 0 x , implies that Y (0) = 0 (since we don't want the trivial solution). Therefore, applying this boundary condition to each of the solutions found above yields Y0 (0) = A0 + 0 = 0 A0 = 0, Dn = 0, Yn (0) = Cn sinh (nDn ) = 0 666 Exercises 10.7
where we have used the fact that sinh x = 0 only when x = 0. By substituting these results into the solutions found above, we obtain Y0 (y) = B0 y, Yn (y) = Cn sinh ny, n = 1, 2, 3 . . . . Combining these solutions with the solutions Xn (x) = an cos nx yields u2,0(x, y) = X0 (x)Y0 (y) = a0 B0 y cos 0 = E0 y , u2,n (x, y) = Xn (x)Yn (y) = an Cn cos nx sinh ny = En cos nx sinh ny , where E0 = a0 B0 and En = an Cn . Thus, by the superposition principle, we find that a formal solution to the second problem is given by u2 (x, y) = E0 y +
n=1 En cos nx sinh ny . By applying the last boundary condition of this second problem, namely u(x, 1) = cos 2x, to these solutions, we see that u2 (x, 1) = E0 +
n=1 En cos nx sinh n = cos 2x .. Therefore, when n = 2, E2 sinh 2 = 1 E2 = 1 , sinh 2 and for all other values of n, En = 0. By substituting these coefficients into the solution u2 (x, y) that we found above, we obtain the formal solution to this second problem u2 (x, y) = cos 2x sinh 2y . sinh 2 By the superposition principle (as noted at the beginning of this problem), a formal solution to the original partial differential equation is the sum of this solution and the solution given in equation (10.28). Thus, the solution that we seek is u(x, y) = cos x sinh(y  1) cos 3x sinh(3y  3) cos 2x sinh 2y  + . sinh(1) sinh(3) sinh 2 667 Chapter 10
11. In this problem, the technique of separation of variables, as in Example 2 on page 642 of the text, leads to the two ODE's r 2 R (r) + rR (r)  R(r) = 0 and T () + T () = 0. Again, as in Example 2, we require the solution u(r, ) to be continuous on its domain. Therefore, T () must again be periodic with period 2. This implies that T () = T () and T () = T (). Thus, as in Example 2, a family of solutions for the second ODE above which satisfies these periodic boundary conditions is T0 () = B and Tn () = An cos n + Bn sin n , n = 1, 2, 3, . . . . In solving this problem, it was found that = n2 , n = 0, 1, 2 . . .. Again, as in Example 2, substituting these values for into the first ODE above leads to the solutions R0 (r) = C + D ln r and Rn (r) = Cn r n + Dn r n , n = 1, 2, 3, . . . . Here, however, we are not concerned with what happens when r = 0. By our assumption that u(r, ) = R(r)T (), we see that solutions of the PDE given in this problem will have the form u0 (r, ) = B(C + D ln r) and un (r, ) = Cn r n + Dn r n (An cos n + Bn sin n) , where n = 1, 2, 3, . . .. Thus, by the superposition principle, we see that a formal solution to this Dirichlet problem is given by u(r, ) = BC + BD ln r +
n=1 Cn r n + Dn r n (An cos n + Bn sin n) , or u(r, ) = a + b ln r +
n=1 cn r n + en r n cos n + dn r n + fn r n sin n , (10.29) where a = BC, b = BD, cn = Cn An , en = Dn An , dn = Cn Bn , and fn = Dn Bn . To find these coefficients, we apply the boundary conditions u(1, ) = sin 4  cos , and u(2, ) = sin ,  . From the first boundary condition, we see that u(1, ) = a +
n=1 [(cn + en ) cos n + (dn + fn ) sin n] = sin 4  cos , 668 Exercises 10.7
which implies that a = 0, d4 + f4 = 1, c1 + e1 = 1, and for all other values of n, cn + en = 0 and dn + fn = 0. From the second boundary condition, we have u(2, ) = a + b ln 2 +
n=1 cn 2n + en 2n cos n + dn 2n + fn 2n sin n = sin , which implies that a = 0, b = 0, 2d1 +21 f1 = 1, and for all other values of n, 2n c1 +2n e1 = 0 and 2n d1 + 2n f1 = 0. By combining these results, we obtain a = 0, b = 0, and the three systems of two equations in two unknowns given by d4 + f4 = 1, 2 d4 + 2 f4 = 0
4 4 and c1 + e1 = 1, 2c1 + 2 e1 = 0
1 and d1 + f1 = 0, 2d1 + 21 f1 = 1, (where the first equation in each system was derived from the first boundary condition and the second equation in each system was derived from the second boundary condition), and for all other values of n, cn = 0 , en = 0, dn = 0, fn = 0. Solving each system of equations simultaneously yields d4 =  1 , 255 f4 =  256 , 255 c1 = 1 , 3 4 e1 =  , 3 d1 = 2 , 3 2 f1 =  . 3 By substituting these values for the coefficients into equation (10.29) above, we find that a solution to this Dirichlet problem is given by u(r, ) = 4 1 r  r 1 cos + 3 3 2 1 4 256 4 2 sin 4 . r  r 1 sin +  r  r 3 3 255 255 15. Here, as in Example 2 on page 642 of the text, the technique of separation of variables leads to the two ODE's given by r 2 R (r) + rR (r)  R(r) = 0 and T () + T () = 0. Since we want to avoid the trivial solution, the boundary condition u(r, 0) = R(r)T (0) = 0 implies that T (0) = 0 and the boundary condition u(r, ) = R(r)T () = 0 implies that T () = 0. Therefore, we seek a nontrivial solution to the ODE T () + T () = 0, with T (0) = 0 and T () = 0. (10.30) 669 Chapter 10
To do this we will consider three cases for . Case 1: = 0. This case leads to the differential equation T () = 0, which has solutions T () = A+B. Applying the first boundary condition yields 0 = T (0) = B. Thus, T () = A. The second boundary condition implies that 0 = T () = A. Hence, A = 0. Therefore, in this case we find only the trivial solution. Case 2: < 0. In this case, the auxiliary equation associated with the linear differential equa tion given in equation (10.30) above is r 2 + = 0, which has the real roots r =  (where  > 0). Thus, the solution to this differential equation has the form T () = C1 e  + C2 e  . Applying the first boundary condition yields 0 = T (0) = C1 + C2 C1 = C2 T () = C2 e  + C2 e  . From the second boundary condition, we obtain 0 = T () = C2 e  + e  C2 e2   1 = 0. Since we are assuming that < 0, the only way that this last expression can be zero is for C2 = 0. Thus, C1 = C2 = 0 and we again obtain the trivial solution. Case 3: > 0. In this case, the roots to the auxiliary equation associated with this differential equation are r = i. Therefore, the solution to the differential equation given in equation (10.30) is T () = C1 sin From the boundary conditions, we see that 0 = T (0) = C2 and 0 = T () = C1 sin 670 T () = C1 sin , + C2 cos . . Exercises 10.7 Since we do not want the trivial solution, this last boundary condition yields sin = 0. This will be true if = n or, in other words, if = n2 , n = 1, 2, 3, . . .. With these values for , we find nontrivial solution for the differential equation given in equation (10.30) above to be Tn () = Bn sin n , n = 1, 2, 3, . . . . Substituting the values for that we have just found into the differential equation r 2 R (r) + rR (r)  R(r) = 0, yields the ODE r 2 R (r) + rR (r)  n2 R(r) = 0, n = 1, 2, 3, . . . . This is the same CauchyEuler equation that was solved in Example 2 on page 642 of the text. There it was found that the solutions have the form Rn (r) = Cn r n + Dn r n , n = 1, 2, 3, . . . . Since we require that u(r, ) to be bounded on its domain, we see that u(r, ) = R(r)T () must be bounded about r = 0. This implies that R() must be bounded. Therefore, Dn = 0 and so the solutions to this CauchyEuler equation are given by Rn (r) = Cn r n , n = 1, 2, 3, . . . . Since we have assumed that u(r, ) = R(r)T (), we see that formal solutions to the original partial differential equation are un (r, ) = Bn Cn r n sin n = cn r n sin n , where cn = Bn Cn . Therefore, by the superposition principle, we obtain the formal solutions to this Dirichlet problem u(r, ) =
n=1 cn r n sin n . The final boundary condition yields u(1, ) = sin 3 =
n=1 cn sin n . 671 Chapter 10
This implies that c3 = 1 and for all other values of n, cn = 0. Substituting these values for the coefficients into the formal solution found above, yields the solution to this Dirichlet problem on the half disk u(r, ) = r 3 sin 3 . 17. As in Example 2 on page 642 of the text, we solve this problem by separation of variables. There it was found that we must solve the two ordinary differential equations r 2 R (r) + rR (r)  R(r) = 0, and T () + T () = 0 with T () = T (n) and T () = T (n). (10.32) (10.31) In Example 2, we found that when = n2 , n = 0, 1, 2, . . ., the linear differential equation given in equation (10.32) has nontrivial solutions of the form Tn () = An cos n + Bn sin n , and equation (10.31) has solutions of the form R0 (r) = C + D ln r and Rn (r) = Cn r n + Dn r n , n = 1, 2, 3, . . . . n = 0, 1, 2, . . . , (Note that here we are not concerned with what happens to u(r, ) around r = 0.) Thus, since we are assuming that u(r, ) = R(r)T (), we see that solutions to the original partial differential equation will be given by u0 (r, ) = A0 (C + D ln r) = a0 + b0 ln r, and un (r, ) = = 672 Cn r n + Dn r n (An cos n + Bn sin n) an r n + bn r n cos n + cn r n + dn r n sin n , Exercises 10.7
where a0 = A0 C, b0 = A0 D, an = Cn An , bn = Dn An , cn = Cn Bn , and dn = Dn Bn . Thus, by the superposition principle, we see that a formal solution to the partial differential equation given in this problem will have the form u(r, ) = a0 + b0 ln r +
n=1 an r n + bn r n cos n + cn r n + dn r n sin n . (10.33) By applying the first boundary condition, we obtain u(1, ) = f () = a0 +
n=1 [(an + bn ) cos n + (cn + dn ) sin n] , where we have used the fact that ln 1 = 0. Comparing this equation with equation (8) on page 594 of the text, we see that a0 , (an + bn ), and (cn + dn ) are the Fourier coefficients of f () (with T = ). Therefore, by equations (9) and (10) on that same page we see that 1 a0 = 2 f () d ,
 1 an + bn = cn + d n = 1 f () cos n d ,
 (10.34) f () sin n d ,
 n = 1, 2, 3 . . . . To apply the last boundary condition, we must find u/r. Hence, we find u(r, ) b0 = + r r n=1 an nr n1  bn nr n1 cos n + cn nr n1  dn nr n1 sin n . Applying the last boundary condition yields b0 u(3, ) = g() = + r 3 n=1 an n3n1  bn n3n1 cos n + cn n3n1  dn n3n1 sin n . Again by comparing this to equation (8) on page 594 of the text, we see that b0 , 3 n3n1 an  n3n1 bn , and n3n1 cn  n3n1 dn 673 Chapter 10
are the Fourier coefficients of g() (with T = ). Thus, by equations (9) and (10) on that same page of the text, we see that 3 b0 = 2 g() d ,
 1 n3n1 an  n3n1 bn = n3n1 cn  n3n1 dn = 1 g() cos n d ,
 (10.35) g() sin n d ,
 n = 1, 2, 3 . . . . Therefore, the formal solution to this partial differential equation will be given by equation (10.33) with the coefficients given by equations (10.34) and (10.35). 674 CHAPTER 11: Eigenvalue Problems and SturmLiouville Equations
EXERCISES 11.2: Eigenvalues and Eigenfunctions, page 671 1. The auxiliary equation for this problem is r 2 + 2r + 26 = 0, which has roots r = 1 5i. Hence a general solution to the differential equation y + 2y + 26y = 0 is given by y(x) = C1 ex cos 5x + C2 ex sin 5x. We will now try to determine C1 and C2 so that the boundary conditions are satisfied. Setting x = 0 and x = , we find y(0) = C1 = 1, y() = C1 e = e . Both boundary conditions yield the same result, C1 = 1. Hence, there is a one parameter family of solutions, y(x) = ex cos 5x + C2 ex sin 5x., where C2 is arbitrary. 13. First note that the auxiliary equation for this problem is r 2 + = 0. To find eigenvalues which yield nontrivial solutions we will consider the three cases < 0, = 0, and > 0. Case 1. < 0: In this case the roots to the auxiliary equation are  (where we note that  is a positive number). Therefore, a general solution to the differential equation y +y = 0 is given by y(x) = C1 e x + C2 e x . By applying the first boundary condition, we obtain y(0) = C1 + C2 = 0 C2 = C1 . 675 Chapter 11
Thus y(x) = C1 e x  e x . In order to apply the second boundary conditions, we need to find y (x). Thus, we have y(x) = C1  e x + e x . Thus y(1) = C1  e  + e  = 0. (11.1) Since  > 0 and e  + e  = 0, the only way that equation (11.1) can be true is for C1 = 0. So in this case we have only the trivial solution. Thus, there are no eigenvalues for < 0. Case 2. = 0: In this case we are solving the differential equation y = 0. This equation has a general solution given by y(x) = C1 + C2 x y (x) = C2 . By applying the boundary conditions, we obtain y(0) = C1 = 0 and y (1) = C1 = 0. Thus C1 = C2 = 0, and zero is not an eigenvalue. Case 3. > 0: In this case the roots to the associated auxiliary equation are r = i. Therefore, the general solution is given by y(x) = C1 cos x + C2 sin x . By applying the first boundary condition, we obtain y(0) = C1 = 0 y(x) = C2 sin x . In order to apply the second boundary conditions we need to find y (x). Thus, we have y (x) = C2 cos x , 676 Exercises 11.2
and so y (1) = C2 cos = 0. Therefore, in order to obtain a solution other than the trivial solution, we must have cos =0 n = 1 n+ 2
