AEM_3e_Chapter01

AEM_3e_Chapter01 - Part I Ordinary Differential Equations...

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Unformatted text preview: 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 x2 ˙ 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 = − 1 e−x/2 . Then 2y + y = −e−x/2 + e−x/2 = 0. 2 12. From y = 6 5 − 6 e−20t we obtain dy/dt = 24e−20t , so that 5 dy + 20y = 24e−20t + 20 dt 6 6 −20t − e 5 5 13. From y = e3x cos 2x we obtain y = 3e3x cos 2x − 2e3x sin 2x and y y − 6y + 13y = 0. = 24. = 5e3x cos 2x − 12e3x sin 2x, so that 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 = − 1 (1 − sin x)−3/2 (− cos x) we have 2 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 1 − 2X − 1 X − 1 4 dX =1 dt 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 -4 -2 2 4 x -2 on (−∞, ∞). The graph of y1 (x) is solid and the graph of y2 is dashed. -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 ) = x 22. Differentiating y = e−x 2 c1 et 1 + c1 et 1− c1 et = P (1 − P ). 1 + c1 et et dt + c1 e−x we obtain 2 2 0 x y = e−x ex − 2xe−x 2 2 2 x et dt − 2c1 xe−x = 1 − 2xe−x 2 2 2 0 et dt − 2c1 xe−x . 2 2 0 Substituting into the differential equation, we have x y + 2xy = 1 − 2xe−x 2 2 2 0 23. From y = c1 e2x + c2 xe2x we obtain x et dt − 2c1 xe−x + 2xe−x 2 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 −4 + 4y = (4c1 + 4c2 − 8c1 − 4c2 + 4c1 )e2x + (4c2 − 8c2 + 4c2 )xe2x = 0. 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 , 2 x , x<0 x≥0 we obtain y = −2x, 2x, x<0 so that xy − 2y = 0. 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 + 1 et and y = − cos 2t − sin 2t − 1 et we obtain 5 5 and dx 1 = −2 sin 2t + 2 cos 2t + et dt 5 and dy 1 = 2 sin 2t − 2 cos 2t − et dt 5 1 d2 x = −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 dx = x=π d (sin x) dx = cos x x=π = cos π = −1, 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 = particular solution is y = 15 13 sin t − 10 13 15 13 and B = − 10 . A 13 cos t. 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 = − 3 . Thus, the solution curve 2 P = (−3/2)et −3et = 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 y = 1 2 + 2 1 + 3x6 x and y = 1 2−2 x 1 + 3x6 , 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 and c1 = y − xy /2, so y= y − xy 2 y 6x x+ 1 x3 = xy − x2 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 d2 y d = dx2 dx dy dx = 2 2 d e−x = −2xe−x . 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 Solving, we find c1 = √ √ 3 1 1 c1 + c2 = 2 2 2 √ 1 3 − c1 + c2 = 0. 2 2 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 = − 1 . The solution of the initial-value problem is y = 3 ex − 1 e−x . 2 2 2 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 = − 1 e2 . The solution of the initial-value problem is 2 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 = 1 x/y . Thus, the differential equation will have a unique solution in any 2 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) = y ∂f 1 we have = . Thus, the differential equation will have a unique solution in any region x ∂y x where x = 0. 