CÁLCULO VECTORIAL, ANÁL. DE FOURIER Y ANÁL. COMPLEJO (Solucionario) - Dennis G. Zill.pdf

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Unformatted text preview: SIGUENOS EN: LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS GRATIS EN DESCARGA DIRECTA VISITANOS PARA DESARGALOS GRATIS. Table of Contents Part I Ordinary Differential Equations 1 Introduction to Differential Equations 1 2 First-Order Differential Equations 22 3 Higher-Order Differential Equations 99 4 The Laplace Transform 198 5 Series Solutions of Linear Differential Equations 252 6 Numerical Solutions of Ordinary Differential Equations 317 Part II Vectors, Matrices, and Vector Calculus 7 Vectors 339 8 Matrices 373 9 Vector Calculus 438 Part III Systems of Differential Equations 10 Systems of Linear Differential Equations 551 11 Systems of Nonlinear Differential Equations 604 Part IV Fourier Series and Partial Differential Equations 12 Orthogonal Functions and Fourier Series 634 13 Boundary-Value Problems in Rectangular Coordinates 680 14 Boundary-Value Problems in Other Coordinate Systems 755 15 Integral Transform Method 793 16 Numerical Solutions of Partial Differential Equations 832 Part V Complex Analysis 17 Functions of a Complex Variable 854 18 Integration in the Complex Plane 877 19 Series and Residues 896 20 Conformal Mappings 919 Appendices Appendix II Gamma function 942 Projects 3.7 Road Mirages 944 3.10 The Ballistic Pendulum 946 8.1 Two-Ports in Electrical Circuits 947 8.2 Traffic Flow 948 8.15 Temperature Dependence of Resistivity 949 9.16 Minimal Surfaces 950 14.3 The Hydrogen Atom 952 15.4 The Uncertainity Inequality in Signal Processing 955 15.4 Fraunhofer Diffraction by a Circular Aperture 958 16.2 Instabilities of Numerical Methods 960 Part I Ordinary Differential Equations Introduction to Differential Equations 1 EXERCISES 1.1 Definitions and Terminology 1. Second order; linear 2. Third order; nonlinear because of (dy/dx)4 3. Fourth order; linear 4. Second order; nonlinear because of cos(r + u) 5. Second order; nonlinear because of (dy/dx)2 or  1 + (dy/dx)2 6. Second order; nonlinear because of R2 7. Third order; linear 8. Second order; nonlinear because of x˙ 2 9. Writing the differential equation in the form x(dy/dx) + y 2 = 1, we see that it is nonlinear in y because of y 2 . However, writing it in the form (y 2 − 1)(dx/dy) + x = 0, we see that it is linear in x. 10. Writing the differential equation in the form u(dv/du) + (1 + u)v = ueu we see that it is linear in v. However, writing it in the form (v + uv − ueu )(du/dv) + u = 0, we see that it is nonlinear in u. 11. From y = e−x/2 we obtain y  = − 12 e−x/2 . Then 2y  + y = −e−x/2 + e−x/2 = 0. 12. From y = 6 5 − 65 e−20t we obtain dy/dt = 24e−20t , so that dy + 20y = 24e−20t + 20 dt  6 6 −20t − e 5 5  = 24. 13. From y = e3x cos 2x we obtain y  = 3e3x cos 2x − 2e3x sin 2x and y  = 5e3x cos 2x − 12e3x sin 2x, so that y  − 6y  + 13y = 0. 14. From y = − cos x ln(sec x + tan x) we obtain y  = −1 + sin x ln(sec x + tan x) and y  = tan x + cos x ln(sec x + tan x). Then y  + y = tan x. 15. The domain of the function, found by solving x + 2 ≥ 0, is [−2, ∞). From y  = 1 + 2(x + 2)−1/2 we have (y − x)y  = (y − x)[1 + (2(x + 2)−1/2 ] = y − x + 2(y − x)(x + 2)−1/2 = y − x + 2[x + 4(x + 2)1/2 − x](x + 2)−1/2 = y − x + 8(x + 2)1/2 (x + 2)−1/2 = y − x + 8. 