2 2 , = n+ 1 2 , n = 0, 1, 2, . . . n = 0, 1, 2, . . . . For the eigenvalue n , we have the corresponding eigenfunctions, yn (x) = Cn sin n+ 1 2 x , n = 0, 1, 2, . . . , where Cn is an arbitrary nonzero constant. 19. The equation (xy ) + x1 y = 0 can be rewritten as the CauchyEuler equation x2 y + xy + y = 0, x > 0. (11.2) Substituting y = xr gives r 2 + = 0 as the auxiliary equation for (11.2). Again we will consider the three cases < 0, = 0, and > 0. Case 1. < 0: Let = 2 , for > 0. The roots of the auxiliary equation are r = and so a general solution to (11.2) is y(x) = C1 x + C2 x . We first find y (x). y (x) = C1 x1  C2 x1 = C1 x1  C2 x1 . Substituting into the first boundary condition gives y (1) = (C1  C2 ) = 0. Since > 0, C1  C2 = 0 C1 = C2 y(x) = C1 x + x . 677 Chapter 11
Substituting this into the second condition yields y (e ) = C1 e + e = 0. (11.3) Since e + e = 0 the only way that equation (11.3) can be true is for C1 = 0. So in this case we have only the trivial solution. Thus, there is no eigenvalue for < 0. Case 2. = 0: In this case we are solving the differential equation (xy ) = 0. This equation can be solved as follows: xy = C1 y = C1 x y(x) = C2 + C1 ln x . By applying the boundary conditions, we obtain y (1) = C1 = 0 and y (e ) = C2 + C1 ln (e ) = C2 + C1 = 0. Solving these equations simultaneously yields C1 = C2 = 0. Thus, we again find only the trivial solution. Therefore, = 0 is not an eigenvalue. Case 3. > 0: Let = 2 , for > 0. The roots of the auxiliary equation are r = i and so a general solution (11.2) is y(x) = C1 cos ( ln x) + C2 sin ( ln x) . We next find y (x). y (x) = C1 sin ( ln x) + C2 cos ( ln x) . x x By applying the first boundary condition, we obtain y (1) = C2 = 0 C2 = 0. Applying the second boundary condition, we obtain y (e ) = C1 cos ( ln(e )) = C1 cos () = 0. Therefore, in order to obtain a solution other than the trivial solution, we must have cos () = 0 678 = n+ 1 2 n = 0, 1, 2, . . . Exercises 11.2 =n+ 1 2 n = n+ 1 2
2 , n = 0, 1, 2, . . . . Corresponding to the eigenvalues, n 's, we have the eigenfunctions. yn (x) = Cn cos n+ 1 2 ln x , n = 0, 1, 2, . . . , where Cn is an arbitrary nonzero constant. 25. As in Problem 13, the auxiliary equation for this problem is r 2 + = 0. To find eigenvalues which yield nontrivial solutions we will consider the three cases < 0, = 0, and > 0. Case 1. < 0: The roots of the auxiliary equation are r =  and so a general solution to the differential equation y + y = 0 is given by y(x) = C1 e x + C2 e x . By applying the first boundary condition we obtain y(0) = C1 + C2 = 0 Thus C2 = C1 . y(x) = C1 e x  e x . Applying the second boundary conditions yields y 1 + 2 = C 1 e Multiplying by e (1+2 ) (1+2 )  e (1+2 ) = 0. yields C1 e2 (1+2 )  1 = 0. Now either C1 = 0 or e2 (1+2 ) =1 (1 + 2 ) = 0  = 0. Since < 0, we must have C1 = 0 and hence there are no eigenvalues for < 0. 679 Chapter 11
Case 2. = 0: In this case we are solving the differential equation y = 0. This equation has a general solution given by y(x) = C1 + C2 x. By applying the boundary conditions, we obtain y(0) = C1 = 0 and y 1 + 2 = C1 + C2 1 + 2 = 0. Solving these equations simultaneously yields C1 = C2 = 0. Thus, we find that is = 0 not an eigenvalue. Case 3. > 0: The roots of the auxiliary equation are r = i and so a general solution is y(x) = C1 cos x + C2 sin x . Substituting in the first boundary condition yields y(0) = C1 cos 0 + C2 sin 0 = C1 = 0. x , we obtain By applying the second boundary condition to y(x) = C2 sin y 1 + 2 = C2 sin (1 + 2 ) = 0. Therefore, in order to obtain a solution other than the trivial solution, we must have sin (1 + 2 ) = 0 (1 + 2 ) = n, n = 1, 2, 3, . . . . n (1 + 2 ) = n; and the n Hence choose the eigenvalues n , n = 1, 2, 3, . . . , such that corresponding eigenfunctions are yn (x) = Cn sin n x , n = 1, 2, 3, . . . , where the Cn 's are arbitrary nonzero constants. 33. (a) We assume that u(x, t) = X(x)T (t). Then utt = X(x)T (t), 680 ux = X (x)T (t), and uxx = X (x)T (t). Exercises 11.2
Substituting these functions into utt = uxx + 2ux , we obtain X(x)T (t) = X (x)T (t) + 2X (x)T (t). Separating variables yields X (x) + 2X (x) T (t) =  = , X(x) T (t) where is some constant. The first equation in (11.4) gives X (x) + 2X (x) + X(x) = 0. Let's now consider the boundary conditions. From u(0, t) = 0 and u(, t) = 0, t > 0, we conclude that X(0)T (t) = 0 and X()T (t) = 0, t > 0. (11.4) Hence either T (t) = 0 for all t > 0, which implies u(x, t) 0, or X(0) = X() = 0. Ignoring the trivial solution u(x, t) 0, we obtain the boundary value problem X (x) + 2X (x) + X(x) = 0, X(0) = X() = 0. 1  . (b) The auxiliary equation for this problem, r 2 + 2r + = 0, has roots r = 1 1  < 0, 1  = 0, and 1  > 0. Case 1, 1  < 0 ( > 1): Let = (1  ) = To find eigenvalues which yield nontrivial solutions, we will consider the three cases  1. In this case the roots to the auxiliary equation are r = 1 i (where is a positive number). Therefore, a general solution to the differential equation is given by X(x) = C1 ex cos x + C2 ex sin x. By applying the boundary conditions, we obtain X(0) = C1 = 0 and X() = e (C1 cos + C2 sin ) = 0 . 681 Chapter 11
Solving these equations simultaneously yields C1 = 0 and C2 sin = 0. Therefore, in order to obtain a solution other than the trivial solution, we must have sin = 0 Since =  1, 1=n = n2 + 1, n = 1, 2, 3, . . . . = n = n, n = 1, 2, 3, . . . . Thus the eigenvalues are given by n = n2 + 1, n = 1, 2, 3, . . . . Corresponding to the eigenvalue n , we obtain the solutions Xn (x) = Cn ex sin nx, where Cn = 0 is arbitrary. Case 2, 1  = 0 ( = 1): In this case the associated auxiliary equation has double root r = 1. Therefore, the general solution is given by X(x) = C1 ex + C2 xex . By applying the boundary conditions we obtain X(0) = C1 = 0 and X() = e (C1 + C2 ) = 0 . n = 1, 2, 3, . . . , Solving these equations simultaneously yields C1 = C2 = 0. So in this case we have only the trivial solution. Thus, = 1 is not an eigenvalue. Case 3, 1  > 0 ( < 1): Let = 1  . In this case the roots to the auxiliary equation are r = 1 (where is a positive number). Therefore, a general solution to the differential equation is given by X(x) = C1 e(1)x + C2 e(1+)x . 682 Exercises 11.3
By applying the first boundary condition we find X(0) = C1 + C2 = 0 So we can express X(x) as X(x) = C1 e(1)x  e(1+)x . Thus the second condition gives us X() = C1 e(1)  e(1+) . Since e(1)  e(1+) = 0, C1 = 0, and again in this case we have only the trivial solution. Thus, there are no eigenvalues for < 1. Therefore, the eigenvalues are n = n2 + 1, n = 1, 2, 3, . . . , with corresponding eigenfunctions Xn (x) = Cn ex sin nx, n = 1, 2, 3, . . . , where Cn is an arbitrary nonzero constant. EXERCISES 11.3: Regular SturmLiouville Boundary Value Problems, page 682 C2 = C1 . 3. Here A2 = x(1  x) and A1 = 2x. Using formula (4) on page 673 of the text, we find (x) = 1 1 1 e [A1 (x)/A2 (x)]dx = e [2x/x(1x)]dx = e2 x(1  x) x(1  x) x(1  x) 1 1 1x e2 ln(1x) = (1  x)2 = . = x(1  x) x(1  x) x
dx/(1x) Multiplying the original equation by (x) = (1  x)/x, we get (1  x)2 y (x)  2(1  x)y (x) + (1  x)2 y (x) + 1x y(x) = 0 x 1x y(x) = 0. x 9. Here we consider the linear differential operator L[y] := y +y; y(0) = y(), y (0) = y (). We must show that (u, L[v]) = (L[u], v), 683 Chapter 11
where u(x) and v(x) are any functions in the domain of L. Now (u, L[v]) =
0 u(x) [v (x) + v(x)] dx =
0 u(x)v (x) dx + 0 u(x)v(x) dx and (L[u], v) =
0 [u (x) + u(x)] v(x) dx =
0 u (x)v(x) dx + 0 u(x)v(x)dx. Hence it suffices to show that 0 0 u(x)v (x) dx =
0 u (x)v(x) dx. To do this we start with u (x)v(x) dx and integrate by parts twice. Doing this we obtain u (x)v(x) dx = u (x)v(x)
0 0 u(x)v (x) = 0 0 +
0 u(x)v (x) dx. Hence, we just need to show u (x)v(x) u (x)v(x) 0 0 u(x)v (x) 0. Expanding gives u(x)v (x) = 0 u ()v()  u (0)v(0)  u()v () + u(0)v (0). Since u is in the domain of L, we have u(0) = u(), and u (0) = u (). Hence, u (x)v(x) 0 u(x)v (x) = 0 u () [v() + v(0)]  u() [v () + v (0)] . But v also lies in the domain of L and hence v(0) = v() and v (0) = v (). This makes the expressions in the brackets zero and we have u (x)v(x) Therefore, L is selfadjoint. 17. In Problem 13 of Section 11.2, we found the eigenvalues to be n = 1 n+ 2
2 0 u(x)v (x) = 0 0. 2, n = 0, 1, 2, . . . with the corresponding eigenfunctions yn (x) = Cn sin n+ 1 2 x , n = 0, 1, 2, . . . , where Cn is an arbitrary nonzero constant. 684 Exercises 11.3
(a) We need only to choose the Cn so that
1 2 Cn sin2 0 n+ 1 2 x dx = 1. We compute
1 2 Cn 0 1 sin 2 1 1 2 C n+ x dx = 2 2 n 1 2 C = 2 n (1  cos [(2n + 1)x]) dx
0 Hence, we can take Cn = 1 sin [(2n + 1)x] x (2n + 1) 1 =
0 1 2 C . 2 n 2 which gives 1 2 sin n + 2 x
n=0 , as an orthonormal system of eigenfunctions. (b) To obtain the eigenfunction expansion for f (x) = x, we use formula (25) on page 679 of the text. Thus, cn =
0 1 x 2 sin n+ 1 x dx . 2 n+
1 2 Using integration by parts with u = cn = 2x and dv = sin
1 1 x dx, we find  2x cos[(n + 1/2)x] (n + 1/2) +
0 0 2x cos[(n + 1/2)x] dx (n + 1/2)
1  2 cos[(n + 1/2)] 2 sin[(n + 1/2)x] = + (n + 1/2) (n + 1/2)2 2 2 sin[(n + 1/2)] (1)n 2 = 0+ = . (n + 1/2)2 2 (n + 1/2)2 2 Therefore 0 x=
n=0 cn 2 sin 1 n+ 2 x =
n=0 2(1)n sin (n + 1/2)2 2 n+ n+ 1 2 1 2 x x . 8 = 2 n=0 (1)n sin (2n + 1)2 685 Chapter 11
23. In Problem 19 of Section 11.2, we found the eigenvalues n = 1 n+ 2
2 , n = 0, 1, 2, . . . , with the corresponding eigenfunctions yn (x) = Cn cos n+ 1 2 ln x , n = 0, 1, 2, . . . , where Cn is an arbitrary nonzero constant. (a) We need only to choose the Cn so that
e 2 Cn cos2 1 n+ 1 2 ln x 1 dx = 1. x To compute, we let u = ln x and so du = dx/x. Substituting, we find
e 2 Cn 1 cos 2 1 n+ 2 1 2 dx = Cn ln x x cos2
0 n+ 1 2 u du 1 2 C = 2 n 1 2 = C 2 n Hence, we can take Cn = 2/, which gives 2 cos {1 + cos [(2n + 1)u]} du
0 1 u+ sin [(2n + 1)u] 2n + 1 =
0 2 C . 2 n 1 n+ 2 ln x
n=0 , as an orthonormal system of eigenfunctions. (b) To obtain the eigenfunction expansion for f (x) = x, we use formula (25) on page 679 of the text. Thus, with w(x) = x1 , we have
e cn =
1 x 2 cos n+ 1 2 ln x x1 dx. 686 Exercises 11.4
Let u = ln x. Then du = dx/x, and we have cn = 2 eu cos
0 n+ 1 2 u du 0 = = Therefore, 2 eu cos[(n + 1/2)u] + eu (n + 1/2) sin[(n + 1/2)u] 1 + (n + 1/2)2 2 e (n + 1/2) sin[(n + 1/2)]  1 = 1 + (n + 1/2)2 n 2 (1) e (n + 1/2)  1 . 1 + (n + 1/2)2 x = = 2 cn
n=0 2 cos n+ 1 2 ln x n+ 1 2 ln x . n=0 (1)n e (n + 1/2)  1 cos 1 + (n + 1/2)2 EXERCISES 11.4: Nonhomogeneous Boundary Value Problems and the Fredholm Alternative, page 692 3. Here our differential operator is given by L[y] = 1 + x2 y + 2xy + y. Substituting into the formula (3) page 684 of the text, we obtain L+ [y] = (1 + x2 )y  (2xy) + y = 2xy + (1 + x2 )y  2y  2xy + y = 2y + 2xy + 2xy + (1 + x2 )y  2y  2xy + y = (1 + x2 )y + 2xy + y. 7. Here our differential operator is given by L[y] = y  2y + 10y; Hence L+ [v] = v + 2v + 10v. 687 y(0) = y() = 0. Chapter 11
To find the D (L+ ), we must have P (u, v)(x) = 0 0 (11.5) for all u in D(L) and v in D (L+ ). Using formula (9) page 685 of the text for P (u, v) with A1 = 2 and A2 = 1, we find P (u, v) = 2uv  uv + u v. Evaluating at and 0, condition (11.5) becomes 2u()v()  u()v () + u ()v() + 2u(0)v(0) + u(0)v (0)  u (0)v(0) = 0. Since u in D(L), we know that u(0) = u() = 0. Thus the above equation becomes u ()v()  u (0)v(0) = 0. Since u () and u (0) can take on any value, we must have v(0) = v() = 0 for this equation to hold for all u in D(L). Hence D (L+ ) consists of all function v having continuous second derivatives on [0, ] and satisfying the boundary condition v(0) = v() = 0. 11. Here our differential operator is given by L[y] = y + 6y + 10y; Hence L+ [v] = v  6v + 10v. To find the D (L+ ), we must have P (u, v)(x) = 0 y (0) = y () = 0. 0 (11.6) for all u in D(L) and v in D (L+ ). Again using formula (9) page 685 of the text for P (u, v) with A1 = 6 and A2 = 1, we find P (u, v) = 6uv  uv + u v. 688 Exercises 11.4
Evaluating at and 0, condition (11.6) becomes 6u()v()  u()v () + u ()v()  6u(0)v(0) + u(0)v (0)  u (0)v(0) = 0. Applying the boundary conditions u (0) = u () = 0 to the above equation yields 6u()v()  u()v ()  6u(0)v(0) + u(0)v (0) = 0 u() [6v()  v ()]  u(0) [6v(0)  v (0)] = 0. Since u() and u(0) can take on any value, we must have 6v()v () = 0 and 6v(0)v (0) = 0 in order for the equation to hold for all u in D(L). Therefore, the adjoint boundary value problem is L+ [v] = v  6v + 10v; 6v() = v () and 6v(0) = v (0). 17. In Problem 7 we found the adjoint boundary value problem L+ [v] = v + 2v + 10v; v(0) = v() = 0. (11.7) The auxiliary equation for (11.7) is r 2 + 2r + 10 = 0, which has roots r = 1 3i. Hence a general solution to the differential equation in (11.7) is given by y(x) = C1 ex cos 3x + C2 ex sin 3x. Using the boundary conditions in (11.7) to determine C1 and C2 , we find y(0) = C1 = 0 and y() = C1 e = 0. Thus C1 = 0 and C2 is arbitrary. Therefore, every solution to the adjoint problem (11.7) has the form y(x) = C2 ex sin 3x. It follows from the Fredholm alternative that if h is continuous, then the nonhomogeneous problem has a solution if and only if h(x)ex sin 3x dx = 0.