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 ∂f −3x2 y 2 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 − 9 and ∂f /∂y = y/ y 2 − 9. We see that f and ∂f /∂y are both continuous in the regions of the plane determined by y < −3 and y > 3 with no restrictions on x. In Problems 25–28, we identify f (x, y) = 25. Since 4 > 3, (1, 4) is in the region defined by y > 3 and the differential equation has a unique solution through (1, 4). 26. Since (5, 3) is not in either of the regions defined by y < −3 or y > 3, there is no guarantee of a unique solution through (5, 3). 27. Since (2, −3) is not in either of the regions defined by y < −3 or y > 3, there is no guarantee of a unique solution through (2, −3). 28. Since (−1, 1) is not in either of the regions defined by y < −3 or y > 3, there is no guarantee of a unique solution through (−1, 1). 29. (a) A one-parameter family of solutions is y = cx. Since y = c, xy = xc = y and y(0) = c · 0 = 0. 10 1.2 Initial-Value Problems (b) Writing the equation in the form y = y/x, we see that R cannot contain any point on the y-axis. Thus, any rectangular region disjoint from the y-axis and containing (x0 , y0 ) will determine an interval around x0 and a unique solution through (x0 , y0 ). Since x0 = 0 in part (a), we are not guaranteed a unique solution through (0, 0). (c) The piecewise-defined function which satisfies y(0) = 0 is not a solution since it is not differentiable at x = 0. d 30. (a) Since tan(x + c) = sec2 (x + c) = 1 + tan2 (x + c), we see that y = tan(x + c) satisfies the differential dx equation. (b) Solving y(0) = tan c = 0 we obtain c = 0 and y = tan x. Since tan x is discontinuous at x = ±π/2, the solution is not defined on (−2, 2) because it contains ±π/2. (c) The largest interval on which the solution can exist is (−π/2, π/2). d 1 1 1 31. (a) Since = y 2 , we see that y = − − = is a solution of the differential equation. dx x+c (x + c)2 x+c (b) Solving y(0) = −1/c = 1 we obtain c = −1 and y = 1/(1 − x). Solving y(0) = −1/c = −1 we obtain c = 1 and y = −1/(1 + x). Being sure to include x = 0, we see that the interval of existence of y = 1/(1 − x) is (−∞, 1), while the interval of existence of y = −1/(1 + x) is (−1, ∞). 32. (a) Solving y(0) = −1/c = y0 we obtain c = −1/y0 and y=− 1 y0 = , −1/y0 + x 1 − y0 x y0 = 0. Since we must have −1/y0 + x = 0, the largest interval of existence (which must contain 0) is either (−∞, 1/y0 ) when y0 > 0 or (1/y0 , ∞) when y0 < 0. (b) By inspection we see that y = 0 is a solution on (−∞, ∞). 33. (a) Differentiating 3x2 − y 2 = c we get 6x − 2yy = 0 or yy = 3x. y (b) Solving 3x2 − y 2 = 3 for y we get y = φ1 (x) = 3(x2 − 1) , y = φ2 (x) = − 3(x2 − 1) , y = φ3 (x) = 3(x2 − 1) , y = φ4 (x) = − 3(x2 − 1) , 4 1 < x < ∞, 2 1 < x < ∞, −∞ < x < −1, -4 2 −∞ < x < −1. 4 x 2 -2 4 x -2 -4 (c) Only y = φ3 (x) satisfies y(−2) = 3. y 34. (a) Setting x = 2 and y = −4 in 3x2 − y 2 = c we get 12 − 16 = −4 = c, so the explicit solution is 4 y = − 3x2 + 4 , −∞ < x < ∞. (b) Setting c = 0 we have y = √ √ 3x and y = − 3x, both defined on 2 -4 -2 -2 (−∞, ∞). -4 11 1.2 Initial-Value Problems In Problems 35–38, we consider the points on the graphs with x-coordinates x0 = −1, x0 = 0, and x0 = 1. The slopes of the tangent lines at these points are compared with the slopes given by y (x0 ) in (a) through (f). 35. The graph satisfies the conditions in (b) and (f). 36. The graph satisfies the conditions in (e). 