06 - CHAPTER 6 Differential Equations Section 6.1 Section...

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Unformatted text preview: CHAPTER 6 Differential Equations Section 6.1 Section 6.2 Section 6.3 Section 6.4 Section 6.5 Section 6.6 Slope Fields and Euler’s Method . . . . . . . . . . . . . . 518 Differential Equations: Growth and Decay . . . . . . . . 529 Differential Equations: Separation of Variables . . . . . . 539 The Logistic Equation . . . . . . . . . . . . . . . . . . . 553 First-Order Linear Differential Equations . . . . . . . . . 559 Predator-Prey Differential Equations . . . . . . . . . . . . 570 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574 Review Exercises Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 586 CHAPTER 6 Differential Equations Section 6.1 Slope Fields and Euler’s Method 4y 2. Differential equation: 3y Solution: y 4y Check: 3 2xy x2 y2 y e x 1. Differential equation: y Solution: y Check: y Ce4x 4Ce4x 4y e x e x e x 4e dy dx x 3e xy y2 1 x 4e x e x 3. Differential equation: y Solution: x 2 Check: 2x y2 2yy y y Cy Cy 4. Differential equation: Solution: y 2 Check: 2yy y y y y xy y2 1 2 ln y 2 y y 1 y y x 1 y x2 2x x 2x 2y C 2xy 2y 2 Cy 2y 2 y2 2xy x2 y2 2xy x2 2xy x 2 y2 5. Differential equation: y Solution: y Check: C1 cos x y y y y y 0 C2 sin x C2 cos x C2 sin x C2 sin x 2y x C1 sin x C1 cos x C1 cos x 2y C1 cos x C2 sin x 0 6. Differential equation: y Solution: y Check: C1e x 0 cos x y y C2e C1 2 C1e 2C1e 2 2C1 x x sin x x C2 e sin x sin x 2C2e 2C2e x x x C1 cos x C2 e x cos x y 2y 2y sin x C2 e 2C1 cos x C1 x C1 sin x 2C2 e C2 e x cos x 2C2 2 C1e 2C1 x cos x C2e x x sin x 0 2C2 sin x 2C2 2C1 e cos x 518 S ection 6.1 7. Differential equation: y Solution: y y y tan x tan x tan x sec x tan x sec x sec2 x sin x ln sec x tan x Slope Fields and Euler’s Method 519 cos x ln sec x cos x 1 sec x tan x sec x 1 y cos x sec x tan x tan x sin x ln sec x 1 sec x tan x tan x sec x sin x ln sec x sin x tan x sec2 x cos x ln sec x tan x sin x sec x Check: y y sin x sec x tan x. 8. Solution: y y y Check: y 9. y y 2 3 2 3 2 3 cos x ln sec x cos x ln sec x tan x tan x cos x ln sec x tan x e 2x ex 2x 2e 4e 2 3 2x ex ex 2x 2y 4e ex 2 2 3 2e 2x ex 2e x. sin x cos x sin2 x 1 cos 2 x cos 2 x 2 cos x sin x sin 2 x 2 cos 2 x Differential equation: 2y y 2 sin x cos x 2 sin x cos x 2 sin 2 x Initial condition: y 4 12 2x cos 2 x 1 sin 2x 1 2 cos 2 x sin 2 x 1 sin 4 cos 4 2 cos2 4 2 2 2 2 2 2 2 0 10. y y 4 cos x 4 sin x 11. y y 6e 6e 2 x2 2 x2 x 4x 24 xe 2 x2 Differential equation: y x 4 sin x Differential equation: y 24 xe 2x2 4x 6e 2x2 4 xy Initial condition: y0 1 2 Initial condition: 4 cos 0 2 4 2 2 y0 6e 202 0 2 6e0 61 6 520 12. y y Chapter 6 e e cos x cos x Differential Equations sin x sin x e cos x Differential equation: y sin x e cos x sin x y y sin x Initial condition: y 2 e cos 2 e0 1 In Exercises 13–18, the differential equation is y 4 13. y y 4 16y 0. 14. y y 4 y 4 3 cos x 3 cos x 3 cos x 16 3 cos x 45 cos x 0, 3 cos 2x 48 cos 2x 48 cos 2x 48 cos 2x 0, 16y y 4 16y No 15. y y4 Yes y 4 Yes e 2x 2x 16. y y4 5 ln x 30 x4 30 x4 80 ln x 0, 16e 16e 16y 2x 16e 2x 0, y4 No 16y 17. y y4 y4 Yes 16y C1e2x 16C1e2x 0, C2e 2x C3 sin 2x 2x C4 cos 2x 16C4 cos 2x 16C2e 16C3 sin 2x 18. y y4 y 4 3e2x 48e2x 48e2x 4 sin 2x 64 sin 2x 64 sin 2x 16 3e2x 4 sin 2x 0, 16y Yes In 19–24, the differential equation is xy 19. y xy No x 2, y 2y 2x x 2x 2 x2 0 x3e x, 2y x3ex. 20. y xy Yes x 2e x, y 2y x x 2e x ex x2 2 xe x 2x 2 e x x2 x2e x 2x x 3e x, 21. y xy Yes x2 2 2y ex , y x x2e x x2 ex 2xe x 4x 2x 2 ex 2x2 x 3e x, 22. y xy No sin x, y 2y cos x x cos x 2 sin x x 3e x, 2 x2e x S ection 6.1 1 x x 1 x 2 ln x x 3e x, Slope Fields and Euler’s Method 521 23. y xy No 25. y 3 ln x, y 2y 24. y xy Yes x2e x 2y 5x2, y xxe 2x x2e x 2 xe x 2xe x 10x 10x 2 x 2e x 5x2 x 3e x, Ce Ce 0 x2 passes through 0, 3 . C⇒C 3 3e x2 26. y x 2 20 y 2 C passes through 0, 2 . C⇒C 2 4 y 4 Particular solution: y 27. y 2 16 Particular solution: y x 28. 2x 2 29 y2 16 Cx3 passes through 4, 4 . C 64 ⇒ C 1 4 13 4x C passes through 3, 4 . C⇒C 2x 2 2 y2 2 Particular solution: y2 or 4y 2 x C 0 x3 Particular solution: 29. Differential equation: 4yy General solution: 4 y 2 Particular solutions: C C x2 30. Differential equation: yy General solution: x 2 Particular solutions: C y 2 1 x x C 0 y2 ± 1, C 0, Two intersecting lines ± 4, Hyperbolas 0, Point C 1, C 4, Circles 1 2 31. Differential equation: y General solution: y y 2y C 2e 2x 2y 2x 0 32. Differential equation: 3x General solution: 3x2 2y2 2yy C 0 Ce 2 Ce 3, 3 3e 2x 2x 0 C 6x 2 3x 3x 4yy 2yy 2yy 0 0 0 3: 3 1 3 18 2y2 21 2 Initial condition: y 0 Particular solution: y Ce0 Initial condition: y 1 23 21 2 C Particular solution: 3x2 33. Differential equation: y General solution: y y y y 9y 3C1 cos 3x 9C1 sin 3x 9C1 sin 3x 9 C1 sin 3x 9y 0 C2 cos 3x Initial conditions: y 2 y 1 C1 sin 6 2, y C2 cos 6 1 2 C1 sin 3x 3C2 sin 3x, 9C2 cos 3x 9C2 cos 3x C2 cos 3x 0 2 2 ⇒ C1 3C1 cos 3x 3C1 cos 2 3C2 sin 3x 3C2 sin 1 3 2 sin 3x 1 cos 3x 3 2 3C2 ⇒ C2 Particular solution: y 522 Chapter 6 Differential Equations 1 2 34. Differential equation: xy General solution: y y y xy y x 1 C2 x C2 1 x2 1 x2 C2 C1 y 0 Initial conditions: y 2 0 y 1 2 C1 C2 x C2 ⇒ C2 2 C2 ln 2 0, y 2 C2 ln x 1, C1 ln 2 ln 2 ln x ln x 2 C2 1 x 0 Particular solution: y 35. Differential equation: x 2 y General solution: y y x2y C1 3x y 3C2 x , y 3y 2 3xy C2 x 3 3y 0 C1 x 6C2 x x2 6C2 x 3x C1 4 3C2 x 2 3 C1x C2 x 3 0 Initial conditions: y 2 0 y 4 C1 C1 2C1 C1 C1 4C2 12C2 8C2 3C2 x 2 12C2 0 4 C2 0, y 2 1 2, C1 13 2x 2 Particular solution: y 36. Differential equation: 9y General solution: y y y 9y 12y 4y 2x 12y C1 4y C2 x C2x C2 0 e2x 3 2 2x 3 3e C1 C2e2x 2 3 C2 x 3 e2x e2x 32 3 C1 C2 2 3 C1 2 3 C2x 2 2x 3 2 3e 3 C1 3 2C 32 2 2x 3 2 3e 3 C1 3 2C2 2 3 C2 x 2 3 C2x 9 2e2x 3 3 2 3 C1 2C2 0 2 3 C2 x 12 e2x C2 4 e2x 3 C1 C2 x 0 Initial conditions: y 0 0 4 0 e2 C1 1 C1 e4 2 4, y 3 3C2 0 ⇒ C1 3C2 ⇒ C2 e2x 3 4 4 3 Particular solution: y dy dx y 4 4 3x 37. 3x2 3x 2 dx x3 C 38. dy dx y x3 x3 4x 4x dx x4 4 2x2 C 39. dy dx y u x 1 x 1 1 x2 x2 dx 1 ln 1 2 2x dx x2 C 40. dy dx y ex 1 ex 1 ex ex dx ln 1 ex C x 2, du Section 6.1 dy dx y x dy dx y u dy dx Let u y 2 2 x 5 x x 1 2 2 x dy dx y C x ln x 2 C u dy dx y Slope Fields and Euler’s Method 523 41. 1 2 dx x 42. x cos x 2 x cos x2 dx x 2, du 2x dx 1 sin x2 2 C 2 ln x 43. sin 2x sin 2x dx 2x, du 2 dx 1 cos 2x 2 C 44. tan2 x sec2 x sec2 x 1 dx 1 tan x x C 45. xx x xx u4 3 3 3, then x 3 dx 3u 2 du 52 46. u2 u2 2 u5 5 3 32 dy dx Let u y x5 5 x5 x x, u2 x dx 5 5 x, dx u2 u 10u2 2u du. 2u du 2u4 du 2u5 5 x 32 3 and dx 3 u 2u du u3 C C 2u du. 2x 10u3 3 10 5 3 C 2 5 5 x 52 C 47. dy dx y u xex 2 48. 2 dy dx y 5e 5e x2 xe x dx x 2, du 1 x2 e 2 2x dx C x 2 dx 5 2 10e e x2 x2 1 dx 2 C 49. x y dy d x 4 2 2 2 0 Undef. 0 4 0 2 4 1 2 4 6 2 3 8 8 1 50. x y dy d x 4 2 6 2 0 2 0 4 4 2 4 2 4 6 2 8 8 0 51. x y dy d x 4 2 22 2 0 2 0 4 0 2 4 0 4 6 22 8 8 8 52. x y dy d x 4 2 3 2 0 0 0 4 3 2 4 3 4 6 0 8 8 3 53. dy dx For x cos 2x , dy dx 1. Matches (b). 54. dy dx For x 1 sin x 2 0, dy dx 0. Matches (c). 524 dy dx Chapter 6 Differential Equations dy dx For x 1 x 0, dy is undefined (vertical tangent). Matches (a). dx (c) As x → y 4 55. e 2x 56. dy → 0. Matches (d). dx (c) As x → , y→ , y→ . . As x → , 57. (a), (b) y 58. (a), (b) , y→ , y→ . . (2, 4) 5 As x → As x → −4 x 6 x 4 −5 −4 59. (a), (b) y (c) As x → (1, 1) 4 , y→ , y→ . . 60. (a), (b) y 8 (c) As x → As x → , y→ , y→ . . As x → −4 x 4 −4 −4 x 4 61. (a) y y 1 ,y 1 x (1, 0) 0 62. (a) y y 6 5 4 x 3 2 1 6 1 ,y 0 y (0, 1) 1 3 2 1 −1 −2 −3 x 1 2 3 4 5 6 As x → , y→ . ln x. As x → (b) y y 6 , y→ 1 . Note: The solution is y (b) y y 1 ,y2 x (2, − 1) 1 ,y1 y (1, 1) 1 3 2 5 4 3 1 x −1 −2 −3 6 2 1 x 1 2 3 4 5 6 As x → , y→ . , y→ . As x → S ection 6.1 dy dx dy dx Slope Fields and Euler’s Method 525 63. 0.5y, y 0 12 6 64. 2 y, y 0 9 4 −6 −4 6 −5 −1 5 65. dy dx 0.02y 10 12 y ,y 0 2 66. dy dx 0.2 x 2 10 y ,y 0 9 − 12 −2 48 −5 5 0 67. Slope field for y through 0, 1 8 0.4y 3 x with solution passing 68. Slope field for y through 0, 2 5 1 e 2 x8 sin y with solution passing 4 −2 −2 8 −3 3 −3 69. y y1 y2 n xn yn x y0 y1 0 0 2 y, y0 2, 2 n 10, 0.1 0 h 2 0.1 2.2 2.2 5 0.5 3.332 2.43, etc. 6 0.6 3.715 7 0.7 4.146 8 0.8 4.631 9 0.9 5.174 10 1.0 5.781 hF x0, y0 hF x1, y1 1 0.1 2.2 2 0.2 2.43 2.2 3 0.3 0.1 0.1 4 0.4 2.992 2.693 70. y y1 y2 x y0 y1 y, y0 2, 2 n 20, h 0.05 2 2.1 2.1 2.2075, etc. hF x0, y0 hF x1, y1 0.05 0 2.1 0.05 0.05 The table shows the values for n n xn yn 0 0 2 2 0.1 2.208 4 0.2 2.447 6 0.3 0, 2, 4, . . . , 20. 8 0.4 3.032 10 0.5 3.387 12 0.6 3.788 14 0.7 4.240 16 0.8 4.749 18 0.9 5.320 20 1.0 5.960 2.720 526 71. y y1 y2 n Chapter 6 3x y0 y1 0 0 3 2y, Differential Equations y0 3, 3 2.7 2 0.1 2.438 n 10, h 0.05 23 2.7 2 2.7 5 0.25 1.839 6 0.3 1.693 2.4375, etc. 7 0.35 1.569 8 0.4 1.464 9 0.45 1.378 10 0.5 1.308 hF x0, y0 hF x1, y1 1 0.05 2.7 0.05 3 0 0.05 3 0.05 3 0.15 2.209 4 0.2 2.010 xn yn 72. y y1 y2 n xn yn 0.5x 3 y0 y1 0 0 1 y, y0 1 1 2 0.8 1.16 1, n 5, h 0.4 1 1 5 2.0 2.201 1 1.16, etc. hF x0, y0 hF x1, y1 1 0.4 1 0.4 0.5 0 3 0.4 0.5 0.4 3 3 1.2 1.454 4 1.6 1.825 73. y y1 y2 n xn yn e xy, y0 y1 0 0 1 y0 1, n 1 1.1 10, h 0.1 1.1 1.1 hF x0, y0 hF x1, y1 1 0.1 1.1 2 0.2 0.1 e0 1 0.1 e 0.1 3 0.3 1.339 4 0.4 1.2116, etc. 5 0.5 1.670 6 0.6 1.900 7 0.7 2.213 8 0.8 2.684 9 0.9 3.540 10 1.0 5.958 1.212 1.488 74. y y1 y2 n xn yn cos x y0 y1 0 0 5 sin y, hF x0, y0 hF x1, y1 1 0.1 5.004 y0 5 5, n 10, h sin 5 0.1 5.0041 sin 5.0041 5 0.5 5.007 6 0.6 4.999 5.0078, etc. 7 0.7 4.985 8 0.8 4.965 9 0.9 4.938 10 1.0 4.903 0.1 cos 0 5.0041 2 0.2 5.008 3 0.3 0.1 cos 0.1 4 0.4 5.010 5.010 75. dy dx y, y 3e x, y 0 3 x yx (exact) yx h yx h 0.2 0 3 3 3 0.1 0.2 3.6642 3.6000 3.6300 0.4 4.4755 4.3200 4.3923 0.6 5.4664 5.1840 5.3147 0.8 6.6766 6.2208 6.4308 1.0 8.1548 7.4650 7.7812 S ection 6.1 dy dx 2x ,y y Slope Fields and Euler’s Method 527 76. 