0 689 Chapter 11
21. In Problem 11 we found the adjoint boundary value problem L+ [v] = v  6v + 10v; 6v() = v () and 6v(0) = v (0). (11.8) The auxiliary equation for (11.8) is r 2  6r + 10 = 0, which has roots r = 3 i. Hence a general solution to the differential equation in (11.8) is given by y(x) = C1 e3x cos x + C2 e3x sin x. To apply the boundary conditions in (11.8), we first determine y (x). y (x) = 3C1 e3x cos x  C1 e3x sin x + 3C2 e3x sin x + C2 e3x cos x. Applying the first condition, we have 6C1 e3 = 3C1 e3  C2 e3 Applying the second condition, we have 6C1 = 3C1 + C2 3C1 = C2 . 3C1 = C2 . Thus C2 = 3C1 where C1 is arbitrary. Therefore, every solution to the adjoint problem (11.8) has the form y(x) = C1 e3x (cos x + 3 sin x). It follows from the Fredholm alternative that if h is continuous, then the nonhomogeneous problem has a solution if and only if h(x)e3x (cos x + 3 sin x) dx = 0.
0 EXERCISES 11.5: Solution by Eigenfunction Expansion, page 698 3. In Example 1 on page 696 of the text we noted that the boundary value problem y + y = 0; 690 y(0) = 0, y() = 0, Exercises 11.5
has eigenvalues n = n2 , n = 1, 2, 3, . . . , with corresponding eigenfunctions n (x) = sin nx, n = 1, 2, 3, . . . . Here r(x) 1, so we need to determine coefficients n such that f (x) = f (x) = r(x) n sin nx = sin 2x + sin 8x.
n=1 Clearly 2 = 8 = 1 and the remaining n 's are zero. Since = 4 = 2 and 2 = 1 = 0 there is no solution to this problem. 5. In equation (18) on page 666 of the text we noted that the boundary value problem y + y = 0; y (0) = 0, y () = 0, has eigenvalues n = n2 , n = 0, 1, 2, . . . , with corresponding eigenfunctions n (x) = cos nx, n = 0, 1, 2, . . . . Here r(x) 1, so we need to determine coefficients n such that f (x) = f (x) = r(x) n cos nx = cos 4x + cos 7x.
n=0 Clearly 4 = 7 = 1 and the remaining n 's are zero. Since = 1 = 1 and 1 = 0, (  1 ) c1  1 = 0 is satisfied for any value of c1 . Calculating c4 and c7 , we get c4 = and 1 7 1 = . =  7 1  49 48 Hence a one parameter family of solutions is c7 = 1 4 1 = =  4 1  16 15 (x) =
n=0 cn n (x) = c1 cos x  1 1 cos 4x  cos 7x , 15 48 where c1 is arbitrary. 691 Chapter 11
9. We first find the eigenvalues and corresponding eigenfunctions for this problem. Note that the auxiliary equation for this problem is r 2 + = 0. To find eigenvalues which yield nontrivial solutions we will consider the three cases < 0, = 0, and > 0. Case 1, < 0: Let = , then the roots to the auxiliary equation are r = and a general solution to the differential equation is given by y(x) = C1 sinh x + C2 cosh x. Since y (x) = C1 cosh x + C2 sinh x, by applying the boundary conditions we obtain y (0) = C1 = 0 and y() = C1 sinh + C2 cosh = 0. Hence C1 = 0 and y() = C2 cosh = 0. Therefore C2 = 0 and we find only the trivial solution. Case 2, = 0: In this case the differential equation becomes y = 0. This equation has a general solution given by y(x) = C1 + C2 x. Since y (x) = C2 , by applying the boundary conditions we obtain y (0) = C2 = 0 and y() = C1 + C2 = 0. Solving these equations simultaneously yields C1 = C2 = 0. Thus, we again find only the trivial solution. Case 3, > 0: Let = 2 , for > 0. The roots of the auxiliary equation are r = i and so a general solution is y(x) = C1 cos x + C2 sin x. Since y (x) = C1 sin x + C2 cos x, 692 Exercises 11.5
using the first boundary condition we find y (0) = C1 sin( 0) + C2 cos( 0) = 0 C2 = 0 C2 = 0. Thus substituting into the second boundary condition yields y() = C1 cos = 0. Therefore, in order to obtain a solution other than the trivial solution, we must have cos = 0 =n+ 1 , 2 n = 0, 1, 2, . . . . Hence choose n = (n + 1/2)2 , n = 0, 1, 2, . . ., and yn (x) = Cn cos where the Cn 's are arbitrary nonzero constants. Next we need to choose the Cn so that 2 Cn cos2 0 n+ 1 x , 2 n+ 1 2 x dx = 1. Computing we find 2 Cn 0 cos 2 1 n+ 2 1 2 C x dx = 2 n 1 2 C = 2 n {1 + cos[(2n + 1)x]} dx
0 x+ 1 sin[(2n + 1)x] 2n + 1 =
0 2 C . 2 n An orthonormal system of eigenfunctions is given when we take Cn = 2 cos Now f (x) has the eigenfunction expansion 2/, 1 n+ 2 x
n=0 . f (x) =
n=0 n 2 cos n+ 1 x , 2 693 Chapter 11
where n = 2 f (x) cos
0 n+ 1 x dx. 2 Therefore, with n as described above, the solution to the given boundary value problem has a formal expansion (x) =
n=0 n 1  n 2 cos 1 n+ 2 x =
n=0 n 1  (n + 1/2)2 2 cos n+ 1 2 x . EXERCISES 11.6: Green's Functions, page 706 1. A general solution to the corresponding homogeneous equation, y = 0, is yh (x) = Ax + B. Thus we seek for paricular solutions z1 (x) and z2 (x) of this form satisfying z1 (0) = 0, z2 () = 0. The first equation yields z1 (0) = B = 0. Since A is arbitrary, we choose A = 1 and so z1 (x) = x. Next, from the second equation in (11.9) we get z2 () = A = 0. Taking B = 1, we obtain z2 (x) = 1. With p(x) 1, we now compute C = p(x)W [z1 , z2 ] (x) = (1)[(x)(0)  (1)(1)] = 1. Thus, the Green's function is G(x, s) = z1 (s)z2 (x)/C, 0 s x, z1 (x)z2 (s)/C, x s = s, 0 s x, x, x s . (11.9) 694 Exercises 11.6
3. A general solution to the homogeneous problem, y = 0, is yh (x) = Ax + B, so z1 (x) and z2 (x) must be of this form. To get z1 (x) we want to choose A and B so that z1 (0) = B = 0. Since A is arbitrary, we can set it equal to 1 and z1 (x) = x. Next, to get z2 (x) we need to choose A and B so that z2 () + z2 () = A + B + A = 0. Thus B = (1 + )A. Taking A = 1, we get z2 (x) = x  1  . Now compute C = p(x)W [z1 , z2 ] (x) = (1)[(x)(1)  (1)(x  1  )] = 1 + . Thus, the Green's function is s(x  1  ) , 0 s x, 1+ = x(s  1  ) , x s . 1+ G(x, s) = z1 (s)z2 (x)/C, 0 s x, z1 (x)z2 (s)/C, x s 5. The corresponding homogeneous differential equation, y + 4y = 0, has the characteristic equation r 2 + 4 = 0, whose roots are r = 2i. Hence, a general solution to the homogeneous problem is given by yh (x) = C1 cos 2x + C2 sin 2x. A solution z1 (x) must satisfy the first boundary condition, z1 (0) = 0. Substitution yields z1 (0) = C1 cos(2 0) + C2 sin(2 0) = 0 C1 = 0. Setting C2 = 1, we get z1 (x) = sin 2x. For z2 (x), we have to find constants C1 and C2 such that the second boundary condition is satisfied. Since yh (x) = 2C1 sin 2x + 2C2 cos 2x, 695 Chapter 11
we have z2 () = 2C1 sin(2) + 2C2 cos(2) = 2C2 = 0 With C1 = 1, z2 (x) = cos 2x. Next we find C = p(x)W [z1 , z2 ] (x) = (1)[(sin 2x)(2 sin 2x)  (cos 2x)(2 cos 2x)] = 2. Thus, the Green's function in this problem is given by G(x, s) = z1 (s)z2 (x)/C, 0 s x, z1 (x)z2 (s)/C, x s = (sin 2s cos 2x)/2 , 0 s x, (sin 2x cos 2s)/2 , x s . C2 = 0. 13. In Problem 3 we found the Green's function for this boundary value problem. When f (x) = x, the solution is given by equation (16) on page 702 of the text. Substituting for f (x) and G(x, s) yields
b x y(x) =
a G(x, s)f (s) ds =
0 G(x, s)s ds =
0 s2 (x  1  ) ds + 1+
x xs(s  1  ) ds. 1+ Computing
x 0 (x  1  ) s2 (x  1  ) ds =  1+ 1+ s3 3 x =
s=0 (x  1  ) 1+ x3 3 = x4 x3 + , 3(1 + ) 3 x xs(s  1  ) ds =  1+ 1+ + s3 (1 + )s2  3 2 =
s=x 3 (1 + ) 2 x  1+ 3 2 x x3 (1 + )x2 2x x4 x3 x 3x  + +  , = 1+ 3 2 3(1 + ) 2 3(1 + ) 2 we finally get y(x) = x3 2x x4 x3 x4 3x + + +  +  3(1 + ) 3 3(1 + ) 2 3(1 + ) 2 3 2 3 3 2 (3 + )x x x  . x= + =  + 6 2 3(1 + ) 6 6(1 + )  696 Exercises 11.6