37. The graph satisfies the conditions in (c) and (d). 38. The graph satisfies the conditions in (a). 39. Integrating y = 8e2x + 6x we obtain y= (8e2x + 6x)dx = 4e2x + 3x2 + c. Setting x = 0 and y = 9 we have 9 = 4 + c so c = 5 and y = 4e2x + 3x2 + 5. 40. Integrating y = 12x − 2 we obtain y = (12x − 2)dx = 6x2 − 2x + c1 . Then, integrating y we obtain y= (6x2 − 2x + c1 )dx = 2x3 − x2 + c1 x + c2 . At x = 1 the y-coordinate of the point of tangency is y = −1 + 5 = 4. This gives the initial condition y(1) = 4. The slope of the tangent line at x = 1 is y (1) = −1. From the initial conditions we obtain 2 − 1 + c1 + c2 = 4 or c1 + c2 = 3 6 − 2 + c1 = −1 or c1 = −5. and Thus, c1 = −5 and c2 = 8, so y = 2x3 − x2 − 5x + 8. 41. When x = 0 and y = 1 2 , y = −1, so the only plausible solution curve is the one with negative slope at (0, 1 ), 2 or the black curve. 42. If the solution is tangent to the x-axis at (x0 , 0), then y = 0 when x = x0 and y = 0. Substituting these values into y + 2y = 3x − 6 we get 0 + 0 = 3x0 − 6 or x0 = 2. 43. The theorem guarantees a unique (meaning single) solution through any point. Thus, there cannot be two distinct solutions through any point. 44. When y = 1 4 16 x , y = 1 x3 = x( 1 x2 ) = xy 1/2 , and y(2) = 4 4 y= 1 16 (16) = 1. When 0, x<0 1 4 16 x , x≥0 we have y = and y(2) = 1 16 (16) 0, x<0 1 3 4x , x≥0 =x 0, x<0 1 2 4x , x≥0 = xy 1/2 , = 1. The two different solutions are the same on the interval (0, ∞), which is all that is required by Theorem 1.1. 45. At t = 0, dP/dt = 0.15P (0) + 20 = 0.15(100) + 20 = 35. Thus, the population is increasing at a rate of 3,500 individuals per year. 12 1.3 Differential Equations as Mathematical Models If the population is 500 at time t = T then dP dt = 0.15P (T ) + 20 = 0.15(500) + 20 = 95. t=T Thus, at this time, the population is increasing at a rate of 9,500 individuals per year. EXERCISES 1.3 Differential Equations as Mathematical Models dP dP = kP + r; = kP − r dt dt 2. Let b be the rate of births and d the rate of deaths. Then b = k1 P and d = k2 P . Since dP/dt = b − d, the differential equation is dP/dt = k1 P − k2 P . 1. 3. Let b be the rate of births and d the rate of deaths. Then b = k1 P and d = k2 P 2 . Since dP/dt = b − d, the differential equation is dP/dt = k1 P − k2 P 2 . dP 4. = k1 P − k2 P 2 − h, h > 0 dt 5. From the graph in the text we estimate T0 = 180◦ and Tm = 75◦ . We observe that when T = 85, dT /dt ≈ −1. From the differential equation we then have k= dT /dt −1 = −0.1. = T − Tm 85 − 75 6. By inspecting the graph in the text we take Tm to be Tm (t) = 80 − 30 cos πt/12. Then the temperature of the body at time t is determined by the differential equation dT π = k T − 80 − 30 cos t dt 12 , t > 0. 7. The number of students with the flu is x and the number not infected is 1000 − x, so dx/dt = kx(1000 − x). 8. By analogy, with the differential equation modeling the spread of a disease, we assume that the rate at which the technological innovation is adopted is proportional to the number of people who have adopted the innovation and also to the number of people, y(t), who have not yet adopted it. If one person who has adopted the innovation is introduced into the population, then x + y = n + 1 and dx = kx(n + 1 − x), dt x(0) = 1. 