2x2 4, y 0 2 x yx (exact) yx h yx h 0.2 0 2 2 2 0.1 0.2 2.0199 2.000 2.0100 0.4 2.0785 2.0400 2.0595 0.6 2.1726 2.1184 2.1460 0.8 2.2978 2.2317 2.2655 1.0 2.4495 2.3751 2.4131 77. dy dx x y cos x, y 0 0 0 0.2 0 0.1 1 sin x 2 0.2 0.2200 0.2000 0.2095 cos x 0.4 0.4801 0.4360 0.4568 ex , y 0 0.6 0.7807 0.7074 0.7418 0 0.8 0.1231 0.0140 0.0649 1.0 0.5097 0.3561 0.4273 78. As h increases (from 0.1 to 0.2), the error increases. yx (exact) yx h yx h 79. dy dt (a) t 1 y 2 72 , y 0 0 140 68e 0 140 1 112.7 t 2, 140, h 2 96.4 0.1 3 86.6 80. dy dt (a) t 1 y 2 72 , y 0 0 1 140, h 2 96.7 0.05 3 86.9 Euler (b) y t Exact 72 Euler (b) y 72 t Exact 140 112.98 68e 0 140 t2 exact 1 2 97.016 3 87.173 , exact 1 2 97.016 3 87.173 0.05. 113.24 113.24 The approximations are better using h 81. A general solution of order n has n arbitrary constants while in a particular solution initial conditions are given in order to solve for all these constants. F x, y 82. A slope field for the differential equation y consists of small line segments at various points x, y in the plane. The line segment equals the slope y F x, y of the solution y at the point x, y . 84. y dy dx Cekx Ckekx 83. Consider y F x, y , y x0 y0. Begin with a point x0, y0 that satisfies the initial condition, y x 0 y0. Then using a step size of h, find the point x1, y1 x0 h, y0 h F x0 , y0 . Continue generating the sequence of points xn 1, yn 1 xn h, yn hF xn , yn . Since dy dx 0.07y, we have Ckekx 0.07Cekx. Thus, k 0.07. C cannot be determined. 86. True 85. False. Consider Example 2. y x3 is a solution to xy 3y 0, but y x3 1 is not a solution. 87. True 88. False. The slope field could represent many different differential equations, such as y 2x 4y. 528 dy dx (a) Chapter 6 Differential Equations 89. 2y, y 0 4, y 4e 2 x, solution (b) If h is halved, then the error is approximately halved r 0.5 . (c) When h 0.05, the errors will again be approximately halved. x y y1 y2 e1 e2 r 0 4 4 4 0 0 0.2 2.6813 2.5600 2.4000 0.1213 0.2813 0.4312 0.4 1.7973 1.6384 1.4400 0.1589 0.3573 0.4447 0.6 1.2048 1.0486 0.8640 0.1562 0.3408 0.4583 0.8 0.8076 0.6711 0.5184 0.1365 0.2892 0.4720 1.0 0.5413 0.4295 0.3110 0.1118 0.2303 0.4855 90. dy dx (a) x x y y1 y2 e1 e2 r y, y 0 0 1 1 1 0 0 0.2 1, y x 0.4 0.7406 0.7122 0.6800 0.0284 0.0606 0.47 1 2e x, 0.6 0.6976 0.6629 0.6240 0.0347 0.0736 0.47 solution 0.8 0.6987 0.6609 0.6192 0.0378 0.0795 0.48 1.0 0.7358 0.6974 0.6554 0.0384 0.0804 0.48 (b) If h is halved, then the error is halved r (c) When h 0.05, the error will again be approximately halved. 0.5 . 0.8375 0.8200 0.8000 0.0175 0.0375 0.47 91. (a) L dI dt RI 12 I dI dt Et 3 I 92. y y y e kt ke kt k 2e kt 0 0 0 ±4 dI 4 dt 24 1 24 4 6 3I t→ −3 t 3 12 I −3 y k 2e kt k 2 16y 16e kt 16 k since e kt 0 (b) As t → In fact, I 93. , I → 2. That is, lim I t 2. 2 is a solution to the differential equation. y y y y 16y 0 0 0 A sin A cos A 2 t t t 94. 2f x f x fx fx xg x f x , g x ≥ 0 2 xg x f x 2 xg x f x 2 2 2 2f x f x 2 sin d fx dx For x < 0, For x > 0, Thus, f x 2 for x > 0. fx 2 2 A 2 sin t 16 A sin t 2 2 xg x f x 2 xg x f x ≥ 0. ≤ 0. A sin t 16 If A 0, then ± 4 radians/sec. f x 2 is increasing for x < 0 and decreasing f x 2 f x 2 has a maximum at x 0. Thus, it is bounded by its value at x 0, f 0 2 f 0 2. Thus, f and f is bounded. S ection 6.2 95. Let the vertical line x k cut the graph of the solution y The tangent line at k, t is y Since y t fkx pxy k. q x , we have y t qk pkt x k. Differential Equations: Growth and Decay 529 f x at k, t . For any value of t, this line passes through the point k 1 qk , . pk pk To see this, note that qk pk t ? ? qk qkk qk pk pkt k p k tk t. 1 pk qk pk t k kq k p k kt Section 6.2 1. dy dx y x x 2 2 dx Differential Equations: Growth and Decay 2. x2 2 2x C dy dx y 4 4 x x dx 4x x2 2 C 3. dy dx dy y 1 y ln y y 2 2 dy 2 2 y y dx dx x ex Cex 2 4. dy dx dy 4 1 4 y y dy 4 dx y 5. y yy yy dx 5x y 5x 5x dx 5x dx 52 x 2 C C1 dx x e 4 x C1 C1 ln 4 Ce x y dy 4 y y C1 C1 y dy Ce x 2 Ce x 12 y 2 y2 5x 2 6. y 3yy 3yy dx 3y2 2 9y2 4x3 2 x 3y x x dx 23 x 3 C 2 7. y y y y dx y xy x x dx x dx 23 x 3 e2 2 8. y 1 y 1 1 C1 2 y y y dx x1 x x dx x dx x2 2 ex 2 y C1 dy y ln y y dy y y y y ln 1 1 C1 2 x2 C1 2 3 x3 C1 2 eC1e 2 Ce 2 3 3 x3 x3 2 eC1 e 1 1 2 Ce x 2 530 9. 1 Chapter 6 x2 y 2xy y y y y dx y dy y ln y ln y ln y y Differential Equations 0 1 1 1 2x 1 ln 1 ln 1 ln C 1 C1 x2 2xy x2 2x x2 2x x dx 2 10. xy y y y 100 y 100 y y dx 100x 100x x x dx x dx x2 2 x2 2 e e x2 2 C1e xy x 100 y dx x2 x2 x2 x2 C1 ln C 1 dy 100 y ln 100 ln 100 100 y y y y y C1 C1 C1 x2 2 100 x2 2 100 Ce 11. dQ dt dQ dt dt dQ Q k t2 k dt t2 k t k t C C 12. dP dt dP dt dt dP P k 10 k 10 k 10 2 k 10 2 t t dt t t 2 13. dN ds dN ds ds k 250 k 250 k 250 2 k 250 2 s s ds s s 2 C C dN N C C 2 2 14. 1 L 1 L L ln L L dy dx dy y dx kx L kx kx dx kx2 2 kx2 2 e y 15. (a) 9 y dy dx y dx 1 y dy y y y y L −5 x −1 5 C1 C1 C1 (0, 0) (b) dy dx dy y 6 6 6 y 0, 0 : 0 6 7 x6 x dx x2 2 e 6 x2 2 y, 0, 0 kx2 2 L e Ce C1 e kx2 2 ln y y C C kx2 2 C1e x2 2 x2 2 C1e C1 ⇒ C1 6⇒y 6 6e x2 2 −6 −1 6 S ection 6.2 Differential Equations: Growth and Decay dy dt dy y 1 t, 0, 10 2 1 t dt 2 12 t 4 1 0 4 12 t 4 16 531 16. (a) 4 y 17. (0, 1 ) 2 −4 x 4 C 2 −4 10 (b) dy dx dy y ln y y 0, xy, x dx x2 2 ex 22 C⇒C 10 10 1 0, 2 y C C (0, 10) C1e x 22 −4 −1 4 11 : 22 3 4 3 4 13 t 2 1 0 2 13 t 2 2 C1e0 ⇒ C1 1 ⇒y 2 1 x2 e 2 2 18. dy dt dy y 10 y 15 t, 0, 10 t dt C C⇒C 10 10 19. dy dt dy y ln y y 10 e 1 y, 0, 10 2 1 dt 2 1 t 2 t2 C1 C1 32 eC1 e 10 t2 Ce t2 Ce0 ⇒ C 10e t2 2 y 16 (0, 10) (0, 10) 0 −5 10 −1 −1 10 20. dy dt dy y ln y y 3 y, 0, 10 4 3 dt 4 3 t 4 e3 C1 4 t C1 40 21. dy dx y ky Cekx Ce0 C 1 5 ln 3 2 4eln 5 2 2 (0, 10) −5 −5 5 0, 4 : 4 3, 10 : 10 When x 4e3k ⇒ k 6, y 4e1 4 5 2 3 ln 5 2 6 2 eC1 e 3 4 t 10 y Ce0 ⇒ C 10e3t 4 Ce3t 4 10 25. 532 dN dt N Chapter 6 Differential Equations dV dt V 250 250ek ⇒ k 250e4 ln 8 250 8 5 4 22. kN Cekt 23. kV Cekt 20,000 20,000e4k ⇒ k 4 ln 5 8 6 32 0, 250 : C 1, 400 : 400 When t 0, 20,000 : C ln 400 250 ln 5 4 8 5 4, 12,500 : 12,500 When t 6, V 1 5 ln 4 8 20,000eln 5 8 32 4, N 5 250eln 8 8192 . 5 20,000e1 20,000 5 8 9882.118. 24. dP dt P kP Cekt 5000 5000ek ⇒ k 5000eln 19 5000 19 20 20 5 5 25. y C 19 20 y 5 3868.905. k y Cekt, 1 2 1 kt e 2 1 5k e 2 ln 10 5 1 ln 10 e 2 0, 1 , 5, 5 2 0, 5000 : C 1, 4750 : 4750 When t 5, P ln 5t 1t 10 2 5 or y 1 0.4605t e 2 1 , 4, 5 2 26. y C y 1 2 k y Cekt, 0, 4 , 4 4ekt 4e5k ln 1 8 5 4e 0.4159t 5, 1 2 27. y 1 5 5Cek 5ek Cekt, 1, 1 , 5, 5 Cek Ce5k Ce5k e5k e4k ln 5 4 28. y 1 2 5 2Ce3k 10e3k Cekt, Ce3k Ce4k 1 4k Ce 5 e4k ek ln 10 3, 0.4159 5 k y 1 C y 0.4024 10 k y Ce0.4024t Ce0.4024 0.6687 C 5 14 2.3026 Ce2.3026t Ce2.3026 4 0.0005 0.0005e2.3026t 5 C y 0.6687e0.4024t 29. In the model y Ce kt, C represents the initial value of y when t 0 . k is the proportionality constant. dy dx 1 xy 2 30. y dy dt ky 31. 32. dy dx 12 xy 2 dy > 0 when xy > 0; Quadrants I and III. dx dy > 0 when y > 0; Quadrants I and II. dx S ection 6.2 33. Since the initial quantity is 10 grams, y 10 e . kt Differential Equations: Growth and Decay 533 34. Since the half-life is 1599 years, 1 2 1e k 1599 1 1599 Since the half-life is 1599 years, 5 k Thus, y When t When t 10e k 1599 1 1599 k ln 1 2 . Since there are 1.5 g after 1000 years, . 1599 t. ln 1 2 1.5 C 1599 1000 Ce ln 1 2.314. 2 1599 1000 10e ln 1 2 1000, y 10,000, y 10e ln 1 2 0.13 g. 6.48 g. Hence, the initial quantity is approximately 2.314 g. When t 10,000, y 2.314e ln 1 2 1599 10,000 0.03 g. 35. Since the half-life is 1599 years, 1 2 36. Since the half-life is 5715 years, 1 2 1e k 1599 1 1599 1e k 5715 1 5715 k ln 1 2 . k ln 1 2 . Since there are 0.5 gram after 10,000 years, 0.5 C Ce ln 1 2 1599 10,000 Since there are 2 grams after 10,000 years, 2 C Ce ln 1 6.726. 2 5715 10,000 38.158. Hence, the initial quantity is approximately 38.158 g. When t 1000, y 38.158e ln 1 2 1599 1000 Hence, the initial quantity is approximately 6.726 g. When t 1000, y 6.726e ln 1 2 5715 1000 24.74 g. 5.96 g. 37. Since the initial quantity is 5 grams, C Since the half-life is 5715 years, 2.5 k When t When t 5e k 5715 1 5715 5. 38. Since the half-life is 5715 years, 1 2 1e k 5715 1 5715 k . 5e ln 1 2 5715 1000 5715 10,000 ln 1 2 . 1000 years, ln 1 2 Since there are 3.2 grams when t 4.43 g. 1.49 g. 3.2 C Ce ln 1 2 3.613. 5715 1000 1000 years, y 10,000 years, y 5e ln 1 2 Thus, the initial quantity is approximately 3.613 g. When t 39. Since the half-life is 24,100 years, 1 2 10,000, y 3.613e ln 1 2 5715 10,000 1.07 g. 40. Since the half-life is 24,100 years, 1 2 1e k 24,100 1 24,100 1e k 24,100 1 24,100 k ln 1 2 . k ln 1 2 . Since there are 2.1 grams after 1000 years, 2.1 C Ce ln 1 2 24,100 1000 Since there are 0.4 grams after 10,000 years, 0.4 C Ce ln 1 2 0.533. 