17. A general solution to the corresponding homogeneous problem y  y = 0 is yh (x) = C1 ex + C2 ex . So z1 (x) and z2 (x) must be of this form. To get z1 (x) we want to choose constants C1 and C2 so that z1 (0) = C1 e0 + C2 e0 = C1 + C2 = 0. Let C1 = 1. Then C2 = 1 and so z1 (x) = ex  ex . Likewise, to get a z2 (x), we find C1 and C2 so that z2 (1) = C1 e1 + C2 e1 = 0 C2 = C1 e2 . If we let C1 = 1, then C2 = e2 . Hence z2 (x) = ex  e2 ex = ex  e2x . We now compute C = p(x)W [z1 , z2 ] (x) = (1) ex  ex ex + e2x  ex + ex ex  e2x = 2e2  2. Thus, the Green's function is in this problem is G(x, s) = z1 (s)z2 (x)/C, 0 s x, z1 (x)z2 (s)/C, x s 1 (es  es ) (ex  e2x ) / (2  2e2 ) , 0 s x, (ex  ex ) (es  e2s ) / (2  2e2 ) , x s 1. = Here f (x) = x. Using Green's function to solve the boundary value problem, we find
b x y(x) =
a G(x, s)f (s) ds =
0 (es  es )(ex  e2x )(s) ds + 2  2e2 1 x (ex  ex )(es  e2s )(s) ds. 2  2e2 Computing integrals yields
x 0 (es  es )(ex  e2x )(s) ex  e2x ds =  2  2e2 2  2e2 = 
x 2x x ses  ses ds
0 e e x ses  es + ses + es 0 2 2  2e ex  e2x xex  ex + xex + ex , =  2 2  2e 697 Chapter 11
1 x (ex  ex )(es  e2s )(s) ex  ex ds =  2  2e2 2  2e2 = 
x x 1 ses  se2s ds
x e e 1 ses  es + se2s + e2s x 2 2  2e ex  ex 2e  xex  ex + xe2x + e2x =  2  2e2 . Thus y(x) =  ex  e2x ex  ex xex  ex + xex + ex  2e  xex  ex + xe2x + e2x 2  2e2 2  2e2 (ex  e2x )(xex  ex + xex + ex )  (ex  ex )(2e  xex + ex  xe2x  e2x ) = 2  2e2 2x + 2xe2  2e1+x + 2e1x e1+x  e1x . = = x + 2  2e2 e2  1 25. Substitution y = xr into the corresponding homogeneous CauchyEuler equation x2 y  2xy + 2y = 0, we obtain the auxiliary equation r(r  1)  2r + 2 = 0 or r 2  3r + 2 = (r  1)(r  2) = 0. Hence a general solution to the corresponding homogeneous equation is yh (x) = C1 x + C2 x2 . To get z1 (x) we want to choose C1 and C2 so that z1 (1) = C1 + C2 = 0 C2 = C1 . Let C1 = 1, then C2 = 1 and z1 (x) = x  x2 . Next we find z2 (x) satisfying z2 (2) = 2C1 + 4C2 = 0 C1 = 2C2 . Hence, we let C2 = 1, then C1 = 2 and z2 (x) = 2x  x2 . Now compute (see the formula for K(x, s) in Problem 22) C(s) = A2 (s)W [z1 , z2 ] (s) = 698 s2 (s  s2 )(2  2s)  (1  2s)(2s  s2 ) Exercises 11.6
2s  4s2 + 2s3  2s + 5s2  2s3 = s4 , (s  s2 )(2x  x2 )  , 1 s x, s4 z1 (s)z2 (x)/C(s), 1 s x, = K(x, s) = z1 (x)z2 (s)/C(s), x s 2 (x  x2 )(2s  s2 )  , x s 2. s4 Simplifying yields = K(x, s) = x(2  x)(s3  s2 ), x(1  x)(2s
3 2 s2 1 s x,  s ), x s 2. Hence, a solution to the boundary value problem with f (x) = x is
b x 2 y(x) =
a x K(x, s)f (s) ds =
1 K(x, s)f (s) ds +
x 2 K(x, s)f (s) ds =
1 [x(2  x)(s3  s2 )](s) ds +
x x [x(1  x)(2s3  s2 )](s) ds
2 = = 2x  x 2 1 s s 2 s 1 ds + x  x
x 1 2 x 2s2  s1 ds 2s1  ln s
2 x 2x  x 2 1  ln s + xx 2 = x2 ln 2  x ln 2  x ln x . 29. Let f (x) = (x  s). Let H(x, s) be the solution to 4 H(x, s) = (x  s) x4 that satisfies the given boundary conditions, the jump condition lim + 3 H(x, s) 3 H(x, s)  lim = 1, xs x3 x3 xs and H, H/x, 2 H/x2 are continuous on the square [0, ] [0, ]. We begin by integrating to obtain 3 H(x, s) = u(x  s) + C1 , x3 699 Chapter 11
where u is the unit step function and C1 is a constant. (Recall in Section 7.8 we observed that u (t  a) = (t  a), at least formally.) 3 H/x3 is not continuous along the line x = s, but it does satisfy the jump condition 3 H(x, s) 3 H(x, s)  lim = lim [u(x  s) + C1 ]  lim [u(x  s) + C1 ] lim xs+ xs xs+ xs x3 x3 = (1 + C1 )  C1 = 1. We want H(x, s) to satisfy the boundary condition y () = 0. So we solve 3H (, s) = u(  s) + C1 = 1 + C1 = 0 x3 to obtain C1 = 1. Thus 3 H(x, s) = u(x  s) + 1 . x3 2 H(x, s) = x  u(x  s)(x  s) + C2 . x2 (The reader should verify this is the antiderivative for x = s by differentiating it.) We selected this particular form of the antiderivative because we need 2 H/x2 to be continuous on [0, ] [0, ]. (The jump of u(x  s) when x  s is canceled by the vanishing of this term by the factor (x  s).) Since 2 H(x, s) lim = s + C2 , xs x2 2H (s, s) = s + C2 , x2 We now integrate again with respect to x to obtain we can define and we now have a continuous function. Next, we want y () = 0. Solving we find 0= 2H (, s) =  u(  s)(  s) + C2 =  (  s) + C2 = s + C2 . x2 Thus, we find that C2 = s. Now, 2 H(x, s) = (x  s)  u(x  s)(x  s) . x2 700 Exercises 11.7
We integrate with respect to x again to get H(x, s) x2 (x  s)2 =  sx  u(x  s) + C3 , x 2 2 which is continuous on [0, ] [0, ]. We now want the boundary condition y (0) = 0 satisfied. Solving, we obtain 0= Hence, H s2 (0, s) = u(0  s) + C3 = C3 . x 2 x2 (x  s)2 H(x, s) =  sx  u(x  s) . x 2 2 Integrating once more with respect to x, we have H(x, s) = x3 sx2 (x  s)3   u(x  s) + C4 . 6 2 6 Now H(x, s) is continuous on [0, ][0, ]. We want H(x, s) to satisfy the boundary condition y(0) = 0. Solving, we find 0 = H(0, s) = u(0  s) Hence, (0  s)3 + C4 = C4 . 6 x3 sx2 (x  s)3 H(x, s) =   u(x  s) , 6 2 6 which we can rewrite in the form 2 s (s  3x) , 0 s x, 6 H(x, s) = 2 x (x  3s) , x s . 6 Singular SturmLiouville Boundary Value Problems, page 715 EXERCISES 11.7: 1. This is a typical singular SturmLiouville boundary value problem. Condition (ii) of Lemma 1 on page 710 of the main text holds since
x0+ lim p(x) = lim+ x = 0
x0 701 Chapter 11
and y(x), y (x) remain bounded as x 0+ . Because
x1 lim p(x) = p(1) = 1 and y(1) = 0, the analogue of condition (i) of Lemma 1 holds at the right endpoint. Hence L is selfadjoint. The equation is Bessel's equation of order 2. On page 712 of the text, we observed that the solutions to this boundary value problem are given by yn (x) = cn J2 (2n x) , where n = 2n is the increasing sequence of real zeros of J2 (x), that is, J2 (2n ) = 0. Now to find an eigenfunction expansion for the given nonhomogeneous equation we compute the eigenfunction expansion for f (x)/x (see page 694): f (x) x where an = Therefore, y(x) =
n=1 1 0 an J2 (2n x) ,
n=1 f (x)J2 (2n x) dx
1 0 2 xJ2 (2n x) dx , n = 1, 2, 3, . . . . an J2 (2n x) . 2  2n 3. Again, this is a typical singular SturmLiouville boundary value problem. L is selfadjoint since condition (ii) of Lemma 1 on page 710 of the main text holds at the left endpoint and the analogue of condition (i) holds at the right endpoint. This is Bessel's equation of order 0. As we observed on page 712 of the text, J0 x satisfies the boundary conditions at the origin. At the right endpoint, we want J0 = 0. Now it follows from equation (32) on page 488 of the text, that the zeros of J0 and J1 are the same. So if we let n = 1n , the increasing sequence of zeros of J1 , then J0 (1n ) = 0. Hence, the eigenfunctions are given by yn (x) = J0 (1n x) , 702 n = 1, 2, 3, . . . . Exercises 11.7
To find an eigenfunction expansion for the solution to the nonhomogeneous equation, we first expand f (x)/x (see page 694): f (x) x where bn = Therefore, y(x) =
n=1 1 0 bn J0 (1n x) ,
n=1 f (x)J0 (1n x) dx
1 0 2 xJ0 (1n x) dx , n = 1, 2, 3, . . . . bn J0 (1n x) . 2  1n 11. (a) Let (x) be an eigenfunction for dy 2 d x  y + xy = 0. dx dx x Therefore, 2 d [x (x)]  (x) + x(x) = 0 dx x 2 (x) + x(x) = 0. (x) + x (x)  x Multiplying both side by (x) and integrating both sides from 0 to 1, we obtain
1 1 1 (x) (x) dx +
0 0 x(x) (x) dx 
0 2 [(x)]2 dx + x 1 x[(x)]2 dx = 0.