9. The rate at which salt is leaving the tank is Rout (3 gal/min) · A lb/gal 300 = A lb/min. 100 Thus dA/dt = A/100. The initial amount is A(0) = 50. 10. The rate at which salt is entering the tank is Rin = (3 gal/min) · (2 lb/gal) = 6 lb/min. 13 1.3 Differential Equations as Mathematical Models Since the solution is pumped out at a slower rate, it is accumulating at the rate of (3 − 2)gal/min = 1 gal/min. After t minutes there are 300 + t gallons of brine in the tank. The rate at which salt is leaving is Rout = (2 gal/min) · A lb/gal 300 + t = 2A lb/min. 300 + t The differential equation is dA 2A =6− . dt 300 + t 11. The rate at which salt is entering the tank is Rin = (3 gal/min) · (2 lb/gal) = 6 lb/min. Since the tank loses liquid at the net rate of 3 gal/min − 3.5 gal/min = −0.5 gal/min, after t minutes the number of gallons of brine in the tank is 300 − 1 t gallons. Thus the rate at which salt is 2 leaving is A 3.5A 7A Rout = lb/gal · (3.5 gal/min) = lb/min = lb/min. 300 − t/2 300 − t/2 600 − t The differential equation is dA 7A =6− dt 600 − t dA 7 + A = 6. dt 600 − t or 12. The rate at which salt is entering the tank is Rin = (cin lb/gal) · (rin gal/min) = cin rin lb/min. Now let A(t) denote the number of pounds of salt and N (t) the number of gallons of brine in the tank at time t. The concentration of salt in the tank as well as in the outflow is c(t) = x(t)/N (t). But the number of gallons of brine in the tank remains steady, is increased, or is decreased depending on whether rin = rout , rin > rout , or rin < rout . In any case, the number of gallons of brine in the tank at time t is N (t) = N0 + (rin − rout )t. The output rate of salt is then A A lb/gal · (rout gal/min) = rout lb/min. N0 + (rin − rout )t N0 + (rin − rout )t Rout = The differential equation for the amount of salt, dA/dt = Rin − Rout , is dA A = cin rin − rout dt N0 + (rin − rout )t or dA rout + A = cin rin . dt N0 + (rin − rout )t 13. The volume of water in the tank at time t is V = Aw h. The differential equation is then dh 1 dV 1 −cAh = = dt Aw dt Aw Using Ah = π 2 12 2 = 2gh =− cAh Aw 2gh . π , Aw = 102 = 100, and g = 32, this becomes 36 dh cπ √ cπ/36 √ 64h = − h. =− dt 100 450 14. The volume of water in the tank at time t is V = 1 πr2 h where r is the radius of the tank at height h. From 3 the figure in the text we see that r/h = 8/20 so that r = 2 h and V = 1 π 5 3 4 respect to t we have dV /dt = 25 πh2 dh/dt or dh 25 dV = . dt 4πh2 dt 14 2 2 5h h= 4 3 75 πh . Differentiating with 1.3 Differential Equations as Mathematical Models √ From Problem 13 we have dV /dt = −cAh 2gh where c = 0.6, Ah = π √ −2π h/15 and √ 2π h dh 25 5 − = = − 3/2 . dt 4πh2 15 6h 2 2 12 , and g = 32. Thus dV /dt = 15. Since i = dq/dt and L d2 q/dt2 + R dq/dt = E(t), we obtain L di/dt + Ri = E(t). dq 1 16. By Kirchhoff’s second law we obtain R + q = E(t). dt C dv 17. From Newton’s second law we obtain m = −kv 2 + mg. dt 18. Since the barrel in Figure 1.35(b) in the text is submerged an additional y feet below its equilibrium position the number of cubic feet in the additional submerged portion is the volume of the circular cylinder: π×(radius)2 ×height or π(s/2)2 y. Then we have from Archimedes’ principle upward force of water on barrel = weight of water displaced = (62.4) × (volume of water displaced) = (62.4)π(s/2)2 y = 15.6πs2 y. It then follows from Newton’s second law that w d2 y = −15.6πs2 y g dt2 d2 y 15.6πs2 g + y = 0, dt2 w or where g = 32 and w is the weight of the barrel in pounds. 