1 1.1 Definitions and Terminology An interval of definition for the solution of the differential equation is (−2, ∞) because y  is not defined at x = −2. 16. Since tan x is not defined for x = π/2 + nπ, n an integer, the domain of y   {x  5x = π/2 + nπ} or {x  x =  π/10 + nπ/5}. From y  = 25 sec2 5x we have = 5 tan 5x is y  = 25(1 + tan2 5x) = 25 + 25 tan2 5x = 25 + y 2 . An interval of definition for the solution of the differential equation is (−π/10, π/10). Another interval is (π/10, 3π/10), and so on.   17. The domain of the function is {x  4 − x2 = 0} or {x  x = −2 or x = 2}. From y  = 2x/(4 − x2 )2 we have   y = 2x 1 4 − x2 2 = 2xy. An interval of definition for the solution of the differential equation is (−2, 2). Other intervals are (−∞, −2) and (2, ∞). √ 18. The function is y = 1/ 1 − sin x , whose domain is obtained from 1 − sin x = 0 or sin x = 1. Thus, the domain  is {x  x = π/2 + 2nπ}. From y  = − 12 (1 − sin x)−3/2 (− cos x) we have 2y  = (1 − sin x)−3/2 cos x = [(1 − sin x)−1/2 ]3 cos x = y 3 cos x. An interval of definition for the solution of the differential equation is (π/2, 5π/2). Another one is (5π/2, 9π/2), and so on. 19. Writing ln(2X −1)−ln(X −1) = t and differentiating implicitly we obtain X 2 dX 1 dX − =1 2X − 1 dt X − 1 dt   2 dX 1 − =1 2X − 1 X − 1 dt 4 2 2X − 2 − 2X + 1 dX =1 (2X − 1)(X − 1) dt -4 dX = −(2X − 1)(X − 1) = (X − 1)(1 − 2X). dt Exponentiating both sides of the implicit solution we obtain -2 2 4 t -2 -4 2X − 1 = et X −1 2X − 1 = Xet − et (et − 1) = (et − 2)X X= et − 1 . et − 2 Solving et − 2 = 0 we get t = ln 2. Thus, the solution is defined on (−∞, ln 2) or on (ln 2, ∞). The graph of the solution defined on (−∞, ln 2) is dashed, and the graph of the solution defined on (ln 2, ∞) is solid. 2 1.1 Definitions and Terminology 20. Implicitly differentiating the solution, we obtain y dy dy −2x2 − 4xy + 2y =0 dx dx −x2 dy − 2xy dx + y dy = 0 4 2 2xy dx + (x2 − y)dy = 0. Using the quadratic formula to solve y 2 − 2x2 y − 1 = 0 for y, we get √ √   y = 2x2 ± 4x4 + 4 /2 = x2 ± x4 + 1 . Thus, two explicit solutions √ √ are y1 = x2 + x4 + 1 and y2 = x2 − x4 + 1 . Both solutions are defined on (−∞, ∞). The graph of y1 (x) is solid and the graph of y2 is dashed. -4 -2 2 4 x -2 -4 21. Differentiating P = c1 et / (1 + c1 et ) we obtain dP (1 + c1 et ) c1 et − c1 et · c1 et c1 et [(1 + c1 et ) − c1 et ] = = 2 dt 1 + c1 et 1 + c1 et (1 + c1 et )   c1 et c1 et 1 − = P (1 − P ). = 1 + c1 et 1 + c1 et x 2 2 2 22. Differentiating y = e−x et dt + c1 e−x we obtain 0  −x2 x2 y =e e −x2 x − 2xe −x2 t2 e dt − 2c1 xe −x2 x = 1 − 2xe 0 et dt − 2c1 xe−x . 2 2 0 Substituting into the differential equation, we have x  −x2 t2 −x2 −x2 e dt − 2c1 xe + 2xe y + 2xy = 1 − 2xe 0 23. From y = c1 e2x + c2 xe2x we obtain x et dt + 2c1 xe−x = 1. 2 2 0 dy d2 y = (2c1 + c2 )e2x + 2c2 xe2x and = (4c1 + 4c2 )e2x + 4c2 xe2x , so that dx dx2 d2 y dy + 4y = (4c1 + 4c2 − 8c1 − 4c2 + 4c1 )e2x + (4c2 − 8c2 + 4c2 )xe2x = 0. −4 2 dx dx 24. From y = c1 x−1 + c2 x + c3 x ln x + 4x2 we obtain dy = −c1 x−2 + c2 + c3 + c3 ln x + 8x, dx d2 y = 2c1 x−3 + c3 x−1 + 8, dx2 and d3 y = −6c1 x−4 − c3 x−2 , dx3 so that x3 d3 y d2 y dy + y = (−6c1 + 4c1 + c1 + c1 )x−1 + (−c3 + 2c3 − c2 − c3 + c2 )x + 2x2 2 − x 3 dx dx dx + (−c3 + c3 )x ln x + (16 − 8 + 4)x2 = 12x2 . 25. From y = −x2 , x < 0 we obtain y  = x2 , x≥0 −2x, x < 0 so that xy  − 2y = 0. 