24,100 10,000 2.161. Thus, the initial quantity is approximately 2.161 g. When t 10,000, y 2.161e ln 1 2 24,100 10,000 Thus, the initial quantity is approximately 0.533 g. When t 1000, y 0.533e ln 1 2 24,100 1000 0.52 g. 1.62 g. 534 41. y 1 C 2 k Chapter 6 Ce kt Ce k 1599 1 1 ln 1599 2 100, y Differential Equations 42. y 1 C 2 k Ce ln 1 2 1599 100 Ce kt Ce k 5715 1 1 ln 5715 2 Ce ln 1 2 5715 t When t 0.15C ln 0.15 t 0.9576 C Therefore, 95.76% remains after 100 years. ln 1 2 t 5715 15,641.8 years 20,000e0.055t, the time to double is given by 20,000e0.055t and we have e0.055t 0.055t ln 2 0.055 12.6 years. 20,000e 0.055 10 43. Since A 1000e0.06t, the time to double is given by 2000 1000e0.06t and we have 2 ln 2 t e0.06t 0.06t ln 2 0.06 11.55 years. 1000e 0.06 1500 when t 10 44. Since A 40,000 2 ln 2 t Amount after 10 years: A 45. Since A 750ert and A the following: 1500 r 750e7.75r ln 2 7.75 0.0894 $1822.12 7.75, we have Amount after 10 years: A 46. Since A 10,000ert and A have the following: 20,000 10,000e5r ln 2 5 $34,665.06 5, we 20,000 when t 8.94% 750e0.0894 10 $1833.67 10, we have r 0.1386 13.86% ln 2 5 10 Amount after 10 years: A 47. Since A 500ert and A the following: 1292.85 r 500e10r Amount after 10 years: A 48. Since A 2000ert and A have the following: 5436.56 2000e10r 10,000e $40,000 10, we 1292.85 when t 5436.56 when t ln 1292.85 500 10 0.0950 9.50% r ln 5436.56 2000 10 0.10 10% The time to double is given by 1000 t 500e0.0950t ln 2 0.095 7.30 years. The time to double is given by 4000 t 2000e0.10t ln 2 0.10 6.93 years. 49. 500,000 P P1 0.075 12 12 20 50. 500,000 0.075 12 12 35 240 P1 0.06 12 12 40 500,000 1 $112,087.09 P 500,000 1.005 480 $45,631.04 51. 500,000 P P1 0.08 12 52. 500,000 0.08 12 420 P1 0.09 12 12 25 500,000 1 $30,688.87 P 500,000 1 $53,143.92 0.09 12 300 S ection 6.2 Differential Equations: Growth and Decay 0.07 365 365t 365t 535 53. (a) 2000 2 ln 2 t (b) 2000 2 ln 2 t 1000 1 1.07t t ln 1.07 ln 2 ln 1.07 1000 1 1 0.07 t (c) 2000 2 1000 1 1 0.07 365 10.24 years 0.07 12 12t 12t ln 2 t (d) 2000 2 365t ln 1 0.07 365 ln 2 0.07 365 9.90 years 365 ln 1 0.007 12 1000e 0.07 t e0.07t 0.07t ln 2 0.07 9.90 years 12t ln 1 0.07 12 9.93 years ln 2 t ln 2 12 ln 1 0.07 12 54. (a) 2000 2 ln 2 t (b) 2000 2 ln 2 t 1000 1 1.06t t ln1.06 ln 2 ln 1.06 1000 1 1 0.06 12 0.6 t (c) 2000 2 1000 1 1 0.06 365 0.06 365 365t 365t 11.90 years 0.06 12 12t 12t ln 2 t 365t ln 1 1 365 0.06 365 11.55 years ln 2 0.06 ln 1 365 (d) 2000 0.06 12 11.58 years 1000e0.06t 2 ln 2 t e0.06t 0.06t ln 2 0.06 11.55 years 12t ln 1 1 12 ln 2 0.06 ln 1 12 55. (a) 2000 2 ln 2 t (b) 2000 2 ln 2 t 1000 1 1.085t t ln1.085 ln 2 ln 1.085 1000 1 1 0.085 t (c) 2000 2 1000 1 1 0.085 365 365t 365t 0.085 365 8.50 years 0.085 12 12t 12t ln 2 t 365t ln 1 1 365 0.085 365 8.16 years 0.085 12 ln 2 0.085 ln 1 365 (d) 2000 0.085 12 8.18 years 2 ln 2 t 1000e0.085t e0.085t 0.085t ln 2 0.085 8.15 years 12t ln 1 1 12 ln 2 0.085 ln 1 12 536 Chapter 6 Differential Equations 0.055 365 365t 365t 56. (a) 2000 2 ln 2 t (b) 2000 2 ln 2 t 1000 1 1.055t t ln1.055 ln 2 ln 1.055 1000 1 1 0.055 t (c) 2000 2 1000 1 1 0.055 365 12.95 years 0.055 12 12t 12t ln 2 t 365t ln 1 1 365 0.055 365 12.60 years 0.055 12 ln 2 0.055 ln 1 365 (d) 2000 0.055 12 12.63 years 2 ln 2 t 1000e0.055t e0.055t 0.055t ln 2 0.055 12.60 years 12t ln 1 1 12 ln 2 0.055 ln 1 12 Ce 0.009 t 0.009 1 57. (a) P P1 Ce kt 7.7 C 58. (a) P P1 Ce kt 12.7 C Ce 0.018 t Ce 0.018 1 12.4734 Ce 7.7696 0.009 t P (b) For t 7.7696e 15, P P 0.009 15 12.4734 e0.018 t 15, P 12.4734 e 0.018 15 16.34 million. 7.7696e 6.79 million. (b) For t (c) If k < 0, the population is decreasing. 59. (a) P 5.2 Ce kt P1 C P (b) For t Ce 0.026 t Ce 0.026 1 5.0665 (c) For k > 0, the population is increasing. 60. (a) P P1 Ce kt 3.6 C P 3.6072e 15, P Ce 0.002 t 0.002 1 Ce 3.6072 0.002 t 5.0665e 0.026 t 15, P 5.0665e 0.026 15 7.48 million. (b) For t 3.6072e 0.002 15 3.5 million. (c) For k > 0, the population is increasing. 61. (a) N (b) N 100.1596 1.2455 400 when t t (c) For k < 0, the population is decreasing. 6.3 hours (graphing utility) Analytically, 400 1.2455t t ln 1.2455 t 100.1596 1.2455 400 100.1596 ln 3.9936 ln 3.9936 ln 1.2455 6.3 hours. t 3.9936 S ection 6.2 62. (a) Let y Ce kt. Cek 2 ⇒ C Ce k 4 ⇒ 350 ln 154 1 2 Differential Equations: Growth and Decay 537 At time 2: 125 At time 4: 350 2k k C 125e 2k 125e 2k 2k 125e e4 k ⇒ 14 5 e 2k ln 154 0.5148 125 5 14 625 14 125e 2 1 2 ln 14 5 44.64 Approximately 45 bacteria at time 0. (b) y 625 1 2 ln 14 5 t 14 e 44.64 e 0.5148 t 625 14 4 14 5 (c) When t (d) 25,000 8, y 625 1 2 ln 14 5 8 14 e 2744. 625 1 2 ln 14 5 t 14 e ⇒t 12.29 hours 63. (a) 19 30e20k k N 30 1 11 e20k 64. (a) 20 30e30k 30 1 10 ln 1 3 30 30 1 30 1 1 6 e30k ln 11 30 20 30 1 30 1 1 6 ln 6 0.0502 181e kt e 0.0502 0.0502t k N (b) e 25 0.0366t ln 3 30 e e 0.0366t 0.0366t 0.0366 (b) e 25 0.0502t e 0.0502t t 36 days t ln 6 0.0366 49 days 65. (a) P1 205 P1 (c) 300 Ce kt (b) Using a graphing utility, P2 1 205 ln 10 181 t 182.3248 1.01091 t 181e10k ⇒ k 181e 0.01245t P1 P2 0.01245 (d) Using the model P2 , 320 320 182.3248 t 182.3248 1.01091 1.01091 t t 181 1.01253 ln 320 182.3248 ln 1.01091 51.8 years, or 2011. 0 150 50 The model P2 fits the data better. 66. (a) R 941.6088 1.0756 941.6088e0.0729 t I (c) 500 t (b) 2500 According to the model, 68.6e 0.0729t. Rt 0.14164 t 4 3.9288t 3 36.599t 2 120.82 t 417.0 0 0 12 (d) P t 0 0 12 1 I R 0 0 12 538 Chapter 6 Differential Equations I 10 16 67. I (a) (b) (c) (d) I 10 log10 , I0 I0 10 10 10 10 14 10 10 10 16 68. 20 decibels 70 decibels 93 6.7 80 8 10 log10 10 log10 I 10 6.7 16 14 16 9 16 6.5 16 4 16 10 log10 log10 I ⇒ I 10 log10 I 10 16 9 10 10 log10 10 10 log10 10 log10 10 log10 I 10 10 8 16 log10 I ⇒ I 6.5 10 10 95 decibels 120 decibels 6.7 Percentage decrease: 10 6.7 8 10 100 95% 4 10 10 69. A t dA dt Vte 0.10t 100,000e0.8 0.10 e0.8 t e 0.10t 100,000e0.8 0 when t t 0.10t 100,000 0.4 t t 0.10t 16. The timber should be harvested in the year 2014, 1998 16 . Note: You could also use a graphing utility to graph A t and find the maximum of A t . Use the viewing rectangle 0 ≤ x ≤ 30 and 0 ≤ y ≤ 600,000. ln I ln I0 ln 10 ln I 0 ,I ln 10 70. R e R ln 10 10R ln I ln I0 ln 10 e 2R ln 10 e 2R ln 10 eR ln 10 2 (a) 8.3 I ln I ln I0 ln 10 108.3 199,526,231.5 (b) 2R I (c) R dR dI ln I ln I0 ln 10 l I ln 10 10R 2 Increases by a factor of e2R ln 10 or 10R. dy dt 80 dy kt dy dt y 160 C. 1500. Thus, C 1120. Thus, ln 1120 ln 1040 142 t 71. Since 1 y ln y ky 80 , k dt 72. ky 20 20 20 e 5k 1 2 ln 5 7 20 e ln 2 20 Ce kt Cek 0 (See Example 6) ⇒C 140 80 0, y 1, y 60 ln 1420. 2 7 k 104 ln . 142 30 1 14 ln 1 14 t 140e k 5 When t When t k1 ln 1420 k 80 ln 1420 0.25055 ln 2 7 t t5 140e 1 5 7t5 Thus, y When t 1420e ln 104 5, y 80. 379.2 . 2 7 t2 ln 57 5 ln 14 ln 2 7 1 5 ln 14 ln 7 2 5 10.53 minutes 5.53 minutes longer. It will take 10.53 73. False. If y 74. True Cekt, y Ckekt constant. 75. True 76. True Section 6.3 Differential Equations: Separation of Variables 539 Section 6.3 1. dy dx y dy y2 2 y2 dr ds dr r x2 x y x dx x2 2 C Differential Equations: Separation of Variables 2. dy dx 3y 2 dy y3 x3 3 x2 2 3y 2 x2 2x 2 dx C 3. dr ds dr r ln r r 0.05r 0.05 ds 0.05s e0.05s C1 C1 C1 Ce0.05s 4. 0.05s 0.05s ds 0.025s 2 C 5. 2 xy dy y ln y 3y 3 2 3 ln 2 ln C 2 x dx x x 2 3 3 6. xy dy y y dx x ln x Cx ln C ln Cx ln C ln y y y Cx 7. yy y dy y2 2 y2 sin x sin x dx cos x 2 cos x C1 C 8. y dy dx 6 cos x 6 cos x dx 6 sin x sin x C1 C y dy y2 2 y2 12 9. 1 4x2 y dy dy x x 1 1 1 8 1 4x2 x 4x2 dx dx 12 10. x2 9 dy dx dy y 5x 5x dx x2 9 5 x2 9 12 C 4x2 4x2 12 8x dx y 1 1 4 C 11. y ln x xy dy y ln y y 0 ln x dx x 1 ln x 2 e1 2 12. u C1 C1 4y dy dx 3ex 3ex dx 3ex C ln x, du dx x 4y dy 2y2 2 ln x 2 Ce ln x 22 540 13. yy Chapter 6 ex y dy y2 2 y2 ex 2ex 0 ex dx Differential Equations 14. x yy y1 2 dy 23 y 3 y3 4, 16 2ex 2 14 C, C 14 2 2 0 x1 2 dx 23 x 3 C 4, 4 2 32 2 2 C1 C C1 x3 2 Initial condition: y 0 Particular solution: y 2 15. y x 1 y dy y ln y y Ce x 2 x Initial condition: y 1 Particular solution: y3 1 9 32 8 1 9 C x3 0 x 1 1 dx 2 16. 2xy ln x2 2x dy dx dy y 0 2 ln x ln x dx x ln x 2 C 1 ln x 2 x2 dy dx dy 12 2 2 C1 122 C Initial condition: y Particular solution: y 2 e1 1, 1 x Ce 12 2 1 2, C x2 e1 2x 2 2 y1 2: 2 y e 2 17. y 1 y 1 1 ln 1 2 ln 1 1 y0 x2 y y2 dy x1 x 1 y2 x2 dx x2 x2 x2 C⇒C 41 3 x2 4x 2 4 C1 ln C ln C 1 x2 18. 1 y1 y2 1 y0 1: x1 1 1 1 1 y2 x2 x2 1 2x 12 1 2y dx C 1 y2 y2 y2 3: 1 1 1 ln 1 2 ln 1 C1 3 y2 y2 y2 0 1 y2 C⇒C x2 1 19. du dv du u ln u u u v sin v 2 v sin v 2 dv 1 cos v 2 2 Ce cos v 2 2 20. dr ds e rdr e r0 r er e 2s 2s ds 2s C1 1 e 2 0: e e 1 r C 1 2 1 e 2 C⇒C 2s 1 2 Initial condition: u 0 Particular solution: u 1, C e1 1 e 12 e1 2 1 2 1 2 cos v 2 2 r 1 e 2 ln ln 2s r r 1 e 2 1 2s 1 2 2s ln 1 e 2 2s 2 e S ection 6.3 21. dP kP dt dP P ln P P 0 k dt kt Cekt P0, P0 P0ekt Ce0 C C1 Differential Equations: Separation of Variables 22. dT kT T ln T T 70 dt dT 70 70 70 0 k dt kt Ce kt 541 C1 Initial condition: P 0 Particular solution: P Initial condition: T 0 140 Particular solution: T 140; 70 70 70 70e Ce0 kt C 70 1 e kt ,T 23. dy dx 16y dy 8y 2 9x 16y 9x dx 92 x 2 C 1, 8 8y 2 16y 2 9x 2 y 2 9 2 C, C 25 2 25 2 24. dy dx 3 dy y ln y3 y3 2y 3x 2 dx x ln x2 Cx2 2, 23 x2, y C 82 , C 12 x 2 3 ln C Initial condition: y 1 Particular solution: Initial condition: y 8 Particular solution: 8y3 1 8 92 x 2 25 25. m dy y ln y y dy dx 0 x 1 dx 2 2 y x 26. m dy y dy dx dx x ln x Cx y x 0 0 y x 1 x 2 Ce x2 C1 ln y y C1 ln x ln C ln Cx 27. f x, y f t x, t y x3 t 3x 3 t 3 x3 4xy 2 y3 t3y3 y3 28. f x, y f t x, ty x3 t 3x3 3x2y2 2y2 2t 2y 2 4 txt 2 y 2 4xy 2 3t 4 x 2 y 2 Not homogeneous Homogeneous of degree 3 x2y2 x2 y2 t 4x 2 y2 tx t 2y 2 22 29. f x, y f tx, t y 30. t3 x 2y 2 x 2 y2 f x, y f t x, t y xy x2 y2 t2 tx ty x2 t 2 y 2 t xy x2 y2 Homogeneous of degree 3 t 2 xy t x2 y2 Homogeneous of degree 1 31. f x, y f t x, t y 2 ln x y 2 ln t xty 2 ln t 2xy 2 ln t2 ln x y 32. f x, y f t x, t y tan x tan t x y ty tan t x y Not homogeneous Not homogeneous 542 Chapter 6 x y tx ty Differential Equations y x ty tx tan y x 33. f x, y f t x, t y 2 ln 2 ln 34. 