19. The net force acting on the mass is F = ma = m d2 x = −k(s + x) + mg = −kx + mg − ks. dt2 Since the condition of equilibrium is mg = ks, the differential equation is m d2 x = −kx. dt2 20. From Problem 19, without a damping force, the differential equation is m d2 x/dt2 = −kx. With a damping force proportional to velocity, the differential equation becomes m d2 x dx = −kx − β 2 dt dt or m d2 x dx + kx = 0. +β 2 dt dt 21. Let x(t) denote the height of the top of the chain at time t with the positive direction upward. The weight of the portion of chain off the ground is W = (x ft) · (1 lb/ft) = x. The mass of the chain is m = W/g = x/32. The net force is F = 5 − W = 5 − x. By Newton’s second law, d x v =5−x dt 32 or x dv dx +v = 160 − 32x. dt dt Thus, the differential equation is x d2 x dx + dt2 dt 2 + 32x = 160. 22. The force is the weight of the chain, 2L, so by Newton’s second law, of chain off the ground is m = 2(L − x)/g, we have d 2(L − x) v = 2L dt g (L − x) or 15 d [mv] = 2L. Since the mass of the portion dt dv dx +v − = Lg. dt dt 1.3 Differential Equations as Mathematical Models Thus, the differential equation is (L − x) d2 x dx − 2 dt dt 2 = Lg. 23. From g = k/R2 we find k = gR2 . Using a = d2 r/dt2 and the fact that the positive direction is upward we get d2 r k gR2 = −a = − 2 = − 2 dt2 r r or d2 r gR2 + 2 = 0. dt2 r 24. The gravitational force on m is F = −kMr m/r2 . Since Mr = 4πδr3 /3 and M = 4πδR3 /3 we have Mr = r3 M/R3 and F = −k Mr m r3 M m/R3 mM = −k = −k 3 r. 2 r r2 R Now from F = ma = d2 r/dt2 we have m d2 r mM = −k 3 r 2 dt R d2 r kM = − 3 r. 2 dt R or dA = k(M − A). dt dA 26. The differential equation is = k1 (M − A) − k2 A. dt 27. The differential equation is x (t) = r − kx(t) where k > 0. 25. The differential equation is 28. By the Pythagorean Theorem the slope of the tangent line is y = −y s2 − y 2 . 29. We see from the figure that 2θ + α = π. Thus y y 2 tan θ . = tan α = tan(π − 2θ) = − tan 2θ = − −x 1 − tan2 θ Since the slope of the tangent line is y = tan θ we have y/x = 2y [1 − (y )2 ] or y − y(y )2 = 2xy , which is the quadratic equation y(y )2 + 2xy − y = 0 in y . Using the quadratic formula, we get y = −2x ± 4x2 + 4y 2 −x ± x2 + y 2 = . 2y y (x,y) θ θα θ Since dy/dx > 0, the differential equation is dy −x + x2 + y 2 = dx y or y dy − dx x α y φ x x2 + y 2 + x = 0. 30. The differential equation is dP/dt = kP , so from Problem 37 in Exercises 1.1, P = ekt , and a one-parameter family of solutions is P = cekt . 31. The differential equation in (3) is dT /dt = k(T − Tm ). When the body is cooling, T > Tm , so T − Tm > 0. Since T is decreasing, dT /dt < 0 and k < 0. When the body is warming, T < Tm , so T − Tm < 0. Since T is increasing, dT /dt > 0 and k < 0. 32. The differential equation in (8) is dA/dt = 6 − A/100. If A(t) attains a maximum, then dA/dt = 0 at this time and A = 600. If A(t) continues to increase without reaching a maximum, then A (t) > 0 for t > 0 and A cannot exceed 600. In this case, if A (t) approaches 0 as t increases to infinity, we see that A(t) approaches 600 as t increases to infinity. 