2x, x≥0 3 1.1 Definitions and Terminology 26. The function y(x) is not continuous at x = 0 since lim y(x) = 5 and lim y(x) = −5. Thus, y  (x) does not x→0− x→0+ exist at x = 0. 27. (a) From y = emx we obtain y  = memx . Then y  + 2y = 0 implies memx + 2emx = (m + 2)emx = 0. Since emx > 0 for all x, m = −2. Thus y = e−2x is a solution. (b) From y = emx we obtain y  = memx and y  = m2 emx . Then y  − 5y  + 6y = 0 implies m2 emx − 5memx + 6emx = (m − 2)(m − 3)emx = 0. Since emx > 0 for all x, m = 2 and m = 3. Thus y = e2x and y = e3x are solutions. 28. (a) From y = xm we obtain y  = mxm−1 and y  = m(m − 1)xm−2 . Then xy  + 2y  = 0 implies xm(m − 1)xm−2 + 2mxm−1 = [m(m − 1) + 2m]xm−1 = (m2 + m)xm−1 = m(m + 1)xm−1 = 0. Since xm−1 > 0 for x > 0, m = 0 and m = −1. Thus y = 1 and y = x−1 are solutions. (b) From y = xm we obtain y  = mxm−1 and y  = m(m − 1)xm−2 . Then x2 y  − 7xy  + 15y = 0 implies x2 m(m − 1)xm−2 − 7xmxm−1 + 15xm = [m(m − 1) − 7m + 15]xm = (m2 − 8m + 15)xm = (m − 3)(m − 5)xm = 0. Since xm > 0 for x > 0, m = 3 and m = 5. Thus y = x3 and y = x5 are solutions. In Problems 29–32, we substitute y = c into the differential equations and use y  = 0 and y  = 0 29. Solving 5c = 10 we see that y = 2 is a constant solution. 30. Solving c2 + 2c − 3 = (c + 3)(c − 1) = 0 we see that y = −3 and y = 1 are constant solutions. 31. Since 1/(c − 1) = 0 has no solutions, the differential equation has no constant solutions. 32. Solving 6c = 10 we see that y = 5/3 is a constant solution. 33. From x = e−2t + 3e6t and y = −e−2t + 5e6t we obtain dx = −2e−2t + 18e6t dt and dy = 2e−2t + 30e6t . dt Then x + 3y = (e−2t + 3e6t ) + 3(−e−2t + 5e6t ) = −2e−2t + 18e6t = dx dt 5x + 3y = 5(e−2t + 3e6t ) + 3(−e−2t + 5e6t ) = 2e−2t + 30e6t = dy . dt and 34. From x = cos 2t + sin 2t + 15 et and y = − cos 2t − sin 2t − 15 et we obtain and dx 1 = −2 sin 2t + 2 cos 2t + et dt 5 and dy 1 = 2 sin 2t − 2 cos 2t − et dt 5 d2 x 1 = −4 cos 2t − 4 sin 2t + et 2 dt 5 and d2 y 1 = 4 cos 2t + 4 sin 2t − et . 2 dt 5 Then and 1 1 d2 x 4y + et = 4(− cos 2t − sin 2t − et ) + et = −4 cos 2t − 4 sin 2t + et = 2 5 5 dt 4 1.1 Definitions and Terminology 1 1 d2 y 4x − et = 4(cos 2t + sin 2t + et ) − et = 4 cos 2t + 4 sin 2t − et = 2 . 5 5 dt 35. (y  )2 + 1 = 0 has no real solutions because (y  )2 + 1 is positive for all functions y = φ(x). 36. The only solution of (y  )2 + y 2 = 0 is y = 0, since if y = 0, y 2 > 0 and (y  )2 + y 2 ≥ y 2 > 0. 37. The first derivative of f (x) = ex is ex . The first derivative of f (x) = ekx is kekx . The differential equations are y  = y and y  = ky, respectively. 38. Any function of the form y = cex or y = ce−x is its own second derivative. The corresponding differential equation is y  − y = 0. Functions of the form y = c sin x or y = c cos x have second derivatives that are the negatives of themselves. The differential equation is y  + y = 0.   √ 39. We first note that 1 − y 2 = 1 − sin2 x = cos2 x = | cos x|. This prompts us to consider values of x for which cos x < 0, such as x = π. In this case     dy  d  = = cos xx=π = cos π = −1, (sin x)   dx  dx x=π x=π but   √ 1 − y 2 |x=π = 1 − sin2 π = 1 = 1.  