0 (11.10) Now integrating by parts with u = (x) (x) and dv = dx, we have
1 1 1 0 (x) (x) dx = x(x) (x)
0 
0 1 x [ (x) (x) + (x) (x)] dx
1 = x(x) (x) 1 0 
0 x [ (x)] dx 
0 2 x(x) (x) dx. 703 Chapter 11
Since (1) = 0, we have x(x) (x)
1 1 = 0 0,
1 1 (x) (x) dx = 
0 0 x [ (x)] dx 
0 2 x(x) (x) dx. Thus equation (11.10) reduces to
1 1 1 
0 x [ (x)] dx 
0 2 x(x) (x) dx +
0 x(x) (x) dx
1 1 
1 1 2 2 0 x [(x)] dx + 0 1 1 2 x[(x)]2 dx = 0 
0 x [ (x)] dx  2
0 x1 [(x)]2 dx + 0 x[(x)]2 dx = 0. (11.11) (b) First note that each integrand in (11.11) is nonnegative on the interval (0, 1), hence each integral is nonnegative. Moreover, since (x) is an eigenfunction, it is a continuous function which is not the zero function. Hence, the second and third integrals are strictly positive. Thus, if > 0, then must be positive in order for the lefthand side of (11.11) to sum to zero. (c) If = 0, then only the first and third terms remain on the left hand side of equation (11.11). Since the first integral need only be nonnegative, we only need to be nonnegative in order for equation (11.11) to be satisfied. To show = 0 is not an eigenvalue, we solve Bessel's equation with = 0, that is, we solve xy + y = 0, which is the same as the CauchyEuler equation x2 y + xy = 0. 704 Exercises 11.8
Solving this CauchyEuler equation, we find a general solution y(x) = c1 + c2 ln x. Since limx0+ y(x) =  if c2 = 0, we take c2 = 0. Now y(x) = c1 satisfies the boundary condition (17) in the text. The right endpoint boundary condition (18) is y(1) = 0. So we want 0 = y(1) = c1 . Hence the only solution to Bessel's equation of order 0 that satisfies the boundary conditions (17) and (18) is the trivial solution. Hence = 0 is not an eigenvalue. EXERCISES 11.8: Oscillation and Comparison Theory, page 725 5. To apply the Sturm fundamental theorem to y + (1  ex ) y = 0, 0 < x < , (11.12) we must find a q(x) and a function (x) such that q(x) 1  ex , 0 < x < , and (x) is a solution to y + q(x)y = 0, 0 < x < . (11.13) Because, for x > 0, 1  ex < 0, we choose q(x) 0. Hence equation (11.13) becomes y = 0. The function (x) = x + 4 is a nontrivial solution to this differential equation. Since (x) = x + 4 does not have a zero for x > 0, any nontrivial solution to (11.12) can have at most one zero in 0 < x < . To use the Sturm fundamental theorem to show that any nontrivial solution to y + (1  ex ) y = 0,  < x < 0, (11.14) has infinitely many zeros, we must find a q(x) and a function (x) such that q(x) 1  ex , x < 0, and (x) is a solution to y + q(x)y = 0,  < x < 0. Because 1  e1 0.632, we choose q(x) 1/4 and only consider the interval (, 1). Hence, we obtain y + 1 y = 0, 4 705 Chapter 11
which has nontrivial solution (x) = sin(x/2). Now the function (x) has infinitely many zeros in (, 1) and between any two consecutive zeros of (x) any nontrivial solution to (11.14) must have a zero; hence any nontrivial solution to (11.14) will have infinitely many zeros in (, 1). 9. First express y + x2 y + 4  ex y = 0, in StrumLiouville form by multiplying by the integrating factor e1/x : e1/x y + e1/x x2 y + e1/x 4  ex y = 0 Now when x gets large, we have p q e1/large e1/large (4  elarge ) 1 (1)(4  small) 1 1 = . 4 2 e1/x y + e1/x 4  ex y = 0. Hence, the distance between consecutive zeros is approximately /2. 11. We apply Corollary 5 with p(x) = 1 + x, q(x) = ex , and r(x) 1 to a nontrivial solution on the interval [0, 5]. On this interval we have pM = 6, pm = 1, qM = 1, qm = e5 , and rM = rm = 1. Therefore, for > max qM qm , ,0 rM rm = 0, the distance between two consecutive zeros of a nontrivial solution (x) to the given equation is bounded between qM pm = + rM 1 1+ and pM = qm + rm e5 6 . + 706 CHAPTER 12: Stability of Autonomous Systems
EXERCISES 12.2: Linear Systems in the Plane, page 753 3. The characteristic equation for this system is r 2 + 2r + 10 = 0, which has roots r = 1 3i. Since the real part of each root is negative, the trajectories approach the origin, and the origin is an asymptotically stable spiral point. 7. The critical point is the solution to the system 4x + 2y + 8 = 0, x  2y + 1 = 0. Solving this system, we obtain the critical point (3, 2). Now we use the change of variables x = u + 3, y = v + 2, to translate the critical point (3, 2) to the origin (0, 0). Substituting into the system of this problem and simplifying, we obtain a system of differential equations in u and v: dx du = = 4(u + 3) + 2(v + 2) + 8 = 4u + 2v, dt dt dv dy = = (u + 3)  2(v + 2) + 1 = u  2v. dt dt The characteristic equation for this system is r 2 + 6r + 6 = 0, which has roots r = 3 3. Since both roots are distinct and negative, the origin is an asymptotically stable improper node of the new system. Therefore, the critical point (3, 2) is an asymptotically stable improper node of the original system. 9. The critical point is the solution to the system 2x + y + 9 = 0, 5x  2y  22 = 0. 707 Chapter 12
Solving this system, we obtain the critical point (4, 1). Now we use the change of variables x = u  4, y = v  1, to translate the critical point (4, 1) to the origin (0, 0). Substituting into the system of this problem and simplifying, we obtain a system of differential equations in u and v: dx du = = 2(u  4) + (v  1) + 9 = 2u + v, dt dt dv dy = = 5(u  4)  2(v  1)  22 = 5u  2v. dt dt The characteristic equation for this system is r 2 + 1 = 0, which has roots r = i. Since both roots are distinct and pure imaginary, the origin is a stable center of the new system. Therefore, the critical point (4, 1) is a stable center of the original system. 15. The characteristic equation for this system is r 2 + r  12 = 0, which has roots r = 4 and r = 3. Since the roots are real and have opposite signs, the origin is an unstable saddle point. To sketch the phase plane diagram, we must first determine two lines passing through the origin that correspond to the transformed axes. To find the transformed axes, we make the substitution y = mx into dy/dt 5x  2y dy = = dx dx/dt x + 2y m= Solving for m yields m(x + 2mx) = 5x  2mx 2m2 + 3m  5 = 0 m= 5 2 or m = 1. 5x  2mx . x + 2mx to obtain So m = 5/2 or m = 1. Hence, the two axes are y = 5x/2 and y = x. On the line y = x one finds dx = 3x, dt so the trajectories move away from the origin. On the line y = 5x/2 one finds dy = 4y, dt 708 Exercises 12.3
so the trajectories move towards the origin. A phase plane diagram is given in Figure B.56 in the answers of the text. 19. The characteristic equation for this system is (r + 2)(r + 2) = 0 which has roots r = 2, 2. Since the roots are equal, real, and negative, the origin is an asymptotically stable point. To sketch the phase plane diagram, we determine the slope of the two lines passing through the origin that correspond to the transformed axes by substituting y = mx into dy dy/dt 2y = = dx dx/dt 2x + y to obtain m= Solving for m yields m(2x + mx) = 2mx m2 = 0 m = 0. 2mx . 2x + mx Since there is only one line (y = 0) through the origin that is a trajectory, the origin is an improper node. A phase plane diagram is given in Figure B.58 in the answers of the text. EXERCISES 12.3: Almost Linear Systems, page 764 5. This system is almost linear since ad  bc = (1)(1)  (5)(1) = 0, and the functions F (x, y) = G(x, y) = y 2 = 0 involve only high order terms in y. Since the characteristic equation for this system is r 2 + 4 = 0 which has pure imaginary roots r = 2i, the origin is either a center or a spiral point and the stability is indeterminant. 7. To see that this system is almost linear, we first express ex+y , cos x, and cos y using their respective Maclaurin series. Hence, the system dx = ex+y  cos x , dt dy = cos y + x  1 , dt 709 Chapter 12
becomes dx (x + y)2 x2 = 1 + (x + y) + +  1 + dt 2! 2! = x + y + (higher orders) = x + y + F (x, y), 2 dy y = 1 + + x  1 = x + (higher orders) = x + G(x, y). dt 2! This system is almost linear since ad  bc = (1)(0)  (1)(1) = 0, and F (x, y), G(x, y) each only involve higher order forms in x and y. The characteristic equation for this system is r 2  r  1 = 0 which has roots r = (1 5)/2. Since these roots are real and have different signs the origin is an unstable saddle point. 9. The critical points for this system are the solutions to the pair of equations 16  xy = 0, x  y 3 = 0. Solving the second equation for x in terms of y and substituting this into the first equation we obtain 16  y 4 = 0 which has solutions y = 2. Hence the critical points are (8, 2) and (8, 2). We consider the critical point (8, 2). Using the change of variables x = u + 8 and y = v + 2, we obtain the system du = 16  (u + 8)(v + 2), dt dv = (u + 8)  (v + 2)3 , dt which simplifies to the almost linear system du = 2u  8v  uv, dt dv = u  12v  6v 2  v 3 . dt The characteristic equation for this system is r 2 +14r +32 = 0, which has the distinct negative roots r = 7 17. Hence (8, 2) is an improper node which is asymptotically stable. 710 Exercises 12.3
Next we consider the critical point (8, 2). Using the change of variables x = u  8 and y = v  2, we obtain the system du = 16  (u  8)(v  2), dt dv = (u  8)  (v  2)3 , dt which simplifies to the almost linear system du = 2u + 8v  uv, dt dv = u  12v + 6v 2  v 3 . dt The characteristic equation for this system is r 2 + 10r  32 = 0, which has the distinct roots r = 5 57. Since these roots are real and have different signs, (8, 2) is an unstable saddle point. 13. The critical points for this system are the solutions to the pair of equations 1  xy = 0, x  y 3 = 0. Solving the second equation for x in terms of y and substituting this into the first equation we obtain 1  y4 = 0 which has solutions y = 1. Hence the critical points are (1, 1) and (1, 1). We consider the critical point (1, 1). Using the change of variables x = u + 1 and y = v + 1, we obtain the almost linear system du = 1  (u + 1)(v + 1) = u  v  uv, dt dv = (u + 1)  (v + 1)3 = u  3v  3v 2  v 3 . dt The characteristic equation for this system is r 2 +4r+4 = 0, which has the equal negative roots r = 2. Hence (1, 1) is an improper or proper node or spiral point which is asymptotically stable. 711 Chapter 12
Next we consider the critical point (1, 1). Using the change of variables x = u  1 and y = v  1, we obtain the almost linear system du = 1  (u  1)(v  1) = u + v  uv, dt dv = (u  1)  (v  1)3 = u  3v + 3v 2  v 3 . dt The characteristic equation for this system is r 2 + 2r  4 = 0, which has roots r = 1 phase plane diagram is given in Figure B.59 in the answers of the text. 21. Case 1: h = 0. The critical points for this system are the solutions to the pair of equations x(1  4x  y) = 0, y(1  2y  5x) = 0. To solve this system, we first let x = 0, then y(1  2y) = 0. So y = 0 or y = 1/2. When y = 0, we must have x(1  4x) = 0. So x = 0 or x = 1/4. And if x = 0 and y = 0, we have the system 1  4x  y = 0, 1  2y  5x = 0, which has the solution x = 1/3, y = 1/3. Hence the critical points are (0, 0), (0, 1/2), (1/4, 0), and (1/3, 1/3). At the critical point (0, 0), the characteristic equation is r 2  2r + 1 = 0, which has equal positive roots r = 1. Hence (0, 0) is an improper or proper node or spiral point which is unstable. From Figure B.61 in the text, we see that (0, 0) is an improper node. Next we consider the critical point (0, 1/2). Using the change of variables y = v + 1/2 and x = u, we obtain the almost linear system 1 1 du = u 1  4u  v  = u  4u2  uv, dt 2 2 1 dv 5 = v+ (1  2v  1  5u) =  u  v  2v 2  5uv. dt 2 2 712 5. Since these roots are real and have different signs, (1, 1) is an unstable saddle point. A Exercises 12.3
The characteristic equation for this system is r 2 + (1/2)r  (1/2) = 0, which has roots r = 1/2 and r = 1. Since these roots are real and have different signs, (0, 1/2) is an unstable saddle point. Now consider the critical point (1/4, 0). Using the change of variables x = u + 1/4 and y = v, we obtain the almost linear system du 1 1 = u+ (1  4u  1  v) = u  v  4u2  uv, dt 4 4 5 1 dv = v 1  2v  5u  =  v  2v 2  5uv. dt 2 4 The characteristic equation for this system is r 2 +(5/4)r+(1/4) = 0, which has roots r = 1/4 and r = 1. Since these roots are distinct and negative, (1/4, 0) is an improper node which is asymptotically stable. At the critical point (1/3, 1/3), we use the change of variables x = u + 1/3 and y = v  1/3 to obtain the almost linear system du 1 = u+ dt 3 dv 1 = v dt 3 4 1 4 1 v+ =  u  v  4u2  uv, 3 3 3 3 2 5 5 2 1  2v +  5u  = u + v  2v 2  5uv. 3 3 3 3 1  4u  The characteristic equation for this system is r 2 + (2/3)r  (1/3) = 0 which has roots r = 1/3 and r = 1. Again since these roots are real and have different signs, (1/3, 1/3) is an unstable saddle point, but not of interest since y < 0. Species x survives while species y dies off. A phase plane diagram is given in Figure B.61 in the answers of the text. Case 2: h = 1/32. The critical points for this system are the solutions to the pair of equations x(1  4x  y)  1 = 0, 32 y(1  2y  5x) = 0. To solve this system, we first set y = 0 and solve x(1  4x)  1/32 = 0, which has solutions x = (2 2)/16. 713 Chapter 12