33. This differential equation could describe a population that undergoes periodic fluctuations. 16 1.3 Differential Equations as Mathematical Models 34. (a) As shown in Figure 1.43(b) in the text, the resultant of the reaction force of magnitude F and the weight of magnitude mg of the particle is the centripetal force of magnitude mω 2 x. The centripetal force points to the center of the circle of radius x on which the particle rotates about the y-axis. Comparing parts of similar triangles gives F cos θ = mg and F sin θ = mω 2 x. (b) Using the equations in part (a) we find tan θ = F sin θ mω 2 x ω2 x = = F cos θ mg g or dy ω2 x = . dx g 35. From Problem 23, d2 r/dt2 = −gR2 /r2 . Since R is a constant, if r = R + s, then d2 r/dt2 = d2 s/dt2 and, using a Taylor series, we get d2 s R2 2gs = −g = −gR2 (R + s)−2 ≈ −gR2 [R−2 − 2sR−3 + · · · ] = −g + 3 + · · · . 2 dt (R + s)2 R Thus, for R much larger than s, the differential equation is approximated by d2 s/dt2 = −g. 36. (a) If ρ is the mass density of the raindrop, then m = ρV and dm dr dV d 4 3 =ρ =ρ πr = ρ 4πr2 dt dt dt 3 dt = ρS dr . dt If dr/dt is a constant, then dm/dt = kS where ρ dr/dt = k or dr/dt = k/ρ. Since the radius is decreasing, k < 0. Solving dr/dt = k/ρ we get r = (k/ρ)t + c0 . Since r(0) = r0 , c0 = r0 and r = kt/ρ + r0 . d (b) From Newton’s second law, [mv] = mg, where v is the velocity of the raindrop. Then dt m dv dm +v = mg dt dt or ρ 4 3 dv 4 πr + v(k4πr2 ) = ρ πr3 g. 3 dt 3 Dividing by 4ρπr3 /3 we get dv 3k + v=g dt ρr dv 3k/ρ v = g, k < 0. + dt kt/ρ + r0 or 37. We assume that the plow clears snow at a constant rate of k cubic miles per hour. Let t be the time in hours after noon, x(t) the depth in miles of the snow at time t, and y(t) the distance the plow has moved in t hours. Then dy/dt is the velocity of the plow and the assumption gives wx dy = k, dt where w is the width of the plow. Each side of this equation simply represents the volume of snow plowed in one hour. Now let t0 be the number of hours before noon when it started snowing and let s be the constant rate in miles per hour at which x increases. Then for t > −t0 , x = s(t + t0 ). The differential equation then becomes dy k 1 . = dt ws t + t0 Integrating, we obtain k [ ln(t + t0 ) + c ] ws where c is a constant. Now when t = 0, y = 0 so c = − ln t0 and y= y= k t ln 1 + ws t0 17 . 1.3 Differential Equations as Mathematical Models Finally, from the fact that when t = 1, y = 2 and when t = 2, y = 3, we obtain 1+ 2 t0 2 = 1+ 1 t0 3 . Expanding and simplifying gives t2 + t0 − 1 = 0. Since t0 > 0, we find t0 ≈ 0.618 hours ≈ 37 minutes. Thus it 0 started snowing at about 11:23 in the morning. dP dA 38. (1): = kP is linear (2): = kA is linear dt dt dx dT (5): (3): = k(T − Tm ) is linear = kx(n + 1 − x) is nonlinear dt dt dX dA A (6): = k(α − X)(β − X) is nonlinear (8): =6− is linear dt dt 100 dh d2 q dq Ah 1 (10): 2gh is nonlinear (11): L 2 + R + q = E(t) is linear =− dt Aw dt dt C d2 s dv (12): = −g is linear (14): m = mg − kv is linear 2 dt dt d2 s ds d2 x 64 (15): m 2 + k − x = 0 is linear = mg is linear (16): dt dt dt2 L (17): linearity or nonlinearity is determined by the manner in which W and T1 involve x. 