2 ln x y f x, y f t x, t y tan tan Homogeneous degree 0 x 2x x 1 2 dx x ln x Cx Cx Cx Cx x x x x 1 1 1 1 y 2 Homogeneous of degree 0 x3 y3 xy2 x3 vx, x 3 35. v x y dv dx y ,y vx 36. y x y 2 dy y vx 2x v v 1 2 v x vx x4 2 2 y3 dx dy vx 3 3 x dv dx 3 v dx dv x dx 2 ln 1 1 1 1 v 2 x dv 3 v dx x v dx x v 2 dv 3 dv 1 v v 2 v dv x dx dx 1 dx x ln x 3 ln x 3x3 ln x 3 v x dx ln C ln Cx v 2 dv v3 3 y x 3 C C Cx3 1 yx x2 x y 2 2 y3 x 37. v v dx x y dv dx y ,y y xv xv v dx v v v dx x ln x vx 38. v 2v dx x y dv dx x2 2xy x2 1 v y2 ,y vx 2x 2 v 2 x2 v dx x dv x dv 2x dv 2v dv 1 1 1 x2 v2 dx x ln x v dx 1 1 2v v v2 dx v2 ln v 2 1 v v2 1 ln v 2 2 v2 y2 x2 y2 2v 1 1 dv 1 1 1 x2 ln C ln C x 2v 2v 2 y x ln C1 ln C1 x v2 y2 x2 y2 C x C x Cx C x2 C x2 C 2xy S ection 6.3 xy x2 2 Differential Equations: Separation of Variables 2x x 2x x 2 ln x2 x2C x2C x2C Cx3 1 x 2v ⇒ ln C 3y 543 39. v v dx x y dv dx y2 ,y vx 40. v x x y dv dx dv dx v v y x y x y ,y 2 vx 3v dv 1 v ln x2C 2 dx x x2 v x x2 v 2 v 1 1 dx x ln x ln C1x v ln C1 y Ce 2xe 2xe x2 2y 2 3vx x dv x dv v2 v v dx v dx v3 1 v 2 2 dx ln 1 1 1 1 v3 1 2v 2 v2 dv ln v 1 2v 2 x2 2y 2 y ln C1 ln C1 x 41. x v dx ev dv ev ey ey x x x dy x dv 2 dx x ln C1 x 2 ln C1 C yx v y dx vx dx 0, y 0 vx 42. x2 v2 dx x2 y 2 dx x2v 1 v v ln v ln x v xx v dx y dy x dv dx x ln ln ev ey x 0, y 0 vx dv ln C1 v ln x 2 ln x 2 0, 1 x C1 x C1 xv Initial condition: y 1 Particular solution: ey C ln x 2 1 C1 vx C1 y y Initial condition: y 1 Particular solution: y 1, 1 e1 Ce yx Ce 1 ⇒C e yx 43. x sec v x sec xv dx y x y dx x v dx sec v x dy x dv v dx 0, y 0 v dx dx x ln x Cesin v C vx x dv cos v dv sin v x Initial condition: y 1 Particular solution: x 0, 1 e sin y x ln C1 Cesin yx Ce0 544 Chapter 6 Differential Equations 44. 2 x 2 Let y 2x 2 y2 dx v x, dy v 2x 2 dx 2x2 xy d y x dv 0 v dx. v dx x3v d v 2v 2 d x 2 dx x 2 ln x ln x x 2 2 x vx x dv 2x2v 2 dx 2 0 0 xv dv v 1 v2 1 ln 1 2 ln 1 C1 C1 C x2 dv v2 v2 v2 y2 x2 y2 12 12 12 C1 ln C 1 x2 1 x y1 0: 1 1 x 1 dy dx y C1 x2 x x2 0 ⇒C y2 y2 1 C2 x x y2 12 12 45. x 4 y 46. dy dx y dy x y x dx x2 2 C C1 −4 −2 −2 4 y x dx 12 x 2 C x 2 2 2 y2 2 y2 x2 x 2 4 4 −4 47. dy dx dy 4 ln 4 4 y y y y 4 dx x e 4 x y 8 y 48. dy dx dy 4 y dy y 4 y 0.25x 4 0.25x dx y 8 4 C1 C1 −4 −3 0.25x dx 1 4 x dx C1 Ce −4 −2 −2 x 2 4 x 1 2 3 4 Ce x ln y y 4 4 y 12 x 8 eC1 4 1 8 x2 1 8 x2 Ce 1 8 x2 S ection 6.3 49. (a) Euler’s Method gives y 1 (b) dy dx dy y ln y y y0 y (c) At x 6xy 6x dx 3x 2 Ce 3x2 Differential Equations: Separation of Variables 50. (a) Euler’s Method gives y 1 (b) dy dx dy y2 1 y 6xy2 6x dx 3x 2 1 3x 2 C 1 3 3 13 3 91 0.2622 1 9x 2 3 10 1 0.3. 1 ⇒C C 1 3x 2 1, y C1 0.2622. 545 0.1602. C1 5⇒C 5e 1, y 3x2 5 y 3 y (c) At x 5e 31 0.2489. 0.0887 Error: 0.2489 0.1602 Error: 0.3 51. (a) Euler’s Method gives y 2 (b) 3y2 y3 y1 2: dy dx 4 dy 4y 23 y3 (c) For x 2, y3 y3 y 3 y2 4y 3y 3 4y 15 5 0.0318 dy dt 22 0 0⇒y 3. 12 2 13 15 (c) At x x2 42 4y 2x 3y2 2x 12 4 12 dx 12x 1 x2 C 12 12x C⇒C 13 13 3.0318. 0.0378 1.7270. 52. (a) Euler’s Method gives y 1.5 (b) dy dx dy 1 y2 arctan y arctan 0 arctan y y 2x 1 2x dx x2 12 x2 C C⇒C 1 1 1 y2 1 tan x 2 1.5, y tan 1.5 2 3.0096. Error: 3.0318 dy dt y0 y0 2 k y When t 53. ky, y0 y C, Cekt initial amount 54. ky, y Cekt 20, y 1 Ce0 20ek ln 4 5 5 Initial conditions: y 0 20 16 16 C y0 e k 1599 1 1 ln 1599 2 Ce ln 1 2 1599 t k Particular solution: y 0.989C or 98.9%. 20et ln 4 25, y When 75% has been changed: 5 1 4 t 20et ln 4 et ln 4 5 5 ln 1 4 ln 4 5 6.2 hr 546 dy dx Chapter 6 Differential Equations dy dx 55. ky 4 0 along y 4; but 56. kx 4 0 along x 4: The direction field satisfies dy dx not along y 0. Matches (a). dy dx The direction field satisfies dy dx Matches (b). dy dx 57. ky y 4 0 along y 0 and 58. ky2 0, and The direction field satisfies dy dx y 4. Matches (c). dw dt dw 1200 w ln 1200 1200 w w w w0 w (a) 1400 0 along y The direction field satisfies dy dx grows more positive as y increases. Matches (d). 59. k 1200 k dt kt e C1 w kt C1 Ce Ce kt kt 1200 60 1200 1200 C⇒C kt 1400 1200 60 1140 1140e 1400 (b) k k k 0.8: t 0.9: t 1.0: t 1.31 years 1.16 years 1.05 years 0 0 10 0 0 10 0 0 10 (c) Maximum weight: 1200 pounds t→ lim w 1200 60. From Exercise 59: w w w0 w 1200 1200 w0 1200 Ce Ce 1200 1200 kt, t k 1 C⇒C w0 e t 1200 w0 61. Given family (circles): x2 2x y2 2yy y C 0 x y 62. Given family (hyperbolas): x2 2x 2y2 4yy y C 0 x 2y Orthogonal trajectory (lines): y dy y ln y y y x dx x ln x Kx ln K −6 4 Orthogonal trajectory: y 6 2y x 2 dx x 2 ln x kx 2 −3 2 dy y ln y y 3 −4 ln k k x2 −2 S ection 6.3 63. Given family (parabolas): x2 2x y Cy Cy 2x C 2x x2 y 2y x Differential Equations: Separation of Variables 64. Given family (parabolas): y2 2yy y 2Cx 2C C y y2 1 2x y y 2x 547 Orthogonal trajectory (ellipses): y 2 y dy y2 x2 2y 2 K x 2y x dx x2 2 K1 −6 4 Orthogonal trajectory (ellipse): y 6 2x y 2x dx x2 K K1 −6 4 y dy y2 2 2x 2 y2 6 −4 −4 65. Given family: y2 2yy y Cx3 3Cx2 3Cx2 2y 3x2 y2 2y x3 3y 2x 66. Given family (exponential functions): y y Cex Cex y Orthogonal trajectory (parabolas): y 4 Orthogonal trajectory (ellipses): y 3 y dy 3y 2 2 3y 2 2x2 K dN dt dN N 500 N 1 500 1 N ln N 1 500 N ln 500 dN N 2x 3y 2 x dx x2 K1 −6 1 y dx x 2x K1 K −6 4 y dy 6 6 y2 2 y2 −4 −4 67. kN 500 k dt k dt 500 kt e500kt C2 N C1 Ce500kt N 500 N N When t When t 0, N 4, N 500Ce500kt 1 Ce500kt 500C 100. Thus, 100 ⇒C 1C 200. Thus, 200 125e0.2452t 1 0.25e0.2452t 1 1 0.25. Thus, N ln 8 3 2000 125e500kt . 1 0.25e500kt 0.00049. 125e2000k ⇒k 0.25e2000k 500 . 4e 0.2452t Therefore, N 548 Chapter 6 Differential Equations 69. The general solution is y 1 kt C . 0, 1 . 45 1 1 45 1 45 . 45kt 0 0 3 68. The differential equation is given by the following. dS dt dS SL S 1 ln S L ln L S S L S S When t Therefore, S C 10 CL e Lkt 45 kS L k dt kt C1 S Since y 45 when t it follows that 45 1 and C C kt Ce Lkt CLe Lkt 1 Ce Lkt 10 . L 10 CL e Therefore, y Since y L kt 4 when t 1 1 2, we have 41 . 360 360 . 41t C 4 Thus, y 0, S 10. Thus, C 45 ⇒k 45k 2 45 41 8 t 8 10 L 10 L 10 L 10 e Lkt Lkt 10 L L 10 e 70. The general solution is y when t 0, we have C y Since y 12 kt 1 1 75 1 1 kt C . Since y 1 75. Therefore, 75 . 75kt 75 71. Since y 100 when t 0, it follows that 100 500e which implies that C ln 5. Therefore, we have kt y 500e ln 5 e . Since y 150 when t 2, it follows that 150 e 2k C, 12 when t 1, we have 7 . 100 80 500e ln 0.3 ln 0.2 ln 5 e 2k y 75 ⇒k 1 75k 300 Thus, we have y 75 1 5.25t 300 . 4 21t 0 k 1 ln 0.3 ln 2 ln 0.2 t 0.1452. Therefore, y is given by y 4 0 15 500e 1.6904e 0.1451t . 72. The general solution is y 5000e Ce . Since y 500 when t 0, it follows that 500 5000e C which 1 implies that C ln 10 ln 10. Therefore, we have ln 10 e k t. Since y y 5000e 625 when t 1, it follows that 625 e k kt 73. From Example 11, the general solution is y Since y 8 Since y 15 1 4 ln t 60e Ce kt . 8 when t 60e C 0, ln 3, 3k ⇒C 15 2 2.0149. 5000e ln 1 8 ln 1 10 ln ln 10 e k 5000 4000 3000 y 15 when t 60e e 2.0149e 3k 2.0149e k ln 1 8 ln 1 10 2000 1000 0.1019. Thus, we have y 5000e 2.3026 e 0.1019 t 1 4 k 2.0149e 1 ln 3 60e 10, y 3k 5 10 15 20 25 ln 1 4 2.0149 0.1246t 0.1246. . Thus, y When t 2.0149e . 34 beavers. S ection 6.3 74. From Example 11, the general solution is y Since y 30 Since y 400e Ce kt Differential Equations: Separation of Variables 75. Following Example 12, the differential equation is dy dt ky 1 y2 y 549 . 0, ln 40 3 2.5903. 30 when t 400e C ⇒C 1, 2.5903e k and its general solution is y2 1 y y y y2 Ce2kt 0⇒ 12 32 1 22 4⇒ C⇒C 15 0.2012. 3 90 when t 90 9 40 400e e k 1 when t 2 0.75 2.5903e 3 when t 4 34 54 1 42 1 ln 5 8 ln 9 40 k 2.5903e ln k 3e2k 4 ⇒ k 0.5519. Hence, the particular solution is y2 1 y y2 3e0.4024t. ln 9 40 2.5903 Thus, y 400e 2.5903e 0.5519t . Finally, when t 3, y 244 rabbits. Using a symbolic algebra utility or graphing utility, you find that when t 10, y2 1 and y y y2 3e0.4024 10 0.92, or 92%. dQ dt dQ Q Q 20 1 dt 20 1 t 20 e C1 Ce 1 20 t 76. Following Example 12, the differential equation is dy dt ky 1 y2 y 77. (a) and its general solution is y2 1 y y y y2 ln Q Ce 2 kt. Q 0.4 1.6 0⇒ 0.6 2 5⇒ ⇒ 0.8 1.2 0.2 2 27 2 e10k 1 27 ln 10 2 0.2603 ln 20 ln 3 5 3 5 t 16 9 24 C 16 2k 5 e 9 1 20 t C1 0.4 when t 0.8 when t Since Q 25 when t 0, we have 25 particular solution is Q 25e 1 20 t. (b) When Q 3 5 15, we have 15 e 1 20 t C; thus, the . 25e 1 20 t ⇒k 1 t 20 t 10.217 minutes Hence, the particular solution is y2 1 y y2 16 0.5205t e . 9 Using a symbolic algebra utility or graphing utility, we find that when t 8, y 0.91. 550 Chapter 6 Differential Equations 1 20 1 5 78. Since Q 20 Q 2 is a first-order linear differential equation with P x the integrating factor u t e 1 20 dt e 1 20 t, and the general solution is 5 0.05t Q e 0.05t dt e 0.05t 50e0.05t C 50 Ce 0.05t. e 2 and R x 5 2, we have Since Q 0 when t Q 38.843 lbs gal. dy dt dy y ln y y (b) y 0 y1 y kt e kt 0, we have C 50 and Q 50 1 e 0.05t . Finally, when t 30, we have 79. (a) ky k dt C1 C1 80. ds dh ds s k h k dh h k ln h C1 k ln Ch Ce kt 20 ln 16 20 ln 4 5 20 ⇒ C 16 Since s 25 when h 2 and s 12 when h 6, it follows that 25 k ln 2C and 12 k ln 6C , which implies C and k 25 ln 2C 13 ln 3 11.8331. 