If y = 0, we have 1  2y  5x = 0. So y = (1/2)  (5/2)x. Substituting, we obtain x 1  4x  Simplifying, we obtain 3 1 1  x2 + x  = 0, 2 2 32 which has the solution x = 1/4 or x = 1/12. When x = 1/4, we have y= And when x = 1/12, we have y= Hence the critical points are 2 2 ,0 , 16 2+ 2 ,0 , 16 1 1 , 4 8 , and 1 7 , 12 24 . 1 5  2 2 1 12 = 7 . 24 1 5  2 2 1 4 1 = . 8 1 5  x 2 2  1 = 0. 32 2 2 2 2 , 0 , we use the change of variables x = u + and y = v At the critical point 16 16 to obtain the almost linear system 2 2 1 2 2 2 1  4u  v  = u v  4u2  uv, 4 32 2 16 dv 10  5 2 6+5 2 = v 1  2v  5u  = v  2v 2  5uv. dt 16 16 du = dt The characteristic equation for this system is 2 r 2 6+5 2 r 16 2 2 ,0 16 = 0, 2 2 u+ 16 which has distinct positive roots. Hence 714 is an unstable improper node. Exercises 12.3
Now consider the critical point 2+ 2 x= u+ to obtain the almost linear system 16 2+ 2 du 2+ 2 1 2+ 2 2 1  4u  = u+ v  = u v  4u2  uv, dt 16 4 32 2 16 10 + 5 2 65 2 dv = v 1  2v  5u  = v  2v 2  5uv. dt 16 16 The characteristic equation for this system is 65 2 2 r r+ 2 16 which has distinct negative roots. Hence node. When the critical point is (1/12, 7/24), the change of variables x = u + 1/12 and y = v + 7/24 leads to the almost linear system 1 du = u+ dt 12 7 dv = v+ dt 24 7 1 1 1 1 v = u v  4u2  uv,  3 24 32 24 12 5 7 7 35  5u  v  2v 2  5uv. 1  2v  =  u 12 12 24 12 1  4u  2+ 2 ,0 16 2+ 2 , 0 , where we use the change of variables y = v and 16 = 0, is an asymptotically stable improper The characteristic equation for this system is r 2 + (13/24)r  (7/48) = 0, which has roots r = (13 505)/48. Since these roots have opposite signs, (1/12, 7/24) is an unstable saddle point. And when the critical point is (1/4, 1/8), the change of variables x = u+1/4 and y = v 1/8 leads to the almost linear system 1 du = u+ dt 4 1 dv = v dt 8 1  4u  1  v + 1  2v + 1 8  1 7 1 =  u  v  4u2  uv, 32 8 4 1 5 = u + v  2v 2  5uv. 8 4 715 5 1  5u  4 4 Chapter 12
The characteristic equation for this system is r 2 + (5/8)r  (1/16) = 0, which has roots r = (5 41)/16. Since these roots have opposite signs, (1/4, 1/8) is an unstable saddle point. But since y < 0, this point is not of interest. Hence, this is competitive exclusion; one species survives while the other dies off. A phase plane diagram is given in Figure B.62 in the answers of the text. Case 3: h = 5/32. The critical points for this system are the solutions to the pair of equations x(1  4x  y)  5 = 0, 32 y(1  2y  5x) = 0. To solve this system, we first set y = 0 and solve x(1  4x)  5 = 0, 32 which has complex solutions. If y = 0, then we must have 1  2y  5x = 0 Substituting we obtain x 1  4x  Simplifying, we obtain 3 1 5  x2 + x  = 0, 2 2 32 which also has only complex solutions. Hence there are no critical points. The phase plane diagram shows that species y survives while the x dies off. A phase plane diagram is given in Figure B.63 in the answers of the text. EXERCISES 12.4: Energy Methods, page 774 1 5  x 2 2  5 = 0. 32 y= 1 5  x. 2 2 3. Here g(x) = x2 /(x  1) = x + 1 + 1/(x  1). By integrating g(x), we obtain the potential function G(x) = 716 x2 + x + ln x  1 + C, 2 Exercises 12.4
and so E(x, v) = Since E(0, 0) = 0 implies C = 0, let E(x, v) = v 2 x2 + + x + ln x  1. 2 2 v 2 x2 + + x + ln x  1 + C. 2 2 Now, since we are interested in E near the origin, we let x  1 = 1  x (because for x near 0, x  1 < 0). Therefore, E(x, v) = v 2 x2 + + x + ln(1  x). 2 2 9. Here we have g(x) = 2x2 + x  1 and hence the potential function G(x) = 2x3 x2 +  x. 3 2 The local maxima and minima of G(x) occur when G (x) = g(x) = 2x2 + x  1 = 0. Thus the phase plane diagram has critical points at (1, 0) and (1/2, 0). Since G(x) has a strict local minimum at x = 1/2, the critical point (1/2, 0) is a center. Furthermore, since x = 1 is strict local maximum, the critical point (1, 0) is a saddle point. A sketch of the potential plane and phase plane diagram is given in Figure B.65 in the answers of the text. 11. Here we have g(x) = x/(x  2) = 1 + 2/(x  2) so the potential function is G(x) = x + 2 ln x  2 = x + 2 ln(2  x), for x near zero. Local maxima and minima of G(x) occur when G (x) = g(x) = x/(x  2) = 0. Thus the phase plane diagram has critical points at (0, 0). Furthermore we note that x = 2 is not in the domain of g(x) nor of G(x). Now G(x) has a strict local maximum at x = 0, hence the critical point (0, 0) is a saddle point. A sketch of the potential plane and phase plane diagram for x < 2 is given in Figure B.66 in the answers of the text. 13. We first observe that vh(x, v) = v 2 > 0 for v = 0. Hence, the energy is continually decreasing along a trajectory. The level curves for the energy function E(x, v) = v 2 x2 x4 +  2 2 4 717 Chapter 12
are just the integral curves for Example 2(a) and are sketch in Figure 12.22 on page 770 of the text. The critical points for this damped system are the same as in the example and moreover, they are of the same type. The resulting phase plane is given in Figure B.67 in the answers of the text. EXERCISES 12.5: Lyapunov's Direct Method, page 782 3. We compute V (x, y) with V (x, y) = x2 + y 2. V (x, y) = Vx (x, y)f (x, y) + Vy (x, y)g(x, y) = 2x y 2 + xy 2  x3 + 2y xy + x2 y  y 3 = 4x2 y 2  2x4  2y 4 = 2 x2  y 2
2 . According to Theorem 3, since V is negative semidefinite, V is positive definite function, and (0, 0) is an isolated critical point of the system, the origin is stable. 5. The origin is an isolated critical point for the system. Using the hint, we compute V (x, y) with V (x, y) = x2  y 2. Computing, we obtain V (x, y) = Vx (x, y)f (x, y) + Vy (x, y)g(x, y) = 2x 2x3  2y 2x2 y  y 3 = 4x4  4x2 y 2 + 2y 2 = 2x4 + x2  y 2
2 , which is positive definite. Now V (0, 0) = 0, and in every disk centered at the origin, V is positive at some point (namely, those points where x > y). Therefore, by Theorem 4, the origin is unstable. 7. We compute V (x, y) with V (x, y) = ax4 + by 2 . V (x, y) = Vx (x, y)f (x, y) + Vy (x, y)g(x, y) = 4ax3 2y  x3 + 2by x3  y 5 = 8ax3 y  4ax6  2bx3 y  2by 6 . 718 Exercises 12.6
To eliminate the x3 y term, we let a = 1 and b = 4, then V (x, y) = 4x6  8y 6, and we get that V is negative definite. Since V is positive definite and the origin is an isolated critical point, according to Theorem 3, the origin is asymptotically stable. 11. Here we set y= Then, we obtain the system dx = y, dt dy =  1  y2 y  x . dt Clearly, the zero solution is a solution to this system. To apply Lyapunov's direct method, we try the positive definite function V (x, y) = ax2 + by 2 and compute V . V (x, y) = Vx (x, y)f (x, y) + Vy (x, y)g(x, y) = 2ax (y) + 2by  1  y 2 y  x = 2axy  2by 2 + 2by 4  2bxy . To eliminate the xy terms, we choose a = b = 1, then V (x, y) = 2y 2 1  y 2 . Hence V is negative semidefinite for y < 1, so by Theorem 3, the origin is stable. EXERCISES 12.6: 5. We compute r r dr : dt Limit Cycles and Periodic Solutions, page 791 dx dt dy d2 x = 2 . dt dt dx dy dr = x +y = x x  y + x r 3  4r 2 + 5r  3 + y x + y + y r 3  4r 2 + 5r  3 dt dt dt = x2  xy + x2 r 3  4r 2 + 5r  3 + xy + y 2 + y 2 r 3  4r 2 + 5r  3 = r 2 + r 2 r 3  4r 2 + 5r  3 = r 2 r 3  4r 2 + 5r  2 . 719 Chapter 12
Hence dr = r r 3  4r 2 + 5r  2 = r(r  1)2 (r  2). dt Now dr/dt = 0 when r = 0, 1, 2. The critical point is represented by r = 0, and when r = 1 or 2, we have limit cycles of radius 1 and 2. When r lies in (0, 1), we have dr/dt < 0, so a trajectory in this region spirals into the origin. Therefore, the origin is an asymptotically stable spiral point. Now, when r lies in (1, 2), we again have dr/dt < 0, so a trajectory in this region spirals into the limit cycle r = 1. This tells us that r = 1 is a semistable limit cycle. Finally, when r > 2, dr/dt > 0, so a trajectory in this region spirals away from the limit cycle r = 2. Hence, r = 2 is an unstable limit cycle. To find the direction of the trajectories, we compute r 2 r2 d . dt dy dx d = x y = x x + y + y r 3  4r 2 + 5r  3  y x  y + x r 3  4r 2 + 5r  3 dt dt dt = x2 + xy + xy r 3  4r 2 + 5r  3  xy + y 2  xy r 3  4r 2 + 5r  3 = x2 + y 2 = r 2 . Hence d/dt = 1, which tells us that the trajectories revolve counterclockwise about the origin. A phase plane diagram is given in Figure B.74 in the answers of the text. 11. We compute r dr/dt: r 1 dr dx dy = x +y = x y + x sin dt dt dt r 1  xy + y 2 sin = xy + x2 sin r dr = r sin dt isolated critical point. Observe that dr >0 dt 720 for 1 1 <r< , (2n + 1) 2n 1 , r + y x + y sin 1 r = r 2 sin 1 . r 1 r Hence, and dr/dt = 0 when r = 1/(n), n = 1, 2, . . . . Consequently, the origin (r = 0) is not an Exercises 12.6
dr <0 dt for 1 1 <r< . 2n (2n  1) Thus, trajectories spiral into the limit cycles r = 1/(2n) and away from the limit cycles r = 1/[(2n + 1)]. To determine the direction of the spiral, we compute r 2 d/dt. r2 dy dx 1 d = x y = x x + y sin dt dt dt r 1  y 2  xy sin = x2 + xy sin r  y y + x sin 1 r = r 2 . 1 r Hence d/dt = 1, which tells us that the trajectories revolve clockwise about the origin. A phase plane diagram is given in Figure B.77 in the answers of the text. 15. We compute fx + gy in order to apply Theorem 6. Thus fx (x, y) + gy (x, y) = 8 + 3x2 + 7 + 3y 2 = 3 x2 + y 2  5 , which is less than 0 for the given domain. Hence, by Theorem 6, there are no nonconstant periodic solutions in the disk x2 + y 2 < 5. 19. It is easily seen that (0, 0) is a critical point, however, it is not easily shown that it is the only critical point for this system. Using the Lyapunov function V (x, y) = 2x2 + y 2, we compute V (x, y). Thus dx dy V (x, y) = Vx (x, y) + Vy (x, y) dt dt = 4x 2x  y  2x3  3xy 2 + 2y 2x + 4y  4y 3  2x2 y = 8x2  8x4  16x2 y 2 + 8y 2  8y 4 = 8 x2 + y 2  8 x2 + y 2
2 . Therefore, V (x, y) < 0 for x2 + y 2 > 1 and V (x, y) > 0 for x2 + y 2 < 1. Let C1 be the curve 2x2 + y 2 = 1/2, which lies inside x2 + y 2 = 1, and let C2 be the curve 2x2 + y 2 = 3, which lies outside x2 + y 2 = 1. Now V (x, y) > 0 on C1 and V (x, y) < 0 on C2 . Hence, we let R be the region between the curves C1 and C2 . Now, any trajectory that enters R is contained in R. So by Theorem 7, the system has a nonconstant periodic solution in R. 721 Chapter 12
25. To apply Theorem 8, we check to see that all five conditions hold. Here we have g(x) = x and f (x) = x2 (x2  1). Clearly, f (x) is even, hence condition (a) holds. Now
x F (x) =
0 s 2 x5 x3  . s  1 ds = 5 3
2 Hence F (x) < 0 for 0 < x < x > 5/3 and F (x) > 0 for x > 5/3. Therefore, condition (b) holds. Furthermore, condition (c) holds since F (x) + as x +, monotonically for 5/3. As stated above, g(x) = x is an odd function with g(x) > 0 for x > 0, thus condition (d) holds. Finally, since
x G(x) =
0 s ds = x2 , 2 we clearly have G(x) + as x +, hence condition (e) holds. It follows from Theorem 8, that the Lienard equation has a unique nonconstant periodic solution. EXERCISES 12.7: Stability of HigherDimensional Systems, page 798 5. From the characteristic equation (r  1) r 2 + 1 = 0, we find that the eigenvalues are 1, i. Since at least one eigenvalue, 1, has a positive real part, the zero solution is unstable. 9. The characteristic equation is r2 + 1 r 2 + 1 = 0, which has eigenvalues i, i. Next we determine the eigenspace for the eigenvalue i. Computing we find i 1 1 1 0 0 722 i 0 0 0 1 i 0 0 0 0 1 0 0 0 0 1 0 0 0 0 . 0 1 i 1 1 i Exercises 12.7
Hence the eigenspace is degenerate and by Problem 8(c) on page 798 of the text, the zero solution is unstable. Note: it can be shown that the eigenspace for the eigenvalue i is also degenerate. 13. To find the fundamental matrix for this system we first recall the Taylor series ex , sin x, and cos x. These are x2 x3 + + , 2! 3! x3 x5 sin x = x  +  , 3! 5! x2 x4 +  . cos x = 1  2! 4! ex = 1 + x + Hence x2 dx1 x2 x2 x2 1 = 1  x1 + 1  + 1  2 +  2 = x1 +  +  2 + , dt 2! 2! 2! 2! 3 3 x x dx2 = x2 + x3  3 + = x2  x3 +  3 + , dt 3! 3! (x2 + x3 )2 dx3 (x2 + x3 )2 = 1  1 + (x2 + x3 ) + + = x2  x3  + . dt 2! 2! Thus, A= Calculating the eigenvalues, we have 1  r A  rI = 0 0 0 1 = 0. 0 0 1 1 . 0 1 1 0 1 0 1  r 1 1  r Hence, the characteristic equation is (r + 1)(r 2 + 2r + 2) = 0. Therefore, the eigenvalues are 1, 1 i. Since the real part of each is negative, the zero solution is asymptotically stable. 723 Chapter 12
15. Solving for the critical points, we must have x1 + 1 = 0, 2x1  x2 + 2x3  4 = 0, 3x1  2x2  x3 + 1 = 0. Solving this system, we find that the only solution is (1, 2, 2). We now use the change of variables x1 = u + 1, x2 = v  2, x3 = w + 2 to translate the critical point to the origin. Substituting, we obtain the system du = u, dt dv = 2u  v + 2w, dt dw = 3u  2v  w. dt Here A is given by 1 0 0 A = 2 1 2 . 3 2 1 Finding the characteristic equation, we have (r + 1)(r 2 + 2r + 5) = 0. Hence the eigenvalues are 1, 1 2i. Since each eigenvalue has a negative real part, the critical point (1, 2, 2) is asymptotically stable. 724 CHAPTER 13: Existence and Uniqueness Theory
EXERCISES 13.1: Introduction: Successive Approximations, page 812 1. In this problem, x0 = 1, y0 = y(x0 ) = 4, and f (x, y) = x2  y. Thus, applying formula (3) on page 807 of the text yields
x x x x y(x) = y0 +
x0 f (t, y(t)) dt = 4 +
1 t  y(t) dt = 4 +
1 2 t dt 
1 2 y(t) dt . Since x t2 dt =
1 t3 3 x =
1 x3 1  , 3 3 the equation becomes x3 1   y(x) = 4 + 3 3
x 1 11 x3 +  y(t) dt = 3 3 x y(t) dt .