39. At time t, when the population is 2 million cells, the differential equation P (t) = 0.15P (t) gives the rate of increase at time t. Thus, when P (t) = 2 (million cells), the rate of increase is P (t) = 0.15(2) = 0.3 million cells per hour or 300,000 cells per hour. 40. Setting A (t) = −0.002 and solving A (t) = −0.0004332A(t) for A(t), we obtain A(t) = A (t) −0.002 = ≈ 4.6 grams. −0.0004332 −0.0004332 CHAPTER 1 REVIEW EXERCISES d dy c1 ekx = c1 kekx ; = ky dx dx d dy dy 2. (5 + c1 e−2x ) = −2c1 e−2x = −2(5 + c1 e−2x − 5); = −2(y − 5) or = −2y + 10 dx dx dx d 3. (c1 cos kx + c2 sin kx) = −kc1 sin kx + kc2 cos kx; dx d2 (c1 cos kx + c2 sin kx) = −k 2 c1 cos kx − k 2 c2 sin kx = −k 2 (c1 cos kx + c2 sin kx); dx2 d2 y d2 y = −k 2 y or + k2 y = 0 dx2 dx2 d (c1 cosh kx + c2 sinh kx) = kc1 sinh kx + kc2 cosh kx; 4. dx d2 (c1 cosh kx + c2 sinh kx) = k 2 c1 cosh kx + k 2 c2 sinh kx = k 2 (c1 cosh kx + c2 sinh kx); dx2 1. 18 CHAPTER 1 REVIEW EXERCISES d2 y = k2 y dx2 d2 y − k2 y = 0 dx2 or 5. y = c1 ex + c2 xex ; y = c1 ex + c2 xex + c2 ex ; y = c1 ex + c2 xex + 2c2 ex ; y + y = 2(c1 ex + c2 xex ) + 2c2 ex = 2(c1 ex + c2 xex + c2 ex ) = 2y ; y − 2y + y = 0 6. y = −c1 ex sin x + c1 ex cos x + c2 ex cos x + c2 ex sin x; y = −c1 ex cos x − c1 ex sin x − c1 ex sin x + c1 ex cos x − c2 ex sin x + c2 ex cos x + c2 ex cos x + c2 ex sin x = −2c1 ex sin x + 2c2 ex cos x; y − 2y = −2c1 ex cos x − 2c2 ex sin x = −2y; 7. a,d 8. c y − 2y + 2y = 0 10. a,c 9. b 11. b 12. a,b,d 13. A few solutions are y = 0, y = c, and y = ex . 14. Easy solutions to see are y = 0 and y = 3. 15. The slope of the tangent line at (x, y) is y , so the differential equation is y = x2 + y 2 . 16. The rate at which the slope changes is dy /dx = y , so the differential equation is y = −y or y + y = 0. 17. (a) The domain is all real numbers. (b) Since y = 2/3x1/3 , the solution y = x2/3 is undefined at x = 0. This function is a solution of the differential equation on (−∞, 0) and also on (0, ∞). 18. (a) Differentiating y 2 − 2y = x2 − x + c we obtain 2yy − 2y = 2x − 1 or (2y − 2)y = 2x − 1. (b) Setting x = 0 and y = 1 in the solution we have 1 − 2 = 0 − 0 + c or c = −1. Thus, a solution of the initial-value problem is y 2 − 2y = x2 − x − 1. (c) Solving y 2 − 2y − (x2 − x − 1) = 0 by the quadratic formula we get y = (2 ± 4 + 4(x2 − x − 1) )/2 √ = 1± x2 − x = 1± x(x − 1) . Since x(x−1) ≥ 0 for x ≤ 0 or x ≥ 1, we see that neither y = 1+ x(x − 1) nor y = 1 − x(x − 1) is differentiable at x = 0. Thus, both functions are solutions of the differential equation, but neither is a solution of the initial-value problem. 19. Setting x = x0 and y = 1 in y = −2/x + x, we get 1=− 2 + x0 x0 or x2 − x0 − 2 = (x0 − 2)(x0 + 1) = 0. 0 Thus, x0 = 2 or x0 = −1. Since x = 0 in y = −2/x+x, we see that y = −2/x+x is a solution of the initial-value problem xy + y = 2x, y(−1) = 1, on the interval (−∞, 0) and y = −2/x + x is a solution of the initial-value problem xy + y = 2x, y(2) = 1, on the interval (0, ∞). 20. From the differential equation, y (1) = 12 + [y(1)]2 = 1 + (−1)2 = 2 > 0, so y(x) is increasing in some neighborhood of x = 1. From y = 2x + 2yy we have y (1) = 2(1) + 2(−1)(2) = −2 < 0, so y(x) is concave down in some neighborhood of x = 1. 