1 e 2 25 13 ln 3 20ek ⇒ k 5 0.0605 20et ln 4 When 75% has changed: 5 1 4 t 20et ln 4 et ln 4 5 5 Therefore, s is given by the following. s ln 1 4 ln 4 5 6.2 hours 13 h ln e ln 3 2 13 h ln ln 3 2 1 h 13 ln ln 3 2 25 13 ln h 2 , ln 3 1 dx dt x dt 1 dx x ln x Cx C1 ln Cx 25 13 ln 3 25 ln 3 13 25 ln 3 2 ≤ h ≤ 15 81. The general solution is y Ce kt. Since y 0.60C when t 1, we have 0.60C Ce k ⇒ k ln 0.60 0.5108. Thus, y Ce 0.5108t. When y 0.20C, we have 0.20C ln 0.20 t 1 kP 1 ln kP k kP N Ce 0.5108t 82. 1 dy dt y dt 1 dy y ln y y 0.5108t 3.15 hours. dy dx y y0 N P C2e kt 83. dP N t dt C1 84. 0.2y Ce 0.2 x 29.92 ⇒ C 29.92 ⇒ y 29.92e 0.2 x (a) 8364 feet N k y 1.5841 (b) 20,320 feet y 3.8485 1.5841 miles 21.80 inches 3.8485 miles 13.86 inches Ce kt S ection 6.3 1 rA 1 ln rA r rA P Differential Equations: Separation of Variables P rt e r 551 85. dA P P A t dt C1 86. (a) A 1 1 100,000 0.06 5 e 0.06 $583,098.01 P P. (b) A 250,000 0.05 10 e 0.05 $3,243,606.35 Ce rt 1 Ce rt r 1 Since A 0 when t Therefore, we have A P rt e r 1. 0, it follows that C 87. From Exercise 85, A Since A have P P rt e r 1. 8 and r 0.0725, we 88. From Exercise 85, A P rt e r 1, 120,000,000 when t Ar er t 1 120,000,000 0.0725 e 0.0725 8 1 $11,068,161.12. which implies that e rt 1 e rt rt t Since A t Ar P Ar P ln Ar P 1 1 1. 75,000, and r 1 0.08, you have 7.71 years. 1 Ar ln r P 800,000, P 800,000 0.08 1 ln 0.08 75,000 89. dy dt (a) 0.02y ln 5000 5000 y (b) As t → (d) 5000 , y→L 5000. 0 0 300 0 0 300 (c) Using a computer algebra system or separation of variables, the general solution is y 5000e Ce kt 5000e Ce 0.02t . 500, you obtain 2.3026. The graph is concave upward on 0, 41.7 and concave downward on 41.7, . Using the initial condition y 0 500 5000e C ⇒C 2.3026e ln 10 0.02t Thus, y 5000e . 552 dy dt (a) Chapter 6 Differential Equations 1000 y 90. 0.05y ln 1200 91. A differential equation can be solved by separation of variables if it can be written in the form Mx Ny dy dx 0. To solve a separable equation, rewrite as 0 0 100 M x dx N y dy (b) As t → , y→L 1000. and integrate both sides. (c) Using a computer algebra system or separation of variables, the general solution is y 1000e Ce kt 1000e Ce 0.05t . Using the initial condition y 0 100 Thus, y (d) 1200 100, you obtain ln 10 2.3026. 1000e 1000e C ⇒C 0.05t 2.3026e . 0 0 100 (e) The graph is concave upward on 0, 16.7 and . concave downward on 16.7, 92. M x, y dx N x, y dy 0, where M and N are homogeneous functions of the same degree. See Example 4. dv dt dv W ln W v v v Initial conditions: W C 20, v 20, k 0 when t ln 3 4 0, and v 5 when t 1. 93. Two families of curves are mutually orthogonal if each curve in the first family intersects each curve in the second family at right angles. 94. (a) kW k dt kt W v (b) s 20 1 20 t e 0.2877t dt C 3.4761e 0, C 0.2877t Since s 0 C1 Ce kt 69.5 and we have 0.2877t s 20t 69.5 e 1. Particular solution: v 20 1 eln 3 4t 20 1 3 4 t or v 20 1 e 0.2877t 95. False. y dy x is separable, but dx y 0 is not a solution. 96. True dy dx x 2y 1 97. False f t x, t y t 2x 2 t 2 f x, y t 2xy 2 S ection 6.4 98. True x2 y2 dy dx x C y K y 2Cy x C x y Kx Cy 2Kx 2Cy x x2 y2 x2 2 The Logistic Equation Product Rule 553 99. x2 y2 dy dx x2 y2 2x2 2y2 y y2 x2 y2 2 fg f fg g Need f x 1. 2 f f f g g gf gf f g ex 2 fg 0 0 2xe x f f x f 2 2Kx K y x 1 2 2x ex 2 0, so avoid 1 2x 1 2xex 2 2x 1 ex 1 12 1 C1 2x 2y2 1 2 ln g x gx 1 ln 2x 2 1 Ce x 2x Hence, there exists g and interval a, b , as long as 1 a, b . 2 Section 6.4 1. y 12 e The Logistic Equation 2. y 6, it matches (c) or (d). 12 3e 1 x 1 x Since y 0 Since y 0 Since (d) approaches its horizontal asymptote slower than (c), it matches (d). 12 12e 12 32 12 e 12 4 3, it matches (a). 3. y 1 x 4. y 8, it matches (b). 1 2x Since y 0 Since y 0 6, it matches (c) or (d). 12 faster for (c), it matches (c). 1; 4t Since y approaches L 1; 2t 5. y y 41 41 2 1 e 2t L 2 4, k 2e 2t 2, b 1 6. y y 51 51 4 1 3e 4t L 2 5, k 12e 4t 4, b 3 e 4 e 1 2t 3e 5 3e 1 51 y 5 4t e 1 1 e 4 2t 2t 3e 1 1 3e 4t 4t e 2t 3e 4t 2y 1 2y 1 2y 1 y0 4 1 1 4y 1 4y 1 4y 1 41 y 4 2 e 2t 5 3e 4t y0 5 1 3 5 4 554 7. Chapter 6 y y 12 1 12 1 1 y1 y1 y1 y0 12 1 6 12 6e 1 6e Differential Equations t t 1; L 2 12, k 6e t t 1, b 6 8. y y 14 1 14 1 3 1 5e 3t 3t 1 ;L 2 14, k 15e 3t 3t 3, b 5 6e t 5e 3t 6e 1 1 6e 12 t 6e t 14 5e 1 5e 1 1 5e 3t 5e 3t 3y 1 3y 1 3y 1 12 1 y 12 12 7 6e t 14 1 y 14 5 7 3 14 5e 3t y0 14 1 9. P t (a) k 1 1500 24 e 0.75t (b) L 60 1500 24 e 0.75t 1500 10. P t (a) k 1 5000 39e 0.2t 0.75 0.2 (b) L 5000 1 39 5000 125 5000 39e (c) P 0 (d) 1 1500 1 24 750 24 e e 0.75t (c) P 0 (d) 1 1 2 1 24 ln 2500 39e e 0.2 t 1 2 1 39 ln 0.2t 0.75t 0.2 t 0.75t t (e) dP dt 0.75P 1 1 24 ln 24 4.2374 P0 60 (e) dP dt 0.2t t 0.2P 1 1 39 ln 39 18.3178 P0 125 ln 24 0.75 P , 1500 ln 39 0.2 P , 5000 11. P t (a) k 1 0.8 6000 4999e 0.8t 12. P t 6000 6 5 1 2 1 4999 ln 1 4999 ln 4999 10.65 years (e) dP dt 6000 4999e 0.8t 1 1000 8e 0.2t (b) L 1000 18 500 1 2 1 8 ln 1 8 ln 8 10.40 years 1000 1000 9 1000 8e 0.2t (b) L 1 6000 4999 3000 (a) k (c) P 0 (d) 1 0.2 (c) P 0 (d) 1 4999e e 0.8t 8e e 0.2t 0.8t 0.2t 0.8t t (e) dP dt 0.8P 1 0.2t t 0.2P 1 ln 4999 0.8 P 6000 ln 8 0.2 P 1000 S ection 6.4 dP dt (a) k (b) L (c) 120 100 80 60 40 20 t 1 2 3 4 5 P The Logistic Equation 555 13. 3P 1 3 100 P 100 (d) d 2P dt 2 3P 1 3 3P 1 9P 1 9P 1 d 2P dt 2 0 for P 50 P 100 P 100 P 100 P 100 1 1 3P 1 P 100 P 100 P 100 2P 100 P 100 3P 3P 1 100 P 100 50, and by the first Derivative Test, this is a maximum. L 2 15. 100 2 dP dt 0.1P 0.0004 P 2 0.1P 1 0.1P 1 (a) k (c) 300 Note: P 14. dP dt (a) k (c) 300 250 200 150 100 50 0.5P 1 0.5 y P 250 (b) L 250 0.004 P P 250 250 0.1 y 1 10 (b) L x 4 8 12 16 20 (d) dP is a maximum for P dt (see Exercise 13). 250 2 125 − 20 x 100 (d) P 250 2 125 (same argument as in Exercise 13) 16. dP dt (a) k (b) L (d) 0.4P 0.4 1600 0.00025P2 0.4P 1 (c) 1600 y P 1600 17. dy dt k y y1 1, L 1 L be 8: 8 y , y0 40 40 kt 8 dP is a maximum for dt 1600 800 P 2 (see Exercise 13). 1200 800 400 x 4 8 12 16 20 1 40 1 1 40 be b t y0 ⇒b 4 Solution: y y5 40 4e 1 t 40 4e 1 5 38.95 40.0 y 100 40 4e 100 556 dy dt k y y0 Chapter 6 Differential Equations y , y0 8 8 dy dt k 1 8 1 1 1 1 b 8 be 1.2 t 18. 1.2y 1 1.2, L 1 L be 5: 5 kt 5 19. 4y 5 4 5 1 y2 150 0.8, L L be 4 y1 5 120 120 be y , 120 y0 8 y b 8 ⇒b 5 3 5 y0 kt 1 0.8 t ⇒1 8: 8 Solution: y y5 8 35e 120 ⇒b 1b 1 1 1 120 14 e 0.8t 14 1.2 t Solution: y 1.2 5 8 35e 7.99 y5 8.0 y 100 y 100 8 35e 120 14e 0.8 5 95.51 120.0 1.2 100 120 14e 0.8 100 20. dy dt k y y0 3y 20 3 ,L 20 1 L be y2 1600 240 3 y1 20 y ; y0 240 15 21. L 250 and y 0 350 Matches (c). kt 1 240 be 3 20 t 15 15: 15 240 ⇒b 1b 1 1 1 100 240 15e 3 20 t Solution: y y5 240 15e 3 20 5 29.68 240.0 y 100 240 15e 3 20 100 22. L 100 and y 0 23. L 250 and y 0 50 24. L 100 and y 0 50 Matches (d). dy dt (a) 1000 800 600 400 200 Matches (b). y 1000 (b) k 0.2, L y y0 1 1000 1000 be 0.2t 1000 1b 200 21 8.524 1000 179 21 e 0 0 1000 Matches (a). 25. 0.2y 1 y (0, 500) 105 1000 105 179 21 1 (0, 105) t 10 20 30 40 50 60 1 b b y 0.2t S ection 6.4 dy dt (a) 300 240 180 120 60 t 2 4 6 8 10 The Logistic Equation 557 26. 0.9y 1 y y 200 (b) k 0.9, L y y0 1 b b y 1 240 200 240 1 6 1 200 1 6e 0.9t 200 200 be 0.9t 300 200 1b 5 6 0 0 10 27. dy dt (a) 0.6y 1 y 1000 900 800 700 600 500 400 300 200 100 y 700 (b) k 0.6, L y y0 t 700 700 be 0.6t 1000 1 1000 700 1000 3 10 1 700 1b 7 10 0.3 0 0 10 1 2 3 4 5 6 7 8 9 10 1 b b y 700 0.3e 0.6t 28. dy dt (a) 0.4y 1 y 700 560 420 280 140 y 500 (b) k 0.4, L y y0 t 500 500 be 0.4t 700 1 375 500 375 1 3 1 500 1b 4 3 0 0 10 2 4 6 8 10 1 b b y 500 1 3e 0.4t 29. L represents the value that y approaches as t tends to infinity. L is the carrying capacity. 30. No, it is not possible to determine b. However, L 2500 and k 0.75. You need an initial condition to determine b. 558 Chapter 6 L be 25 39 1 7e e 2k Differential Equations 31. (a) y 1 kt , L 200, y 0 7 25 (b) For t (c) 1 7e 5, y 100 0.264t 70 panthers. 1 2 ln 1 7 200 7e 0.264 t 200 ⇒b 1b 1 200 39 23 39 1 23 ln 2 39 1 200 7e 0.2640t 200 7e k2 0.264t t (d) 1 39 ln 2 23 0.2640 dy dt ky 1 y L 7.37 years 0.264y 1 y , 200 y0 5. 100, 25 2k k y Using Euler’s Method, y 65.6 when t 200 2 (e) y is increasing most rapidly where y corresponds to t 7.37 years. 32. (a) y 1 L be 27 y2 kt , L 200, y 0 27 173 27 k2 (b) For t (c) 100 (d) dy dt 5, y 1 ky 1 78 panthers. 200 173 27 e y L 0.2812t 200 ⇒b 1b 43 200 43 4239 7439 ln 4239 7439 1 ⇒t 6.6 years 27 200 173 27 e 0.2812y 1 y , y0 200 1 173 e 27 e Using Euler’s Method, y 5 77.33. 200 2 2k (e) y is increasing most rapidly when y corresponding to 6.6 years. 100, 2k 2k k y 0.2812 1 200 173 27 e 0.2812t 33. (a) y 1 L be 1 2 kt , L 10, y 0 ⇒b 9 1 (b) For t (c) 8 1 5, y 4.58 grams. t ln 3 2 10 1 1 5 4 9 b 10 9e 10 ⇒ 72e 9e t ln 3 2 ⇒ 3 2 t 2 36 8.84 hours 2k 1 9e e 2k ⇒t (d) 4 1 ln 2 9 9 1 ln 2 4 3 2 t 0.4055 Exact 1.0 1.4286 2.0 dy dt ln 0 3 y1 2 1 y 10 2 2k 0.4055y 1 3 2.7273 4 y 10 5 4.5763 k y Note: y ln 3.6 1 1 10 9e 0.40547t 10 9e t ln 3 2 1 10 93 2 t Euler 1.0 1.3649 1.8428 2.4523 3.2028 4.0855 10 92 3 For t t 5, Euler’s Method gives y 4.09 grams. 1 (e) The weight is increasing most rapidly when y 10 2 5, corresponding to t 5.42 hours. S ection 6.5 L be First-Order Linear Differential Equations 1.25 0.25e 0.1792 5 559 34. (a) y 1 kt , L 1.25, y 0 1.25 ⇒b 1b 1 125 120 1 6 1.25 0.25e 25 24 k 10 1, y 10 0.25 1.2 (b) y 5 1 1.13 g 1.25 . 