1 3. In the initial conditions, x0 = 1 and y0 = 3. Also, f (x, y) = (y  x)2 = y 2  2xy + x2 . Therefore,
x x y(x) = y0 +
x0 f (t, y(t)) dt = 3 +
1 y 2 (t)  2ty(t) + t2 dt. Using the linear property of integrals, we find that
x x x x y 2(t)  2ty(t) + t2 dt =
1 1 x y 2(t) dt  2
1 ty(t) dt +
1 t2 dt =
1 y 2 (t)  2ty(t) dt + x3 1  , 3 3 725 Chapter 13
we have x3 10 y(x) =  + 3 3
x x y (t) dt  2
1 1 2 ty(t) dt. (13.1) Note that we can rewrite the last integral using integration by parts in terms of integrals of the function y(x) alone. Namely,
x t t=x x t x x t ty(t) dt = t
1 1 y(s) ds
t=1 
1 1 y(s) ds dt = x
1 y(s) ds 
1 1 y(s) ds dt Thus, another form of the answer (13.1) is x3 10  + y(x) = 3 3 5. In this problem, we have g(x) = 1 2 x+ 3 x .
x x x t y (t) dt + x
1 1 2 y(t) dt 
1 1 y(s) ds dt . Thus the recurrence formula (7) on page 807 of the text becomes xn+1 = g (xn ) = 1 2 xn + 3 xn , n = 0, 1, . . . . With x0 = 3 as an initial approximation, we compute x1 = 1 2 x0 + 3 x0 = 1 2 3+ 3 3 = 2.0 , x2 = 1 2 x1 + 3 x1 = 1 2 2+ 3 2 = 1.75 , and so on. The results of these computations is given in Table 13A. 1 2 3 . x Table 13A: Approximations for a solution of x =
x0 = 3.0 x1 = 2.0 x2 = 1.75 x3 = 1.732142857 x4 = 1.732050810 x5 = 1.732050808 x+ We stopped iterating after x5 because x4  x5 < 108 . Hence x 1.73205081. 726 Exercises 13.1
7. Since g(x) = 1/ (x2 + 4), we have the recurrence formula xn+1 = g (xn ) = x2 n 1 , +4 n = 0, 1, . . . with an initial approximation x0 = 0.5. Hence x1 = 1 1 4 = = 0.2352941176, 2+4 +4 (0.5) 17 1 1 0.2465870307 , x2 = 2 x1 + 4 (0.2352941176)2 + 4 x2 0 etc. See Table 13B. We stopped iterating after x7 because the error x6  x7 < 109 . Hence x 0.24626617. 1 . +4 Table 13B: Approximations for a solution of x =
x0 x1 x2 x3 = 0.5 = 0.2352941176 = 0.2465870307 = 0.2462565820 x4 x5 x6 x7 x2 = 0.2462664586 = 0.2462661636 = 0.2462661724 = 0.2462661721 9. To start the method of successive substitutions, we observe that g(x) = 5x 3
1/4 . Therefore, according to equation (7) on page 807 of the text, we can find the next approximation from the previous one by using the recurrence relation xn+1 = g (xn ) = 5  xn 3
1/4 . We start the procedure at the point x0 = 1. Thus, we obtain x1 = 5  x0 3
1/4 = 51 3 1/4 = 4 3 1/4 1.074569932 , 727 Chapter 13
x2 = x3 = 5  x1 3 5  x2 3
1/4 1/4 5  1.074569932 3 5  1.069526372 3 1/4 1.069526372 ,
1/4 1.069869749 . By continuing this process, we fill in Table 13C below. Noticing that x7  x6 < 108 , we stopped the procedure after seven steps. So, x 1.06984787. 5x 3
1/4 Table 13C: Approximations for a solution of x =
x0 x1 x2 x3 = 1.0 = 1.074569932 = 1.069526372 = 1.069869749 x4 x5 x6 x7 . = 1.069846382 = 1.069847972 = 1.069847864 = 1.069847871 11. First, we derive an integral equation corresponding to the given initial value problem. We have f (x, y) = y, x0 = 0, y0 = y(0) = 2, and so the formula (3) on page 807 of the text yields y(x) = 2 +
0 x x [y(t)] dt = 2 
0 y(t) dt . Thus, Picard's recurrence formula (15) becomes
x yn+1 (x) = 2 
0 yn (t) dt, n = 0, 1, . . . . Starting with y0 (x) y0 = 2, we compute
x x y1 (x) = 2 
0 x y0 (t) dt = 2 
0 x 2 dt = 2  2t t=x = t=0 2  2x ,
t=x = t=0 y2 (x) = 2 
0 y1 (t) dt = 2 
0 (2  2t) dt = 2 + (t  1)2 2  2x + x2 . 728 Exercises 13.1
13. In this problem, f (x, y) = 3x2 , x0 = 1, y0 = y(1) = 2, and so Picard's iterations to the solution of the given initial value problem are given by
x yn+1(x) = 2 +
1 3t2 dt = 2 + t3 t=x = t=1 x3 + 1 . Since the righthand side does not depend on n, the sequence of iterations yk (x), k = 1, 2, . . ., is a constant sequence. That is, yk (x) = x3 + 1 In particular, y1 (x) = y2 (x) = x3 + 1. (In this connection, note the following. If it happens that one of the iterations, say, yk (x), obtained via (15) matches the exact solution to the integral equation (3), then all the subsequent iterations will give the same function yk (x). In other words, the sequence of iterations will become a constant sequence starting from its kth term. In the given problem, the first application of (15) gives the exact solution, x3 + 1, to the original initial value problem and, hence, to the corresponding integral equation (3).) 15. We first write this differential equation as an integral equation. Integrating both sides from x0 = 0 to x and using the fact that y(0) = 0, we obtain
x x for any k 1. y(x)  y(0) =
0 y(t)  et dt y(x) =
0 y(t)  et dt . Hence, by equation (15) on page 811 of the text, the Picard iterations are given by
x yn+1 (x) =
0 yn (t)  et dt . Thus, starting with y0 (x) y0 = 0, we calculate
x x y1 (x) =
0 y0 (t)  e dt = 
t 0 et dt = 1  ex , 729 Chapter 13
x x y2 (x) =
0 y1 (t)  e dt =
t 0 1  2et dt = 1  ex = 2 + x  2ex . 17. First of all, remark that the function f (x, y(x)) in the integral equation (3), that is,
x y(x) = y0 +
x0 f (t, y(t)) dt , is a continuous function as the composition of f (x, y) and y(x), which are both continuous by our assumption. Next, if y(x) satisfies (3), then
x0 y (x0 ) = y0 +
x0 f (t, y(t)) dt = y0 , because the integral term is zero as a definite integral of a continuous function with equal limits of integration. Therefore, y(x) satisfies the initial condition in (1). We recall that, by the fundamental theorem of calculus, if g(x) is a continuous function on an interval [a, b], then, for any fixed c in [a, b], the function G(x) := for g(x) on (a, b), i.e.,
x x a g(t) dt is an antiderivative G (x) =
a g(t) dt = g(x). Thus, y (x) = y0 + x f (t, y(t)) dt = f (t, y(t))
x0 t=x = f (x, y(x)), and so y(x) satisfies the differential equation in (1). 19. The graphs of the functions y = (x2 + 1)/2 and y = x are sketched on the same coordinate axes in Figure 13A. By examining this figure, we see that these two graphs intersect only at (1, 1). We can find this point by solving the equation x= 730 x2 + 1 , 2 Exercises 13.1
y= x2 + 1 2 P1 Q1 y=x 4 3
P0 Q0 2 1
P1 P0 Q1 Q0 0 1 2 3 4 Figure 13A: The method of successive substitution for the equation x = x2 + 1 . 2 for x. Thus, we have 2x = x2 + 1 x2  2x + 1 = 0 (x  1)2 = 0 x = 1. Since y = x, the only intersection point is (1, 1). To approximate the solution to the equation x = (x2 + 1)/2 using the method of successive substitutions, we use the recurrence relation xn+1 = x2 + 1 n . 2 Starting this method at x0 = 0, we obtain the approximations given in Table 13D. These approximations do appear to be approaching the solution x = 1. However, if we start the process at the point x = 2, we obtain the approximations given in Table 13E. We observe that these approximations are getting larger and so do not seem to approach a fixed point. This also appears to be the case if we examine the pictorial representation for the 731 Chapter 13
x2 + 1 starting at x0 = 0. Table 13D: Approximations for a solution of x = 2
x1 = 0.5 x2 = 0.625 x3 = 0.6953125 x4 = 0.7417297 x5 = 0.7750815 x10 = 0.8610982 x15 x20 x30 x40 x50 x99 = 0.89859837 = 0.91988745 = 0.94337158 = 0.95611749 = 0.96414507 = 0.98102848 Table 13E: Approximations for a solution of x =
x1 = 2.5 x2 = 3.625 x3 = 7.0703125 x2 + 1 starting at x0 = 2. 2 x4 = 25.4946594 x5 = 325.488829 x6 = 52971.9891 method of successive substitutions given in Figure 13A. By plugging x0 = 0 into the function (x2 + 1)/2, we find P0 to be the point (0, 0.5). Then by moving parallel to the xaxis from the point P0 to the line y = x, we observe that Q0 is the point (0.5, 0.5). Next, by moving parallel to the yaxis from the point Q0 to the curve y = (x2 + 1)/2, we find that P1 is the point (0.5, 0.625). Continuing this process moves us slowly in a step fashion to the point (1, 1). However, if we start this process at x0 = 2, we observe that this method moves us through larger and larger steps away from the point of intersection (1, 1). Note that for this equation, the movement of the method of successive substitutions is to the right. This is because the term (x2 + 1)/2, in the recurrence relation, is increasing for n x > 0. Thus, the sequence of approximations {xn } is an increasing sequence. Starting at a nonnegative point less than 1 moves us to the fixed point at x = 1, but starting at a point larger that 1 moves us to ever increasing values for our approximations and, therefore, away from the fixed point. 732 Exercises 13.2
EXERCISES 13.2: Picard's Existence and Uniqueness Theorem, page 820 1. In order to determine whether this sequence of functions converges uniformly, we find yn  y . Since yn (x)  y(x) = 1  we have
x[1,1] x x 1= , n n 1 x = . x[1,1] n n yn  y = max yn (x)  y(x) = max Thus 1 =0 n n n and {yn (x)} converges to y(x) uniformly on [1, 1]. lim yn  y = lim 3. In order to determine whether this sequence of functions converges uniformly, we must find
n lim yn  y . Therefore, we first compute yn  y = yn = max
x[0,1] nx nx = max , 2 x2 x[0,1] 1 + n2 x2 1+n where we have removed the absolute value signs because the term (nx)/(1 + n2 x2 ) is nonnegative when x [0, 1]. We will use calculus methods to obtain this maximum value. Thus, we differentiate the function yn (x) = (nx)/(1 + n2 x2 ) to obtain yn (x) = n(1  n2 x2 ) . (1 + n2 x2 )2 Setting yn (x) equal to zero and solving yields n 1  n2 x2 = 0 n2 x2 = 1 1 x= . n Since we are interested in the values of x on the interval [0, 1], we will only examine the critical point x = 1/n. By the first derivative test, we observe that the function yn (x) has a local maximum value at the point x = 1/n. At this point, we have yn 1 n = n (n1 ) 1 = . 2 (n1 )2 2 1+n 733 Chapter 13
Computing n(0) = 0, 1 + n2 (0)2 n(1) n 1 1 = < yn (1) = 2 (1)2 2 1+n 1+n n 2 yn (0) = we conclude that
x[0,1] for n 2, max yn (x) = 1 . 2 Therefore, 1 1 = = 0. n n 2 2 Thus, the given sequence of functions does not converge uniformly to the function y(x) 0 lim yn  y = lim on the interval [0, 1]. This sequence of functions does, however, converge pointwise to the function y(x) 0 on the interval [0, 1]. To see this, notice that for any fixed x (0, 1] we have
n lim [yn (x)  y(x)] = lim nx 1 = 0, = lim 2 x2 n 1 + n n 2nx where we have found this limit by using L'Hospital's rule. At the point x = 0, we observe that 0 = 0. 1 Thus, we have pointwise convergence but not uniform convergence. See Figure 13B(a) for
n lim [yn (0)  y(0)] = lim n the graphs of functions y1 (x), y10 (x), y30 (x), and y90 (x). 5. We know (as was stated on page 433 of the text) that for all x such that x < 1 the geometric series, k k=0 x , converges to the function f (x) = 1/(1  x). Thus, for all x [0, 1/2], we have 1 == 1 + x + x2 + + xk + = 1x xk .