21. (a) y 3 y 3 2 2 1 1 -3 -2 -1 -1 1 2 3 x -3 -2 -1 -1 -2 2 -2 -3 1 -3 y = x2 + c1 y = −x2 + c2 19 3 x CHAPTER 1 REVIEW EXERCISES (b) When y = x2 + c1 , y = 2x and (y )2 = 4x2 . When y = −x2 + c2 , y = −2x and (y )2 = 4x2 . (c) Pasting together x2 , x ≥ 0, and −x2 , x ≤ 0, we get y = 22. The slope of the tangent line is y (−1,4) √ = 6 4 + 5(−1)3 = 7. −x2 , x ≤ 0 x2 , x > 0. 23. Differentiating y = x sin x + x cos x we get y = x cos x + sin x − x sin x + cos x and y = −x sin x + cos x + cos x − x cos x − sin x − sin x = −x sin x − x cos x + 2 cos x − 2 sin x. Thus y + y = −x sin x − x cos x + 2 cos x − 2 sin x + x sin x + x cos x = 2 cos x − 2 sin x. An interval of definition for the solution is (−∞, ∞). 24. Differentiating y = x sin x + (cos x) ln(cos x) we get − sin x cos x y = x cos x + sin x + cos x − (sin x) ln(cos x) = x cos x + sin x − sin x − (sin x) ln(cos x) = x cos x − (sin x) ln(cos x) and y = −x sin x + cos x − sin x − sin x cos x − (cos x) ln(cos x) sin2 x − (cos x) ln(cos x) cos x 1 − cos2 x = −x sin x + cos x + − (cos x) ln(cos x) cos x = −x sin x + cos x + sec x − cos x − (cos x) ln(cos x) = −x sin x + cos x + = −x sin x + sec x − (cos x) ln(cos x). Thus y + y = −x sin x + sec x − (cos x) ln(cos x) + x sin x + (cos x) ln(cos x) = sec x. To obtain an interval of definition we note that the domain of ln x is (0, ∞), so we must have cos x > 0. Thus, an interval of definition is (−π/2, π/2). 25. Differentiating y = sin(ln x) we obtain y = cos(ln x)/x and y = −[sin(ln x) + cos(ln x)]/x2 . Then x2 y + xy + y = x2 − sin(ln x) + cos(ln x) x2 +x cos(ln x) + sin(ln x) = 0. x An interval of definition for the solution is (0, ∞). 26. Differentiating y = cos(ln x) ln(cos(ln x)) + (ln x) sin(ln x) we obtain y = cos(ln x) =− 1 cos(ln x) − sin(ln x) x + ln(cos(ln x)) − ln(cos(ln x)) sin(ln x) (ln x) cos(ln x) + x x 20 sin(ln x) x + ln x cos(ln x) sin(ln x) + x x CHAPTER 1 REVIEW EXERCISES and y = −x ln(cos(ln x)) 1 sin(ln x) cos(ln x) + sin(ln x) − x cos(ln x) x + ln(cos(ln x)) sin(ln x) = 1 x2 1 sin(ln x) 1 cos(ln x) 1 + x (ln x) − − (ln x) cos(ln x) 2 + x2 x x x2 x sin2 (ln x) 1 − ln(cos(ln x)) cos(ln x) + + ln(cos(ln x)) sin(ln x) 2 x cos(ln x) − (ln x) sin(ln x) + cos(ln x) − (ln x) cos(ln x) . Then x2 y + xy + y = − ln(cos(ln x)) cos(ln x) + sin2 (ln x) + ln(cos(ln x)) sin(ln x) − (ln x) sin(ln x) cos(ln x) + cos(ln x) − (ln x) cos(ln x) − ln(cos(ln x)) sin(ln x) + (ln x) cos(ln x) + cos(ln x) ln(cos(ln x)) + (ln x) sin(ln x) 2 = sin (ln x) sin2 (ln x) + cos2 (ln x) 1 + cos(ln x) = = = sec(ln x). cos(ln x) cos(ln x) cos(ln x) To obtain an interval of definition, we note that the domain of ln x is (0, ∞), so we must have cos(ln x) > 0. Since cos x > 0 when −π/2 < x < π/2, we require −π/2 < ln x < π/2. Since ex is an increasing function, this is equivalent to e−π/2 < x < eπ/2 . Thus, an interval of definition is (e−π/2 , eπ/2 ). (Much of this problem is more easily done using a computer algebra system such as Mathematica or Maple.) 27. From the graph we see that estimates for y0 and y1 are y0 = −3 and y1 = 0. 28. The differential equation is dh cA0 =− dt Aw 2gh . Using A0 = π(1/24)2 = π/576, Aw = π(2)2 = 4π, and g = 32, this becomes dh c √ cπ/576 √ 64h = h. =− dt 4π 288 21 ...
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