Thus, y = sin x will only be a solution of y  = 1 − y 2 when cos x > 0. An interval of definition is then (−π/2, π/2). Other intervals are (3π/2, 5π/2), (7π/2, 9π/2), and so on. 40. Since the first and second derivatives of sin t and cos t involve sin t and cos t, it is plausible that a linear combination of these functions, A sin t + B cos t, could be a solution of the differential equation. Using y  = A cos t − B sin t and y  = −A sin t − B cos t and substituting into the differential equation we get y  + 2y  + 4y = −A sin t − B cos t + 2A cos t − 2B sin t + 4A sin t + 4B cos t = (3A − 2B) sin t + (2A + 3B) cos t = 5 sin t. Thus 3A − 2B = 5 and 2A + 3B = 0. Solving these simultaneous equations we find A = 10 particular solution is y = 15 13 sin t − 13 cos t. 15 13 and B = − 10 13 . A 41. One solution is given by the upper portion of the graph with domain approximately (0, 2.6). The other solution is given by the lower portion of the graph, also with domain approximately (0, 2.6). 42. One solution, with domain approximately (−∞, 1.6) is the portion of the graph in the second quadrant together with the lower part of the graph in the first quadrant. A second solution, with domain approximately (0, 1.6) is the upper part of the graph in the first quadrant. The third solution, with domain (0, ∞), is the part of the graph in the fourth quadrant. 43. Differentiating (x3 + y 3 )/xy = 3c we obtain xy(3x2 + 3y 2 y  ) − (x3 + y 3 )(xy  + y) =0 x2 y 2 3x3 y + 3xy 3 y  − x4 y  − x3 y − xy 3 y  − y 4 = 0 (3xy 3 − x4 − xy 3 )y  = −3x3 y + x3 y + y 4 y = y 4 − 2x3 y y(y 3 − 2x3 ) . = 2xy 3 − x4 x(2y 3 − x3 ) 44. A tangent line will be vertical where y  is undefined, or in this case, where x(2y 3 − x3 ) = 0. This gives x = 0 and 2y 3 = x3 . Substituting y 3 = x3 /2 into x3 + y 3 = 3xy we get 5 1.1 Definitions and Terminology   1 1 x3 + x3 = 3x x 2 21/3 3 3 3 x = 1/3 x2 2 2 x3 = 22/3 x2 x2 (x − 22/3 ) = 0. Thus, there are vertical tangent lines at x = 0 and x = 22/3 , or at (0, 0) and (22/3 , 21/3 ). Since 22/3 ≈ 1.59, the estimates of the domains in Problem 42 were close. √ √ 45. The derivatives of the functions are φ1 (x) = −x/ 25 − x2 and φ2 (x) = x/ 25 − x2 , neither of which is defined at x = ±5. 46. To determine if a solution curve passes through (0, 3) we let t = 0 and P = 3 in the equation P = c1 et /(1+c1 et ). This gives 3 = c1 /(1 + c1 ) or c1 = − 32 . Thus, the solution curve P = −3et (−3/2)et = 1 − (3/2)et 2 − 3et passes through the point (0, 3). Similarly, letting t = 0 and P = 1 in the equation for the one-parameter family of solutions gives 1 = c1 /(1 + c1 ) or c1 = 1 + c1 . Since this equation has no solution, no solution curve passes through (0, 1). 47. For the first-order differential equation integrate f (x). For the second-order differential equation integrate twice. In the latter case we get y = ( f (x)dx)dx + c1 x + c2 . 48. Solving for y  using the quadratic formula we obtain the two differential equations   1 1 2 + 2 1 + 3x6 2 − 2 1 + 3x6 , and y  = y = x x so the differential equation cannot be put in the form dy/dx = f (x, y). 49. The differential equation yy  − xy = 0 has normal form dy/dx = x. These are not equivalent because y = 0 is a solution of the first differential equation but not a solution of the second. 50. Differentiating we get y  = c1 + 3c2 x2 and y  = 6c2 x. Then c2 = y  /6x      y xy   y= y − x+ x3 = xy  − 2 6x and c1 = y  − xy  /2, so 1 2  x y 3 and the differential equation is x2 y  − 3xy  + 3y = 0. 