k=0 Therefore, we see that
n yn  y = max yn (x)  y(x) = max
x[0,1/2] x[0,1/2] x 
k k=0 k=0 x k = max x[0,1/2] x =
k=n+1 k=n+1 k 1 2 k 734 Exercises 13.2
0.5 y1 (x) 20 y20 (x) y15 (x)
10 y10 (x) y10 (x) y30 (x) y90 (x)
1 y5 (x)
0 0.2 0.4 0.6 0.8 1 0 (a) (b) Figure 13B: Graphs of functions in Problems 3 and 7. and so
n lim yn  y = lim n k=n+1 1 2 k . Since have k k=n+1 (1/2) is the tail end of a convergent series, its limit must be zero. Hence, we n lim yn  y = lim n k=n+1 1 2 k = 0. Therefore, the given sequence of functions converges uniformly to the function y(x) = 1/(1x) on the interval [0, 1/2]. 7. Let x [0, 1] be fixed. If x = 0, then yn (0) = n2 (0) = 0 for any n and so limn yn (0) = limn 0 = 0. For x > 0, let Nx := [2/x] + 1 with denoting the interger part of a number. Then, for 2 2 2 +1> x> x x n and so, in evaluating yn (x), the third line in its definition must be used. This yields yn (x) = 0 n for all n Nx , which implies that limn yn (x) = limn 0 = 0. 735 n Nx , one has Chapter 13
Hence, for any fixed x [0, 1], limn yn (x) = 0 = y(x). On the other hand, for any n, the function yn (x) is a continuous piecewise linear function, which is increasing on [0, 1/n], decreasing on (1/n, 2/n), and zero on [1/n, 1]. Thus it attains its maximum value at x = 1/n, which is yn Therefore,
n 1 n = n2 1 n = n. lim yn  y = lim yn = lim n = ,
n n and the sequence does not converge uniformly on [0, 1]. See Figure 13B(b) for the graphs of y5 (x), y10 (x), y15 (x), and y20 (x). 9. We need to find an h > 0 such that h < min (h1 , 1 /M, 1/L). We are given that R1 = {(x, y) : x  1 1, y 1} = {(x, y) : 0 x 2, 1 y 1} , and so h1 = 1 and 1 = 1. Thus, we must find values for M and L. In order to find M, notice that, as was stated on page 816 of the text, we require that M satisfy the condition f (x, y) = y 2  x M, for all (x, y) in R1 . To find this upper bound for f (x, y), we must find the maximum and the minimum values of f (x, y) on R1 . (Since f (x, y) is a continuous function on the closed and bounded region R1 , it will have a maximum and a minimum there.) We will use calculus methods to find this maximum and this minimum. Since the first partial derivatives of f (x, y), given by fx (x, y) = 1, and fy (x, y) = 2y, are never both zero, the maximum and minimum must occur on the boundary of R1 . Notice that R1 is bounded on the left by the line x = 0, on the right by the line x = 2, on the top by the line y = 1, and on the bottom by the line y = 1. Therefore, we will examine the behavior of f (x, y) (and, thus, of f (x, y)) on each of these lines. 736 Exercises 13.2
Case 1: On the left side of R1 where x = 0, the function f (x, y) becomes the function in the single variable y, given by f (0, y) = F1 (y) = y 2  0 = y 2 , y [1, 1]. This function has a maximum at y = 1 and a minimum at y = 0. Thus, on the left side of R1 we see that f reaches a maximum value of f (0, 1) = 1 and a minimum value of f (0, 0) = 0. Case 2: On the right side of R1 where x = 2, the function f (x, y) becomes the function in the single variable y, given by f (2, y) = F2 (y) = y 2  2, y [1, 1]. This function also has a maximum at y = 1 and a minimum at y = 0. Thus, on the right side of R1 , the function f (x, y) reaches a maximum value of f (2, 1) = 1 and a minimum value of f (2, 0) = 2. Case 3: On the top and bottom of R1 where y = 1, the function f (x, y) becomes the function given by f (x, 1) = F3 (x) = (1)2  x = 1  x, x [0, 2]. This function also has a maximum at x = 0 and a minimum at x = 2. Thus, on both the top and bottom of the region R1 , the function f (x, y) reaches a maximum value of f (0, 1) = 1 and a minimum value of f (2, 1) = 1. From the above cases we see that the maximum value of f (x, y) is 1 and the minimum value is 2 on the boundary of R1 . Thus, we have f (x, y) 2 for all (x, y) in R1 . Hence, we choose M = 2. To find L, we observe that L is an upper bound for f = 2y = 2y, y on R1 . Since y [1, 1] in this region, we have y 1. Hence, we see that f = 2y 2, y 737 Chapter 13
for all (x, y) in R1 . Therefore, we choose L = 2. Now we can choose h 0 such that h < min h1 , 1 1 , M L = min 1, 11 22 = 1 . 2 Thus, Theorem 3 guarantees that the given initial value problem will have a unique solution on the interval [1  h, 1 + h], where 0 < h < 1/2. 11. We are given that the recurrence relation for these approximations is yn+1 = T [yn ]. Using the definition of T [y], we have
x yn+1 = x  x + 1
0 3 (u  x)yn (u) du. Thus, starting these approximations with y0 (x) = x3  x + 1, we obtain
x x y1 (x) = x3  x + 1 +
0 x (u  x)y0 (u) du = x3  x + 1 +
0 (u  x) u3  u + 1 du = x3  x + 1 +
0 (u4  u2 + u  xu3 + xu  x) du x5 x3 x2 x5 x3  +  +  x2 . 5 3 2 4 2 = x3  x + 1 + By simplifying, we obtain y1 (x) =  1 5 7 3 1 2 x + x  x  x + 1. 20 6 2 Substituting this result into the recurrence relation yields
x y2 (x) = x3  x + 1
0 x 3 (u  x)y1 (u) du 1 5 7 3 1 2 u + u  u  u + 1 du 20 6 2 = x x+1
0 (u  x)  738 Exercises 13.2
= x3  x + 1 +  1 7 7 5 1 4 1 3 1 2 x + x  x  x + x 140 30 8 3 2 1 7 7 5 1 4 1 3 x  x + x + x  x2 . + 120 24 6 2 When simplified, this yields y2 (x) = 1 7 7 5 1 4 7 3 1 2 x  x + x + x  x  x + 1. 840 120 24 6 2 13. Using properties of limits and the linear property of integrals, we can rewrite the statement that
b n a b lim yn (x) dx =
a y(x) dx in an equivalent form n b b y(x) dx = 0 n a b lim a yn (x) 
a lim [yn (x)  y(x)] dx = 0. (13.2) The sequence {yn } converges uniformly to y on [a, b], which means, by the definitionof uniform convergence, that yn  y Since
b b C[a,b] := max yn (x)  y(x) 0 as n .
x[a,b] [yn (x)  y(x)] dx a a yn (x)  y(x) dx (b  a) yn  y C[a,b] 0 as n , we conclude that
b n a lim [yn (x)  y(x)] dx = 0, and (13.2) follows. (Recall that limn an = 0 if and only if limn an  = 0.) 739 Chapter 13
15. (a) In the given system, x (t) = y 2 (t), y (t) = z(t), x(0) = 0; y(0) = 1; (13.3) z (t) = x(t)y(t), z(0) = 0, replacing t by s, integrating the differential equations from s = 0 to s = t, and using the fundamental theorem of calculus we obtain
t 0 t 0 t x (s) ds =  y 2 (s) ds;
0 t t x(t)  x(0) =  y(t)  y(0) = z(t)  z(0) =
t 0 t t 0 y 2(s) ds; y (s) ds =
0 t z(s) ds; x(s)y(s) ds
0 z(s) ds; x(s)y(s) ds.
0 z (s) ds =
0 By the initial conditions in (13.3), x(0) = 0, y(0) = z(0) = 1. Substituting these values into the above system yields x(t) = 
t 0 t y 2(s) ds; z(s) ds;
0 t y(t)  1 = z(t)  1 = (13.4) x(s)y(s) ds,
0 which is equivalent to the given system of integral equations. Thus, (13.3) implies (13.4). Conversely, differentiating equations in (13.4) and using the fundamental theorem of calculus (its part regarding integrals with variable upper bound), we conclude that solutions x(t), y(t), and z(t) to (13.4) also satisfy differential equations in (13.3). Clearly, x(0) =  y 2 (s) ds = 0;
0 0 y(0)  1 = z(t)  1 = 0 z(s) ds = 0;
0 0 x(s)y(s) ds = 0,
0 and the initial conditions in (13.3) are satisfied. Therefore, (13.4) implies (13.3). 740 Exercises 13.3
(b) With starting iterations x0 (t) x(0) = 0, y0 (t) y(0) = 1, and z0 (t) z(0) = 1, we compute x1 (t), y1 (t), and z1 (t).
2 x1 (t) =  y0 (s) ds =  (1)2 ds = t; 0 0 t t t t y1 (t) = 1 + z1 (t) = 1 + 0 t 0 z0 (s) ds = 1 + (1) ds = 1 + t;
0 t 0 x0 (s)y0 (s) ds = 1 + (0) ds = 1. Applying given recurrence formulas again yields
2 x2 (t) =  y1 (s) ds =  (1 + s)2 ds = (1 + s)3 /3 0 0 t t t t t = 0 t  t2  t3 ; 3 t2 t3 1  . 2 3 y2 (t) = 1 + z2 (t) = 1 + EXERCISES 13.3: 1. In this problem, 0 t 0 z1 (s) ds = 1 + 1 ds = 1 + t;
0 t 0 x1 (s)y1 (s) ds = 1  s(1 + s) ds = 1  (s /2 + s /3) 2 3 t = 0 Existence of Solutions of Linear Equations, page 826 A(t) = cos t t t3 1 , f(t) = tan t et . t is continuous on In A(t), functions cos t, t3 , and 1 are continuous on (, ) while [0, ). Therefore, A(t) is continuous on [0, ). In f(t), the exponential function is continuous everywhere, but tan t has infinite discontinuities at t = (k + 1/2), k = 0, 1, 2, . . . . The largest interval containing the initial point, t = 2, where tan t and, therefore, f(t), is continuous is (/2, 3/2). Since A(t) is also continuous on (/2, 3/2), by Theorem 6, given initial value problem has a unique solution on this interval. 3. By comparing this problem to the problem given in (14) on page 825 of the text, we see that in this case p1 (t) =  ln t, p2 (t) 0, p3 (t) = tan t, and g(t) = e2t . 741 Chapter 13
We also observe that t0 = 1. Thus, we must find an interval containing t0 = 1 on which all of the functions p1 (t), p2 (t), p3 (t), and g(t) are simultaneously continuous. Therefore, we note that p2 (t) and g(t) are continuous everywhere; p1 (t) is continuous on the interval (0, ); and the interval which contains t0 = 1 on which p3 (t) is continuous is (/2, /2). Hence, these four functions are simultaneously continuous on the interval (0, /2) and this interval contains the point t0 = 1. Therefore, Theorem 7 given on page 825 of the text guarantees that we will have a unique solution to this initial value problem on the whole interval (0, /2). 5. In this problem, we use Theorem 5. Since f(t, x) = we have f (t, x) = x1 0 3 , f (t, x) = x2 cos x2 0 . sin x2 3x1 , Vectors f, f/x1 , and f/x2 are continuous on R = { < t < ,  < x1 < ,  < x2 < } (which is the whole space R3 ) since their components are. Moreover, f (t, x) = 3, x1 f (t, x) =  cos x2  1 x2 for any (t, x), and the condition (3) in Theorem 5 is satisfied with L = 3. Hence, given initial value problem has a unique solution on the whole real axis  < t < . 7. The equation y (t)  (sin t)y (t) + et y(t) = 0 is a linear homogeneous equation and, hence, has a trivial solution, y(t) 0. Clearly, this solution satisfies the initial conditions, y(0) = y (0) = y (0) = 0. All that remains to note is that the coefficients,  sin t and et , are continuous on (, ) and so, by Theorem 7, the solution y 0 is unique. 742 Exercises 13.4
EXERCISES 13.4: Continuous Dependence of Solutions, page 832 3. To apply Theorem 9, we first determine the constant L for f (x, y) = ecos y + x2 . To do this, we observe that f (x, y) = ecos y sin y. y Now on any rectangle R0 , we have f (x, y) = ecos y sin y = ecos y  sin y e. y (More detailed analysis shows that this function attains its maximum at y = ( 5  1)/2, and this maximum equals to 1.4585 . . . .) Thus, since h = 1, we have by Theorem 9, (x, y0 )  (x, y0) y0  y0  ee . Since we are given that y0  y0  102 , we obtain the result (x, y0 )  (x, y0 ) 102 ee 0.151543 . 9. We can use inequality (18) in Theorem 10 to obtain the bound, but first must determine the constant L and the constant . Here f (x, y) = sin x + (1 + y 2 )1 and F (x, y) = x + 1  y 2 . Now, 2y f (x, y) = y (1 + y 2 )2 and F (x, y) = 2y 2. y To find an upper bound for f /y on R0 , we maximize 2y/(1 + y 2)2 . Hence, we obtain 2y (1 + y 2)2 = 2(1 + y 2)2  2y 2(1 + y 2 )2y 2(1 + y 2 )  8y 2 2  6y 2 = = . (1 + y 2 )4 (1 + y 2)3 (1 + y 2)3 Setting this equal to zero and solving for y, we obtain 2  6y 2 =0 (1 + y 2)3 2  6y 2 = 0 1 y = . 3 743 Chapter 13 Since 2y/(1 + y 2 )2 is odd, we need only use y = 1/ 3. Thus f 2/ 3 3 3 (x, y) , = y (1 + 1/3)2 8 and so L = 3 3/8. To obtain we seek an upper bound for f (x, y)  F (x, y) = sin x + 1 1  x  1 + y 2 sin x  x +  1 + y2 . 2 2 1+y 1+y x3 cos , 3! Using Taylor's theorem with remainder we have sin x = x  where 0 x. Thus for 1 x 1 we obtain sin x  x = x  1 x3 cos x3 cos x = . 3! 3! 6 Applying Taylor's theorem with remainder to 1/(1 + y 2 )  1 + y 2 , we obtain g(y) = (1 + y 2 )1  1 + y 2 , g (y) = 2y(1 + y 2 )2 + 2y , g (y) = 2(1 + y 2 )2 + 2(1 + y 2)3 (2y)2 + 2 , g (y) = 4(1 + y 2)3 (2y)  6(1 + y 2)4 (2y)3 + 2(1 + y 2)3 (8y) . Since g(0) = g (0) = g (0) = 0, we have (1 + y 2 )1  1 + y 2 = where 0 y. Thus, we obtain (1 + y 2)1  1 + y 2 = Hence 1 73 + 12 = . 6 6 It now follows from inequality (18) in Theorem 10 that f (x, y)  F (x, y) (x)  (x) for x in [1, 1]. 744 73 33/8 e 23.294541 , 6 8 + 48 + 16 g () = 12. 3! 6 g () , 3! ...
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 Spring '08
 comech
 Math, Derivative, Tn, Linear Differential Equations, isoclines, Almost Linear Systems

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