1 1.2 1 0.25e e 10k (c) The culture will never reach 8 grams L (d) dy dt 0.1792y 1 y , y0 1.25 1 1.1349 g. Using Euler’s Method, y 5 (e) y is increasing most rapidly when y L 2 1.25 2 0.625. 7.7. 10k 10k k y ln 1 6 The corresponding t-value is t 0.1792 1 1.25 0.25e 0.1792t 35. False. If y > L, then dy dt < 0 and the population decreases. 36. True. If 0 < y < L, then dy dt > 0 and the population increases. 1 be 1 1 1 37. dy dt d 2y dt 2 ky 1 ky 1 y , y0 <L L y L y L 2 38. y y y L 1 kt ky y L ky 1 1 be k be k be k be y kt 2 bke be kt kt kt k2y 1 ky 1 1 1 1 be be kt kt L y L y L 1 kt kt be 1 k2 1 k2 1 Hence, d 2y dt 2 y y L y y1 L 1 1 ky 1 kt 1 be kt 2y L 2y L 0⇒y L . 2 0 when 1 By the First Derivative Test, this is a maximum. Section 6.5 1. x 3y y xy 1 y x2 ex 1x e x3 First-Order Linear Differential Equations 1 1 2. ln x y y Linear 2xy 1 1 y ln x 2x y 2x y ln x y 0 0 Linear 560 Chapter 6 Differential Equations 1 y 1 y y 3xy y 3. y y cos x xy 2 xy2-term. 4. 3x 3xy 1 Not linear, because of the Linear dy dx 1 y x dy dx 2 y x 5. 3x 4 1 x dx 6. e ln x 2x 2 x C 3x 2 2 x dx Integrating factor: e xy y x2 x 3x 2x 4 dx C x Integrating factor: e x2y y x 2 3x 32 x 4 2xy 2 x 3 4x e ln x 34 x 4 2 x2 2x3 3 C x3 2 dx C x2 7. y y 10 1 dx x 8. y e x Integrating factor: e e xy e xy ye x Integrating factor: e ye x 2 2x dx ex 2 2 10e 4 xe x dx 2 Ce x2 2 2e x C 10e 10 dy y x dx 10e x C y y 9. y 1 cos x dx y y cos x y Ce x 10. 1 cos x y cos x cos x y 1 sin x dx y dy 0 sin x e cos x e cos x C sin x y sin x dx cos x cos x dx sin xy sin x Integrating factor: e e sin x Integrating factor: e ye sin x ye cos x sin x sin xe cos x dx 1 Ce cos x cos x e ye cos x e y cos x e e y 1 sin x sin x dx C sin x Ce 11. y x 1y 1 x 1 y y x2 x 1 1 1x 1 dx 12. y 3y e3x 3 dx Integrating factor: e ye3x x C1 1 y 1 3x e 6 Ce 3x e 3x 1 6x e 6 C Integrating factor: e yx 1 y x3 3x x2 3x e ln x x 1 e3xe3x dx e6x dx 1 dx C 1 13 x 3 S ection 6.5 3 First-Order Linear Differential Equations y cos x 1 dx 561 13. y 3x2y ex 14. y 3x2 dx Integrating factor: e ye x3 e x3 Integrating factor: e C ye x e 1 e 2 x x ex e x 3 x3 dx 3 dx x e x cos x dx cos x cos x sin x C y C ex y 1 sin x 2 Ce x 15. (a), (c) 5 y (b) dy dx dy dx y exy ye x ex y dx e x Integrating factor: e e 2x e 2x d x 1 2x e 2 C 1 2 ex −4 −3 6 x 4 exy ye x −6 −2 6 y0 ye x y 1⇒1 1 2x e 2 1x e 2 1 2 1 e 2 C⇒C 1 2 x 1x e 2 1 , Qx x e x 16. (a), (c) 4 y (b) y ux x 1 y x e y yx sin x 2, P x 1 x dx sin x 2 eln x 2 x yx x sin x −4 4 x sin x 2 dx 1 x 1 1 x 1 cos x 2 2 1 cos 2 1 cos x 2 2 x3y y 1 2 C 1 cos x 2 2 C −4 4 y 0 y −4 4 C ⇒C 1 2 −4 17. y cos 2 x y y 1 0 sec 2 x sec2 x dx 18. 2y 2 y x3 e1 x2 x2 2 x3 dx 1 x2 sec 2 x y Integrating factor: e ye tan x etan x e tan x 11 e x3 Integrating factor: e sec xe 1 Ce 2 tan x e C1 dx C ye 1 x2 y tan x 1 dx x3 e1 x2 1 2x 2 Initial condition: y 0 Particular solution: y 5, C 1 4e 4 tan x y Cx 2 1 2x 2 e, C e1 x 2 Initial condition: y 1 Particular solution: y 3 3x 2 1 2x 2 562 19. y Chapter 6 y tan x Differential Equations sec x cos x tan x dx 20. y e ln sec x sec x tan x x C y sec x sec x sec x dx Integrating factor: e y sec x y Integrating factor: e y sec x tan x e ln sec x tan x sec x tan x sec x sec x sin x x cos x cos x dx C cos x 1, 1 sin x C x sec x sec x tan x sec x dx tan x C C tan x 1 C 1 0 ,C 3 Initial condition: y 0 Particular solution: y y 1 cos x 1 sec x 4, 4 Initial condition: y 0 Particular solution: y 1 3 sec x tan x 1 1 3 cos x sin x 21. y 1 y x 0 1 x dx 22. y e ln x x 2x 1y 0 2x 1 dx Integrating factor: e Integrating factor: e ye x 2 ex 2 x x Separation of variables: dy dx 1 dy y ln y ln xy xy y x 1 dx x ln x ln C C 2, C 4 2 dx 2 y x 1 2 x 4 ln C C Ce x x2 y Separation of variables: 1 dy y ln y ln C1 y C1 y x ex Ce x 1 x2 x2 x2 2x dx Initial condition: y 2 Particular solution: xy 23. x dy dy dx dy dx ux y x 1 y x e 1 x x y y x 2 x Initial condition: y 1 Particular solution: y 2, 2 2e x x2 C 24. 2xy dy dx ux y x1 y 1 y 2x e 2 x3 x2 2 1 2x dx x 1 , 2 Linear 1 x1 2 x1 2 1 1 x dx Linear 1 x x C Cx C⇒C 12 x 12 1 x 2 dx x2 x2 2 x5 2 5 x 11 dx 2 x1 2 x1 2 Cx C x3 2 2 x 12 2 dx 21 dx xx 2 x x ln x 2 x ln x x1 2 x3 5 y4 y x3 5 2 x ln x 2 y1 y 10 2 64 5 x 17 5 4 x 2C ⇒ C 17 5 S ection 6.5 25. y n y 2e y y 3x 2 y 3, Q x2y3 x 2, P 3x 2 2 x 2e 2x 2e 1 e 3 1 3 Ce 2x 2x 3 2x 3 2 3x2 dx First-Order Linear Differential Equations xy xy 1 563 26. y n dx y2e x 2 1, Q x, P 2 x, e ex 2 2x dx ex 2 2 3x2 dx 2xe x dx 1 Ce x2 C 2e 2x 3 dx y2 2 e 2x 3 C 3 y 2 C e 2x 3 1 y2 1 y x 2, Q 1 x dx 1 3 1 y x 1 ,Q 2 1 x dx 27. y n e y 1 xy 2 x, P e ln x 28. y x 1 xy x, P e12 x ln x 1 n 1 x 1 e12 x C x x 1 xx x2 1 Cx x2 dx y1 2x1 2 1 12 x x dx 2 1 52 x 5 C1 2 1 y y Cx x5 2 5 C y x5 2 C 25x ex exy 1 29. y e y 2 3 dx e x 3 y, n e 2 3x 1 ,Q 3 e x, P 1 30. yy y n 2y 2 2y 1, Q 2 2 dx e x, P e 4x 2 y2 3e y2 3e 2 3x 2x ee 3 2e 1 2e x 3x 2 3 x dx 21 e 3 3x dx e 2 3x C Ce2x 3 y 2e 4x 2e 2x e 3 4xe x dx 2 e 3 3x C y2 3 y2 Ce 4x 31. (a) −4 5 (b) 4 dy dx 1 y x x2 1 x dx Integrating factor: e 1 y x 1 y x2 1 y x x x dx y x3 2 8 2 8 2 e ln x 1 x −5 (c) −4 5 x2 2 C 4 Cx 2C ⇒ C 4⇒y x3 2 x3 2 2x 4x 12 xx 2 12 xx 2 4 8 −5 2, 4 : 4 (2, 8): 8 2C ⇒ C 2⇒y 564 32. (a) Chapter 6 Differential Equations (b) y 4x3y x3 4x 3 dx 4 6 Integrating factor: e −4 −2 4 ex 4 y ex 4 4x3ye x ye x 4 x 3e x 4 x3e x dx 1 4 Ce x4 4 (c) 1 x4 e 4 C 6 y −4 −2 4 0, 0, 77 : 22 1 : 2 1 4 1 2 C⇒C 1 4 2 cot x dx 13 ⇒y 4 1 4 13 e 4 1 4 x4 C⇒C 3 ⇒y 4 3 e 4 x4 33. (a) −2 3 (b) y 6 cot x y Integrating factor: e y sin x −3 eln sin x sin x cos x y y sin x y 2 sin x 2 sin x dx 2 cot x 2 cos x C csc x 1 2 cot 1 csc 1 sin 1 2 cos 1 C (c) −2 3 6 (1, 1 : 1 −3 2 cot 1 2 cot x 1 y C csc 1 ⇒ C sin 1 y 3, 1: 2 cos 1 csc x 2 cot 3 1 csc 3 2 cos 3 sin 3 2 cot 3 2 cot x C csc 3 ⇒ C 2 cos 3 sin 3 csc x 34. (a) 6 (b) y 2xy xy 2 1, −4 −2 Bernoulli equation, n 2 letting z y 1 2 y 2 2 and 1 xe x dx 1 e x . The solution is: 2 4 you obtain e 2 x dx e x2 y 1e x2 1 e 2 1 2 1 1 1 x2 C 2 (c) 6 1 y 4 Ce x 1 2Ce x 2 2 −4 −2 y 0, 3 : 3 y 0, 1 : 1 y 2 2Ce x 2 2 2C ⇒1 2C 6 3 ex 2C 2 2 ⇒C 3 1 6 2 2 ex 3 2 2C 2 ex2 ⇒1 1 1 2⇒C 1 2 S ection 6.5 dQ dt First-Order Linear Differential Equations 565 35. (a) q kQ kQ, q constant q k, Q t kt (b) Q Let P t Q e q, then the integrating factor is u t e Q0 q k q k C⇒C Q0 q e k Q0 kt kt e kt. qe kt dt 0: Q Q0 Q q kt e k C q k Ce kt When t q k (c) lim Q t→ q k 36. (a) dN dt k 40 1, N 40 20, N 19 30 21 N 10: Ce k (b) N kN 40k (c) For t 10 For t Integrating factor: ekt ⇒ 30 Ce k Nekt N 40 40 ke kt dt Ce kt 40e kt C 19: Ce k 20k 40 e e 20k ⇒ 21 Ce 20k e 19k 1 10 ln 19 7 30ek 0.0188t ln 10 7 30 N 19k ⇒ k Ce 40 k 0.0188 30.5685 ⇒C 30.5685e 37. Let Q be the number of pounds of concentrate in the solution at any time t. Since the number of gallons of solution in the tank at any time t is v0 r1 r2 t and since the tank loses r2 gallons of solution per minute, it must lose concentrate at the rate v0 Q r1 r2 t r2. The solution gains concentrate at the rate r1q1. Therefore, the net rate of change is dQ dt q 1 r1 v0 Q r1 r2 t r2 r2 or dQ dt v0 r2 Q r1 r 2 t q 1 r 1. 38. From Exercise 37, and using r1 dQ dt rQ v0 q1r. r, 566 Chapter 6 Differential Equations r2Q r1 r2 t 39. (a) Q Q0 r1 v0 q0 10, r2 1 dQ Q ln Q Q q 1 r1 0, v0 1 Q 20 200, 0 (b) ln 15 3 5 t 25e 1 20 t 25, q1 10, Q 1 dt 20 1 t 20 Ce 1 t 20 20 ln 1 20 t 3 5 0 10.2 min (c) lim 25e t→ ln C1 1 20 t Initial condition: Q 0 Particular solution: Q 25, C 25e 25 1 20 t 40. (a) Q Q0 r1 v0 q0 10, r2 r2Q r1 r2 t 25, q1 10, Q q 1 r1 0.04, v0 1 Q 20 20 t (b) 15 200, 0.4 7 ln t→ 8 17e 7 17 17e 1 20 t 1 20 t 1 t⇒t 20 8 lbs 20 ln 7 17 17.75 min Integrating factor: e 1 Qe 1 Q Q0 Q 8 20 t (c) lim Q t 8e 1 20 t 0.4e 1 Ce 25 17e 1 20 t 20 t dt C 8 8 C⇒C 1 20 t 17 41. (a) The volume of the solution in the tank is given by v0 (b) Q Q0 v0 q0 r2Q r1 r2 t 0, q1 q1 r1 0.5, v0 3 100 r1 r2 t. Therefore, 100 5 3t 200 at t 50 minutes. 100, r1 50 t 3 2 dt 5, r2 t 32 3, Q 3 100 2t Q 2.5 Integrating factor: e Q 50 t 32 2t dt 2.5 50 50 0 Q Q 50 t 50 t 32 32 t 52 C Q Initial condition: Q 0 Particular solution: C 50 50 50 100 C 50 t ,C t 50 5 2 32 50 5 2 50 32 50 5 2 100 100 25 2 82.32 lbs S ection 6.5 42. (a) The volume of the solution is given by v0 (b) Q Q0 Q v0 q0 r2Q r1 r2 t 0, q1 5 t 5 50 2 50 C 50 52 32 First-Order Linear Differential Equations 5 3t 200 ⇒ t 50 minutes. 567 r1 r2 t 100 q1 r1 1, v0 100, r1 5, r2 3 3Q 100 2t Integrating factor is 50 Q 50 t 32 , as in #41. t 32 dt 2 50 t 32 t 52 C Q Q0 Q 0: 0 2 50 t 50, Q 100 t 32 C 50 ⇒C t 52 32 100 50 32 2 50 52 2 50 200 50 2 50 When t dv dt kv m v g m 100 32 164.64 lbs (double the answer to #41) 43. g mg 1 k e kt m 44. s t , solution 101, v t dt 159.47 1 159.47t e 0.2007t dt C 5794.57 5794.57 32, mg 8 g 8, v 5 794.57e 794.57 794.57e 0.2007t 1 implies that 4 101 8 1 k e 5k 1 4 . s0 st 5000 159.47t C⇒C 0.2007t Using a graphing utility, k v As t → below. 50 0 40 0.050165, and . 159.47 1 , v→ e 0.2007t The graph of s t is shown below. 