51. (a) Since e−x is positive for all values of x, dy/dx > 0 for all x, and a solution, y(x), of the differential equation must be increasing on any interval. 2 2 dy dy (b) lim = lim e−x = 0 and lim = lim e−x = 0. Since dy/dx approaches 0 as x approaches −∞ x→−∞ dx x→−∞ x→∞ dx x→∞ and ∞, the solution curve has horizontal asymptotes to the left and to the right. 2 (c) To test concavity we consider the second derivative   2 d2 y d −x2 d dy = −2xe−x . = e = dx2 dx dx dx Since the second derivative is positive for x < 0 and negative for x > 0, the solution curve is concave up on (−∞, 0) and concave down on (0, ∞). 6 1.1 (d) Definitions and Terminology y x 52. (a) The derivative of a constant solution y = c is 0, so solving 5 − c = 0 we see that c = 5 and so y = 5 is a constant solution. (b) A solution is increasing where dy/dx = 5−y > 0 or y < 5. A solution is decreasing where dy/dx = 5−y < 0 or y > 5. 53. (a) The derivative of a constant solution is 0, so solving y(a − by) = 0 we see that y = 0 and y = a/b are constant solutions. (b) A solution is increasing where dy/dx = y(a − by) = by(a/b − y) > 0 or 0 < y < a/b. A solution is decreasing where dy/dx = by(a/b − y) < 0 or y < 0 or y > a/b. (c) Using implicit differentiation we compute d2 y = y(−by  ) + y  (a − by) = y  (a − 2by). dx2 Solving d2 y/dx2 = 0 we obtain y = a/2b. Since d2 y/dx2 > 0 for 0 < y < a/2b and d2 y/dx2 < 0 for a/2b < y < a/b, the graph of y = φ(x) has a point of inflection at y = a/2b. (d) y y=aêb y=0 x 54. (a) If y = c is a constant solution then y  = 0, but c2 + 4 is never 0 for any real value of c. (b) Since y  = y 2 + 4 > 0 for all x where a solution y = φ(x) is defined, any solution must be increasing on any interval on which it is defined. Thus it cannot have any relative extrema. (c) Using implicit differentiation we compute d2 y/dx2 = 2yy  = 2y(y 2 + 4). Setting d2 y/dx2 = 0 we see that y = 0 corresponds to the only possible point of inflection. Since d2 y/dx2 < 0 for y < 0 and d2 y/dx2 > 0 for y > 0, there is a point of inflection where y = 0. 7 1.1 Definitions and Terminology (d) y x 55. In Mathematica use Clear[y] y[x ]:= x Exp[5x] Cos[2x] y[x] y''''[x] − 20y'''[x] + 158y''[x] − 580y'[x] +841y[x]//Simplify The output will show y(x) = e5x x cos 2x, which verifies that the correct function was entered, and 0, which verifies that this function is a solution of the differential equation. 56. In Mathematica use Clear[y] y[x ]:= 20Cos[5Log[x]]/x − 3Sin[5Log[x]]/x y[x] xˆ3 y'''[x] + 2xˆ2 y''[x] + 20x y'[x] − 78y[x]//Simplify The output will show y(x) = 20 cos(5 ln x)/x − 3 sin(5 ln x)/x, which verifies that the correct function was entered, and 0, which verifies that this function is a solution of the differential equation. EXERCISES 1.2 Initial-Value Problems 1. Solving −1/3 = 1/(1 + c1 ) we get c1 = −4. The solution is y = 1/(1 − 4e−x ). 2. Solving 2 = 1/(1 + c1 e) we get c1 = −(1/2)e−1 . The solution is y = 2/(2 − e−(x+1) ) . 3. Letting x = 2 and solving 1/3 = 1/(4 + c) we get c = −1. The solution is y = 1/(x2 − 1). This solution is defined on the interval (1, ∞). 4. Letting x = −2 and solving 1/2 = 1/(4 + c) we get c = −2. The solution is y = 1/(x2 − 2). This solution is √ defined on the interval (−∞, − 2 ). 5. Letting x = 0 and solving 1 = 1/c we get c = 1. The solution is y = 1/(x2 + 1). This solution is defined on the interval (−∞, ∞). 