6000 159.47 ft sec. The graph of v is shown 0 −500 40 −200 st 0 when t 36.33 sec. 45. L dI dt RI E0, I R I L R L dt E0 L e Rt L L 46. I 0 I C 0 I lim I 1 5 0, E0 E0 R 120 600 1 5 120 volts, R Ce Rt L 600 ohms, L 4 henrys Integrating factor: e I e Rt L E0 Rt L e dt L E0 R Ce Rt L E0 Rt e R C⇒C 1 e 5 150t 1 5 I t→ 1 amp 5 0.18 1 0.1 ln 0.1 ln 0.1 150 0.0154 sec e 1 1 5 150t 0.90 e 150t 0.9 e 150t 150t t 568 dy dx ux 48. y Chapter 6 Differential Equations 47. Pxy e P x dx Qx Standard form Integrating factor Standard form ny n Pxy y1 1 n Q x yn n ny Let z 0, 1 . Multiplying by 1 1 z n P x y1 1 n produces ny 1 1 nQx nQx. Linear 50. y 2y dy y ln y y Matches d. 2x Ce 2x nPxz 49. y 2x dy y 0 2x dx x2 C 0 2 dx C1 Matches c. 51. y 2xy dy y ln y y 0 2x dx 52. y 2y 2xy dy 1 1 1 y x x dx 12 x 2 C2ex 1 2 2 x2 2 Ce x C1 1 ln 2y 2 2y C1 Matches a. Ce x 2 Matches b. 53. e 2x y dx ex y dy 0 54. x 1 dx y2 2y dy 0 Separation of variables: e 2xe y dx e xe y e 1 e 2 C dy 0 dy dy C1 Separation of variables: x 1 dx 12 x 2 3x 2 6x 2y 3 x 6y 2 y2 13 y 3 C 2y dy y2 C1 e x dx ex 2e x 55. y cos x e 2y 2y 2y cos x dx 56. y 2x 1 y2 Separation of variables: cos x dx sin x ln y 1 y y 1 dy 1 1 ln C 1 ln C Separation of variables: 1 1 y2 dy x2 sin x 2x dx C 2 ln y sin x Ce sin x arcsin y y C S ection 6.5 57. 3y 2 4xy dx 2xy vx, dy 2vx 2 First-Order Linear Differential Equations y dx x dy 1 y x 1 1 x dx 569 x 2 dy v dx x 2 0 x dv v dx 2v v2 x dv 1 dv v v v xy 4 58. x 0 Homogeneous: y 3v x 22 Linear: y 0 0 ln C C C 60. y 2 xy dx 4vx dx 2 5 dx x ln x 5 Integrating factor: e y 1 x y 1 dx x x ln x ln x C e ln x 1 1 x ln v 2 x5 y v 2 C x3 2 59. 2y e x dx x dy 2 y x 0 1x e x 2 x dx x 2 dy 0 v dx x dv x dv 1 dx x ln x y x dv 0 0 1 dv v2 1 v C C x ln x Linear: y Homogeneous: y v2x2 e ln x 1 2 vx, dy x2 v dx v2 dx vx 2 dx Integrating factor: e yx 2 y 1 x 2 e x dx x ex x x2 1 x2 ex x C x2 C 61. x 2y 4 y 1 dx 1 y x x x 3 y 3 dy 3y 3 0 62. y dx 3x 4y dy 0 v dy 0 4 dy y ln y 4 C C ln C y dv Homogeneous: x y v dy y dv x 3, P e ln x 3 4 vy, dx 3vy v 4y dy 1 1 ln v y4 v dv 1 1 y 0 Bernoulli: n e y4x4 x 4y4 3, Q 4 x dx x 1, x4 2x 2 C 4x C x 4 dx 2x 2 yx x dy 3y 12x 3 x dx 3 63. 3 y 4x 2 dx x y dy dx 3 y x 64. x dx 12x 2 y ey x2 1 dy Separation of variables: x x2 e3 ln x x3 ln x 2 1 1 ln x 2 2 y2 1 dx 1 2e y C y 12 y 2 e y dy ey C1 Integrating factor: e y x3 33 xy x yx3 y 12x x3 12x 4 dx 12 2 x 5 C x3 12x 4 12 5 x 5 C 65. False. The equation contains y. 66. True. y x ex y 0 is linear. 570 Chapter 6 Differential Equations Section 6.6 1. dx dt dy dt x x, y y ax my bxy nxy Predator-Prey Differential Equations 0.7x 0.05xy 0.4y 0.007xy 0, 0 and 2. dx dt dy dt x x, y y ax my bxy nxy 0.5x 0.004xy 0.3y 0.01xy 0, 0 and 0 when x, y ma , nb 0 when x, y ma , nb 0.4 0.7 , 0.007 0.05 400 , 14 7 57.14, 14.0 . 0.3 0.5 , 0.01 0.004 30, 125 . 3. dx dt dy dt x x, y ax my y bxy nxy 0.3x 0.006xy 0.5y 0.009xy 0, 0 and 4. dx dt dy dt x x, y ax my y bxy nxy 0.6x 0.02xy 0.6y 0.01xy 0, 0 and 0 when x, y ma , nb 0 when x, y ma , nb 0.5 0.3 , 0.009 0.006 500 , 50 9 55.56, 50.0 . 0.6 0.6 , 0.01 0.02 60, 30 . 5. 40 30 y 6. 10 y 7. (a) The initial conditions are x0 40 and y 0 20. (b) 100 80 y (150, 30) 8 6 20 4 10 x 80 160 240 320 400 4 8 12 16 20 2 (15, 3) x 60 40 20 (40, 20) x 20 40 60 80 100 8. (a) The initial conditions are x0 60 and y 0 10. (b) 100 80 60 40 20 x 20 40 60 80 100 y 9. Critical points are x, y and x, y ma , nb 0, 0 10. 150 x(t) 0 36 0 0.3 0.8 , 0.006 0.04 (60, 10) y(t) 50, 20 . S ection 6.6 11. 12. 50 Predator-Prey Differential Equations 571 50 13. Critical points are x, y and x, y ma , nb 0.4 0.1 , 0.00004 0.00008 10,000, 1250 . 0, 0 0 0 150 0 0 150 (55, 10) 14. 25,000 15. x(t) 5,000 16. 5,000 0 0 y(t) 240 0 0 25,000 0 0 (4000, 1000) 25,000 17. Using x 0 solutions x 60 50 and y 0 20, you obtain the constant 50 and y 20. 18. Using x 0 10,000 and y 0 1250, you obtain the constant solutions x 10,000 and y 1250. 15,000 x x(t) y y(t) 0 0 36 0 0 240 The slope field is the same, but the solution curve reduces to a single point at 50, 20 . 50 The slope field is the same, but the solution curve reduces to a single point at 10,000, 1250 . 2,500 0 0 150 0 0 20,000 (50, 20) (10,000, 1,250) 19. dx dt dy dt x x, y x, y x, y x, y x y y 2x 2 2y 2 xy xy 20. dx dt dy dt x x, y 2x 5y y x2 4y 2 xy xy 0 when 0, 0 0, m n 0, 1 2 0 when 0, 0 0, m n 0, 5 4 x, y x, y ap cp 11 ,. 33 x, y a ,0 b an bn 1 ,0 2 mc bm , cp bn a ,0 b an bn 2, 0 mc bm , cp bn ap cp 33 , 33 1, 1 . 572 dx dt dy dt x Chapter 6 Differential Equations dx dt dy dt x x, y 0, 0.1 0.8 0, 1 8 x, y x, y 31 , . 17 17 x, y 21. 0.1x 0.1y y 0.4x 2 0.8y 2 0 when 0.5xy 0.3xy 22. 0.05x 0.06y y 0.2x 2 0.9y 2 0.4xy 0.2xy 0 when 0, 0 0, m n 0, 0.06 0.9 0, 1 15 x, y x, y x, y x, y 0, 0 0, m n a ,0 b an bn 0.1 ,0 0.4 mc bm , cp bn ap cp 1 ,0 4 0.03 0.01 , 0.17 0.17 a ,0 b an bn 0.05 ,0 0.2 mc bm , cp bn ap cp 1 ,0 4 0.021 0.002 , 0.1 0.1 23. a 0.8, b 0.4, c 0.1, m 0.3, n 0.6, p 0.1 24. 10 21 1 . , 100 50 Four critical points: 0, 0 0, m n 0, 0.3 0.6 0, 1 2 0 0 x(t) y(t) 36 Both species survive. a ,0 b an bn 25. a 0.8 ,0 0.4 mc bm , cp bn ap cp 2, 0 0.45 0.04 , 0.23 0.23 1, m 0.3, n 45 4 , 23 23 0.6, p 1 26. 10 0.8, b 0.4, c Four critical points: 0, 0 0, m n 0, 0.3 0.6 0, 1 2 0 0 x(t) y(t) 36 One species (the trout) becomes extinct. a ,0 b an bn 0.8 ,0 0.4 mc bm , cp bn ap cp 2, 0 0.18 , 0.76 0.68 0.76 9 17 , 38 19 28. Assuming the initial conditions are the critical points x 0 ,y 0 0, 1 2 27. Assuming the initial conditions are the critical points x 0 ,y 0 45 4 23 , 23 you obtain constant solutions. 3 you obtain constant solutions. 2 x x(t) y 0 0 36 0 0 y(t) 36 S ection 6.6 29. Solve the equations dx dt dy dt ax my bxy nxy 0 0 30. As in Exercise 29, using any of the four critical points as initial conditions will yield constant solutions. Predator-Prey Differential Equations 31. False 573 to obtain the critical points 0, 0 and ma ,. nb The solutions will be constant for these initial conditions. 32. False 33. False. The predator-prey equations are a special case of the competing species equations. 0, then ax 1 x , L y 0 0 34. False 35. (a) If y dx dt (c) 100 x which is a logistic equation. (b) dx dt dy dt 0.4x 1 0.3y x 100 0.005xy 0.01xy (d) 72 80 0, 0 is a critical point. If y 0, then x 100 and 100, 0 is a critical point. If x, y 0, then 0.4 1 0.3 Thus, x 0.4 1 x 100 0.005x 0.3 0.005 60 100 0.01y 0. 60 and 0.01y ⇒ y 16. 0 0 100 (e) 80 0 0 100 The third critical point is 60, 16 . dx dt dy dt 36. (b) ax my bxy nx 0.4x 0.01xy 0.005xy (d) 100 0.3y Critical numbers: 0, 0 , 60, 40 (c) 160 0 0 140 (e) x(t) 100 0 0 y(t) 72 0 0 140 (40, 80) 574 Chapter 6 Differential Equations Review Exercises for Chapter 6 1. y x2y x 3, y 3y 3x 2 x 2 3x 2 3 x3 3 x4 x3 6x 3 2. y y y y y 8y 2 sin 2 x 4 cos 2x 8 sin 2 x 16 cos 2 x 16 cos 2 x 8 2 sin 2 x 0 Not a solution Not a solution dy dx y dy dx y 3. 2x 2 2x2 5 5 dx 2x 3 3 5x C 4. x3 x3 2x 2x dx x4 4 x2 C 5. dy dx y cos 2x cos 2 x dx 1 sin 2 x 2 C 6. dy dx y 2 sin x 2 sin x dx 2 cos x C 7. dy dx y Let u 2x x 2x x x 7 7 dx dx, x 7 u1 2 du 28 3 2 u 3 7 52 8. dy dx y 3e x3 3e x3 dx 9e x3 C 7, du 2u u 7. y 4 52 u 5 4 x 5 4 x 15 dy dx x y dy dx 2x y 4 2 4 C 7 32 28 x 3 3x 14 C 7 32 C 9. 10. 2 0 Undef. 0 4 0 2 4 1 4 6 43 8 8 2 dy dx x y x sin y 4 4 2 4 2 0 0 0 4 0 2 4 0 4 6 4 8 8 0 dy dx R eview Exercises for Chapter 6 11. y x 2, y 2 4 575 1, 1 12. y 2x 2 y x, 0, 2 13. y 12 4x y 1 3 x, (0, 3) 4 0, 3 (− 1, 1) x 1 −4 4 x −4 x 4 −3 −4 −4 14. y y 3x, y 2, 1 15. y xy x2 y 4 4 , 0, 1 16. y y x2 y 1 1 , 0, 2 4 (0, 1) x 5 −4 x 4 −4 4 x −4 −4 −7 17. dy dx y 6 6 6x x x dx x2 2 C 18. dy dx dy y ln y y 6 6 6 y y dx x ex 6 6 19. 3 y 3 Ce x Ce x 3 dy dx 2 dy 1 3 dx x x 3 y 2 C1 C1 y C 1 C 1 x C y y 20. y dy dx 1 2 dy 4 y 4 dx 21. 2 xy 2 xy dy x dx 1 dy y 1 dy y 0 xy x 2 1 x x dx 2 2 2 ln 2 x 2 2y1 2 y1 2 y 4x 2x 2x C1 C C 2 x dx x C1 Ce x 2x 2 ln y y Ce x 2 576 22. x y Chapter 6 x 1y x dy dx dy y ln y y Differential Equations 0 x x x x ln x 1y 1 dx C1 23. y 0, Ce kt 33 : 44 5 20 3 k y C 3 k5 e 4 e 5k 1 20 ln 5 3 5t 5, 5 : Cxe x 3 ln 20 3 e 4 Ce kt 5 3 0.379t e 4 24. y 2, Ce kt 33 : 22 Ce 2k ⇒ C Ce 4 k 3 e 2 2k 25. y 3 e 2 e 4k 2k 0, 5 : C 3 2k e 2 5, 11 : 66 k 9 . 20 y 5e 5e 5k 1 1 ln 5 30 5e 0.680 t 4, 5 : 5 10 3 Hence, C Finally, y e 2k ⇒ k 3 e 2 1 10 ln 2 3 33 2 10 ln 30 5 t ln 30 5 2 1 2 ln 10 3 9 1 2 ln 10 3 t e . 20 dP dh 26. y Ce kt Ce k Ce 6k 27. ⇒C ⇒2 9e 9e k k p, P 0 Ph 30 1, 9 : 9 6, 2 : 2 k Hence, C Finally, y 28. y 2.5 k Cekt 30e kh 30e18,000k ln 1 2 18,000 30e 30e 15 ln 2 18,000 k e6 k 9e 5k P 18,000 k 2 1 ln 5 9 9e 1 5 ln 2 9 0.3008 9 2 9 15 12.15864. Ph P 35,000 h ln 2 18,000 35,000 ln 2 18,000 12.1586e 5ekt 0.3008t. 7.79 inches 5e k 1599 1 1599 ln 1 2 0.000433 5e 0.000433 600 When t 29. S (a) Ce k t 600, y 3.86 g. S 5 t→ 5 when t Ce k C 30e k ln 1 6 30eln 1 6 30 1 (b) When t (c) 30 5, S 20.9646 which is 20,965 units. lim Ce kt 5 k S 1.7918 t 0 0 40 30 1 1t 6 R eview Exercises for Chapter 6 30. S 25 1 e kt ek 1 ⇒1 ⇒e ⇒k (b) 25,000 units t→ k 577 (a) 4 25 1 ek 21 25 4 25 (c) When t (d) 0.1744 25 5, S 14.545 which is 14,545 units. ln 25 21 25 lim S 0 0 8 31. P 2C 2 ln 2 t Ce0.015t Ce0.015t e0.015t 0.015t ln 2 0.015 46.21 years 32. (a) 1 0.012 dy ds dy y s 0.012y, s > 50 ds C1 0.012s 1 ln y 0.012 y When s y (b) Ce 50, y 28 Ce 0.012 50 ⇒C 28e0.6 28e0.6 0.012s, s > 50. 50 55 60 65 70 Speed(s) Miles per gallon ( y) 28 26.4 24.8 23.4 22.0 33. dy dx dy y x2 x x x2 2 2xy dy dx 1 dy y ln y 3 3 dx x 3 ln x C 34. dy dx dy y e 1 1 2x e e 2x 2x e 2x dx e 2x 1 2e 21 e C 2x 2x dx 1 ln 1 2 ey sin x dy dx e y 35. y 0 2xy 2x dx x2 y Ce x2 36. y 0 ey sin x sin x dx cos x 1 cos x ln 1 cos x C C C1 C C1 ln cos x C dy y C1 e 2 ex C1 ey y y 37. x2 Let y y2 dx vx, dy dy dx 2xy dy x dv x2 2xy 0 v dx. y2 , (homogeneous differential equation) —CONTINUED— 578 Chapter 6 Differential Equations 37. —CONTINUED— x2 x2 v2x2 dx v2x2 2x vx x dv 2x2v2 dx x2 1 v dx 2x3v dv 0 0 2x3v dv 2xv dv 2v 1 ln 1 C 1 x2 v2 Cx y2 1 or v2 dv v2 C1 C yx C1 2 x2v2 dx v2 dx dx x ln x x 1 ln 1 v2 ln C Cx2 x2 y2 x y2 x2 38. 