8 1.2 Initial-Value Problems 6. Letting x = 1/2 and solving −4 = 1/(1/4 + c) we get c = −1/2. The solution is y = 1/(x2 − 1/2) = 2/(2x2 − 1). √ √ This solution is defined on the interval (−1/ 2 , 1/ 2 ). In Problems 7–10, we use x = c1 cos t + c2 sin t and x = −c1 sin t + c2 cos t to obtain a system of two equations in the two unknowns c1 and c2 . 7. From the initial conditions we obtain the system c1 = −1 c2 = 8. The solution of the initial-value problem is x = − cos t + 8 sin t. 8. From the initial conditions we obtain the system c2 = 0 −c1 = 1. The solution of the initial-value problem is x = − cos t. 9. From the initial conditions we obtain √ 3 1 1 c1 + c2 = 2 2 2 √ 3 1 c2 = 0. − c1 + 2 2 Solving, we find c1 = √ 3/4 and c2 = 1/4. The solution of the initial-value problem is √ x = ( 3/4) cos t + (1/4) sin t. 10. From the initial conditions we obtain √ √ 2 2 c1 + c2 = 2 2 2 √ √ √ 2 2 c1 + c2 = 2 2 . − 2 2 Solving, we find c1 = −1 and c2 = 3. The solution of the initial-value problem is x = − cos t + 3 sin t. √ In Problems 11–14, we use y = c1 ex + c2 e−x and y  = c1 ex − c2 e−x to obtain a system of two equations in the two unknowns c1 and c2 . 11. From the initial conditions we obtain c1 + c2 = 1 c1 − c2 = 2. Solving, we find c1 = 3 2 and c2 = − 12 . The solution of the initial-value problem is y = 32 ex − 12 e−x . 12. From the initial conditions we obtain ec1 + e−1 c2 = 0 ec1 − e−1 c2 = e. Solving, we find c1 = 1 2 and c2 = − 12 e2 . The solution of the initial-value problem is y= 13. From the initial conditions we obtain 1 x 1 2 −x 1 1 e − e e = ex − e2−x . 2 2 2 2 e−1 c1 + ec2 = 5 e−1 c1 − ec2 = −5. 9 1.2 Initial-Value Problems Solving, we find c1 = 0 and c2 = 5e−1 . The solution of the initial-value problem is y = 5e−1 e−x = 5e−1−x . 14. From the initial conditions we obtain c1 + c2 = 0 c1 − c2 = 0. Solving, we find c1 = c2 = 0. The solution of the initial-value problem is y = 0. 15. Two solutions are y = 0 and y = x3 . 16. Two solutions are y = 0 and y = x2 . (Also, any constant multiple of x2 is a solution.) ∂f 2 = y −1/3 . Thus, the differential equation will have a unique solution in any ∂y 3 rectangular region of the plane where y = 0.  √ 18. For f (x, y) = xy we have ∂f /∂y = 12 x/y . Thus, the differential equation will have a unique solution in any 17. For f (x, y) = y 2/3 we have region where x > 0 and y > 0 or where x < 0 and y < 0. 19. For f (x, y) = where x = 0. ∂f 1 y we have = . Thus, the differential equation will have a unique solution in any region x ∂y x 20. For f (x, y) = x + y we have ∂f = 1. Thus, the differential equation will have a unique solution in the entire ∂y plane. 21. For f (x, y) = x2 /(4 − y 2 ) we have ∂f /∂y = 2x2 y/(4 − y 2 )2 . Thus the differential equation will have a unique solution in any region where y < −2, −2 < y < 2, or y > 2. x2 −3x2 y 2 ∂f = we have 2 . Thus, the differential equation will have a unique solution in 1 + y3 ∂y (1 + y 3 ) any region where y = −1. 22. For f (x, y) = y2 2x2 y ∂f = we have 2 . Thus, the differential equation will have a unique solution in x2 + y 2 ∂y (x2 + y 2 ) any region not containing (0, 0). 23. For f (x, y) = 24. For f (x, y) = (y + x)/(y − x) we have ∂f /∂y = −2x/(y − x)2 . Thus the differential equation will have a unique solution in any region where y < x or where y > x.   y 2 ...
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