3x Let y 3x y dx vx, dy vx dx 3x dy dx x dy x dv 3x x 0 y , (homogeneous differential equation) v dx. v dx x2 dv 0 0 x dv 1 3 1 ln 3 2 C2 3 C3 C 3x x3 2v dv 2v 2v 2v 2y 12 x x dv 2vx dx 3 2v dx 1 dx x ln x x x2 x3 y C1 ln 3 2v 12 ln C2 C3 3Cx 2 y x 2Cy 3Cx 2C 39. y y y x2 y 3xy 3y C1x C1 6C2 x C2 x3 3C2 x2 x2 6C2 x 6C2 x3 3x C1 3C1x 3C2 x2 9C2 x3 4C2 8C2 ⇒ C2 3 C1x 3C1x C2 x3 0 3C2 x3 x x 2, y 2, y 0: 0 4: 4 4 2C1 C1 8C2 ⇒ C1 12C2 4C2 12C2 1 2, C1 2 y 2x 13 2x R eview Exercises for Chapter 6 dv dt (a) 579 40. kv 9.8 dv kv 9.8 9.8 9.8 9.8 v t kt ekt dt C1 C2 C2 (b) lim v t t→ 9.8 k 1 9.8 k 1 9.8t k 9.8t k kv0 1 kv k0 1 kv k2 0 9.8 1 kv k2 0 1 kv k2 0 9.8 ekt dt 9.8 ekt 9.8 ekt C⇒C 9.8 ekt 9.8 ekt s0 1 C s0 1 kv k2 0 1 kv k2 0 s0 9.8 9.8 C 1 ln kv k ln kv kv (c) s t C3ekt C3ekt C3 ⇒ C3 kv0 kv0 9.8 s0 st 1 9.8 k 1 9.8 k 1 9.8 k At t 0, v0 v 1 kv k2 0 9.8t k 9.8t k dy dx dy 2y 3 3 3 3 2y y 9.8 ekt . Note that k < 0 since the object is moving downward. 41. dy dx y dy y2 2 4x2 y2 C 4x y 4 y 42. 3 2y dx y 4x dx 2x2 C1 −4 x 4 x 1 ln 2y 2 ln 2y 2y 4 x 2x C2 e 3 3 2 C1 2C1 2x ellipses −4 C2 e Ce 2x 2x 43. P t (a) k (b) L (c) P 0 (d) 1 1 7200 4 4e 0.55 t 44. P t (a) k (b) L 160 7200 44 e 0.55t (c) P 0 (d) 14 e 1 4800 14e 0.15t 0.55 7200 7200 1 44 3600 4 4e e 0.55t 0.15 4800 4800 1 14 2400 0.15t 320 4800 14e 0.15t 1 2 1 44 1 1 0.55t t 6.88 yrs. (e) dP dt 1 1 ln 0.15 14 P 4800 17.59 yrs. t (e) dP dt 0.55P 1 1 1 ln 0.55 44 P 7200 0.15P 1 580 Chapter 6 Differential Equations 1200, y 1 2000 (b) y 8 (c) 10,000 20,400 ⇒b 1b 1 20,400 16e k 45. (a) L y y0 y1 20,400, y 0 1 20,400 be kt 1200 2000 16e k 17,118 trout 1 20,400 ⇒t 16e 0.553t 4.94 yrs. 16 46 5 ln 23 40 ln 40 23 0.553 k y 1 20,400 16e 0.553t y , y0 20,400 1. L 4 7414 5853 6 12,915 10,869 8 17,117 16,170 ln L L dS dt dS S S S S kt e L 46. dy dt 0.553y 1 1200 47. kL k dt S Use Euler’s Method with h t Exact Euler 0 1200 1200 2 3241 2743 C1 kt C1 Ce kt Euler’s Method gives y 8 16,170 trout. Since S 0 when t 0, we have 0 C L. Thus, S L 1 e kt . L C⇒ 48. The general solution is S (a) Since L 100 and S following. 25 1 4 L1 e kt . 2, we have the (b) Since L 500 and S following. 50 1 10 25 when t 2k 50 when t k 1, we have the 100 1 1 3 4 e 2k 500 1 1 e k e e e 2k e k 9 10 9 ln 10 9 ln 10 2k k ln 3 4 1 2 k ln 3 4 0.1438 100 1 e 0.1438t k . ln 190 0.1054 500 1 e 0.1054t Thus, the particular solution is S Thus, the particular solution is S . 49. The differential equation is given by the following. dP dn 1 PL 1 ln P L ln L P dP P P L P P kn Ce Lkn CLe Lkn Ce Lkn CL e L kn kP L k dn P C1 1 C R eview Exercises for Chapter 6 CL e 581 50. The general solution is P C Lkn . (b) Since L when n 0.25 0.80 and P 0.25 when n 10, we have the following. 0.80C ⇒C C1 5 11 1 5 ln 8 33 0.2359 0, and P 0.60 (a) Since L 1 and P 0.50 when n 0, and P 0.85 when n 4, we have the following. 0.50 0.85 C C 1 1 1 e 1 ⇒C ⇒k 1 . e 0.4337n 1 1 3 ln 4 17 0.4337 4k 0.60 Therefore, P P 0.8 5 11 0.80 ⇒k 5 11 e 8k 5 4 11e 0.1887n Therefore, P P . 1 0.6 0.4 t 1 −1 2 3 4 0.2 t 5 10 15 20 51. (a) 4 3 2 1 − 4 −3 −2 −1 y (b) y x 1 2 3 4 y y ye x ex 2 y dx (c) ex −9 6 ex 2, Integrating factor: e e x 2e x d x 23 e 3 2x 9 e3 C 2x dx −6 y y0 y 2x e 3 1 2x e 3 2y 2e 2x y ye 2x 2 Ce 2 3 x C ⇒C 5 e 3 x 5 3 2 2 1 2e x 3 5e x 52. (a) 4 2 y (b) e 2x y y sin x, Integrating factor: e e 2x sin x 2 dx e2x (c) −6 4 6 e 2x sin x dx −4 −4 −2 −2 −4 x 2 4 ye 2x y0 1 2x e 2 sin x 5 4⇒4 ⇒4 1 0 5 1 5 cos x 1 C C 21 5 C⇒C cos x 21 5 21 e 5 2x ye 2x y 1 2x e 2 sin x 5 1 2 sin x 5 cos x 582 Chapter 6 Differential Equations dy dx dy dx cot x y 53. (a) 3 y (b) csc x csc x y cot x −3 x 3 Integrating factor: e csc x y cot x dx e ln sin x csc x −3 csc x cot x y csc x y csc2 x csc2 x csc2 x dx cos x 1⇒1 ⇒C 1 cot x C sin x cos 1 cos 1 sin 1 C sin 1 1.83 C (c) − 4.5 3 4.5 y csc x y y1 −3 54. (a) 4 y (1, 2) (b) dy dx x 2 dy dx cot x y csc x csc x y cot x 2 −2 −2 Integrating factor: e sin x y cos x y y sin x cot x dx e ln sin x sin x 1 1 x C 1 C 1 0.683 (c) −4 4 y sin x 4 y1 2 ⇒ 2 sin 1 ⇒C 2 sin 1 C csc x −4 y 56. e xy 8 x x csc x 55. y Px ux y y 8 1, Q x e dx 4e xy y 4y 4, Q x e 1 e4x e 4x 4 dx 1 e x 57. y e x 4y 1 y 4 ex y y e x Px ux y 1 x4 e 4 1 , Qx 4 1 x4 e 4 e 14x 1 8e ex ex 8 8e dx C e4x Px ux y e e 1 x e x e 4 x dx 1 3x e 3 Ce C 4x 1 4 dx Ce x 14x 1 x4 ee 4 1 x 4 C Ce x 4 14x dx 1 e 3 x e14x 1 x4 xe 4 R eview Exercises for Chapter 6 dy dx Px ux y 1 e5 x 1 e5 x 1 5 60. x dy dx Px ux y x 1 x x 2 62. dy dx dy tan x y 3 3 2 2 583 58. 5y x2 1 x2 5 , Qx x2 1 x2 e5 x 59. x dy dx Px ux C y 2y 1 x x e 1 x 1 x 2 2 2 1 y y 1 1 x 2 1 x e ln x 2 C x 2 e 5 x2 dx 2 1x , Qx 2 dx 2 x 2 1 5x e dx x2 1 5x e 5 Ce 5x 1 x x 2 dx 3y 2 x 2 x e 1 3 2y 3 3 2x 2x 2x 3 3 2x 2 61. 3y y sin 2x dx 3y sin 2x dy 0 y Integrating factor: e , Qx 3 dx 3 dx e 3x 3 3 ye x 3 2 3x e 1 e 13 3x sin 2 x dx 3 sin 2 x 2 cos 2 x Ce 3x C e 2 ln x 3x 3 2 C 4 3x 2 2x x 3 2 dx y C 1 3 sin 2x 13 2 cos 2x x 3 2 y tan x 2e x tan x dx 2e x dx 63. y 5y e 5x 5 dx Integrating factor: e e ln cos x e x cos x C sec x cos x sin x C ye 5x y e10x dx 1 5x e 10 Ce e 5x C Integrating factor: e y cos x y a y x 1 10 x e 10 5x 2e x cos x dx ex 1 tan x 64. y bx 3 a x dx 65. y e a x4 a ln x y x y2, y Bernoulli equation 1 2 2 Integrating factor: e yx a x C a n y 2, let z 2 y 1, z y z x x, y 2 y. bx 3 x bx 4 4 a a dx Cx a b 4 y y z a Linear equation y ux z y 1 e 1 ex x x dx e xe x x dx e x xe x e x C 1 1 1 Ce x Ce x y 584 Chapter 6 Differential Equations 1 y x 3, let z 2y Linear equation 3 66. y n y 2xy 2, let z 2 x y2, y1 2 xy z Bernoulli equation 2 67. y 2y y3 , x2 Bernoulli equation y1 3 y 1, z 2 y . n y 2, z 3 2y 3 y. y y x x, y 2 xz e x2 x2 1 y x z 2y 2 x2 2 , x2 x x2 2 3 ux z 1 y y e 1 2 e 1 x2 2x dx xe Ce x 1 2 dx ex 2 1 e 2 2 z x e 2 2 ln x Linear equation x2 C ux z x e 1 2 2 x dx 2 x 2 x dx 2x 3 C 12 y 1 y x 2, let z y 2y Ce x2 xy 2 y 2, 1 2 C1e x2 1 y2 2 3x Cx 2 68. xy y n 69. Answers will vary. Bernoulli equation y1 y 2 Sample answer: x 2 Solution: Let y y 2y 3y 2 dx x dv 2xy dy v dx. v dx 2x 3v dv 0 0 0 vx, dy y 1, z 2 . x2 3v 2x 2 dx x2 2 x vx x dv v 2 x 2 dx 1 1 y x z y2 1, y 2 1 z x 1 x x ln x Cx Linear equation v 2 dx dx x ln x 2 xv dv 2v 1 ln 1 C1 C1 v2 v2 v2 y2 x2 y2 dv C1 ux z 1 y y x e 1 x dx 1 dx x x ln x C x Cx 1 x ln x 71. Answers will vary. Sample answer: x 3y y ux y e 1 x2 2 x dx x3 C x2 70. Answers will vary. 2x 2 y 2 y x 1 1 x3 x2 1 ln x x2 C 12 x dx x3 72. Answers will vary. 73. dA dt rA P r For this linear differential equation, we have P t and Q t P. Therefore, the integrating factor is ux e r dt e rt and the solution is A ert Pe rt dt ert P e r rt C A0 P r Cert. Pr Since A A0 when t which implies that A P r A0 0, we have C P rt e. r R eview Exercises for Chapter 6 74. A 0 (a) P A 500,000, r 40,000 40,000 0.10 500,000 40,000 0.10 t e 0.10 100,000 4 e 0.10t 0 0 30 585 0.10 2,500,000 The balance continues to increase. (b) P A 50,000 50,000 0.10 500,000 50,000 0.10 t e 0.10 500,000 2,500,000 The balance remains at $500,000. (c) P A 60,000 60,000 0.10 500,000 60,000 0.10 t e 0.10 100,000 6 ln 6 0.10 e 0.10 t 17.9 years. 0 0 30 2,500,000 The balance decreases and is depleted in t 0 0 30 75. A 0 e0.14t t 200,000 0.14 200,000 10 3 ln 10 3 0.14 50 7 1,000,000 5 200,000 0.14t e 0.14 76. (a) dx dt dy dt ax my y bxy nxy 0.3x 0.02xy 0.4y 0.01xy 0, 0 and 40, 15 . 50 0.14t e 7 (b) x x, y 8.6 years (c) 80 0 when x, y ma , nb 0.4 0.3 , 0.01 0.02 x(t) y(t) 0 0 36 77. (a) dx dt dy dt ax my y bxy nxy 0.4x 0.04xy 0.6y 0.02xy 0, 0 and 30, 10 . 78. (a) dx dt dy dt ax my y bx 2 ny 2 cxy pxy 3x 2y x2 y2 xy 0.5xy (b) x x, y (c) 50 0 when x, y ma , nb (b) x x, y x, y 0 when x, y 0, m n 0, 2 , 3, 0 , mc bm , cp bn 0, 0 , 0.6 0.4 , 0.02 0.04 a ,0 b an bn x(t) y(t) 0 0 24 x, y (c) 4 ap cp 1 12 , 1212 2, 1 . x(t) y(t) 0 0 6 586 Chapter 6 dx dt dy dt Differential Equations 79. (a) ax my y bx 2 ny 2 cxy pxy 15x 17y 2x 2 2y 2 4xy 4xy (b) x x, y x, y x, y (c) 15 0 when x, y 0, m n 0, 17 , 2 0, 0 , a ,0 b an bn 15 ,0 , 2 mc bm , cp bn ap cp 38 , 12 26 12 19 13 , . 66 y(t) x(t) 0 0 4 One species, x, becomes extinct. Problem Solving for Chapter 6 1. (a) y 1.01 dy dt dy y1.01 dt t C1 0.01t 1 0.01t 1 0.01t 100 (b) y 1 dy y y kt k dt C1 kt C 1 C 1 C1 1 y0 . kt 1 y 0.01 0.01 1 y0.01 y0.01 y y0 1: 1 C y0 y0 y ⇒ C1 1 1 C C 1 ⇒C y0 . 1 y0 Hence, y 1 For t → kt 1 ⇒C C100 1 t→T Hence, y For T 1 0.01t 100 . 1 , y→ y0 k 100, lim y . 2. Since dy dt 1 y ln y ky dy 20 , k dt kt Cekt C 20. 52. 52ek 22.35 . 20, ek 28 52 7 13 , 20 20 y When t When t When t 0, y 1, y 5, y 72. Therefore, C 48. Therefore, 48 52e 5 ln 7 13 and k 7 ln 13. Thus, y 52e ln 7 13 t 20. 20 P roblem Solving for Chapter 6 dS dt S dS dt 1 587 3. (a) k1S L L Ce L1 LC ke 1 Ce k L1 k L1 k1S L S is a solution because Ce kt kt 2 kt 2 (b) dS dt d 2S dt 2 k1S 100 k1 S k1 100 0 when S dS dt S 100 dS dt 50 or dS dt 0. S dS dt kt Cke kt 2S L Ce L Ce kt 1 C Le kt Ce kt L 1 k . L 9. And, (d) L Ce kt Choosing S 50 2 ln 1 9 ln 4 9 t S 140 50, we have: 1 1 t 2.7 months (This is the inflection point.) 100 9eln 4 9eln 4 9t 9t kt S , where k1 L S 100. Also, S 10 when t 0 ⇒ C 20 when t 1 ⇒ k ln 4. 9 Particular solution: S (c) 125 1 100 9eln 4 9t 1 100 9e 0.8109t 120 100 80 60 40 20 t 1 2 3 4 0 0 10 (e) Sales will decrease toward the line S L. 4. (a) y x y, y 0 1, h 0.1 (b) 1.0 Using the modified Euler Method, you obtain: x 0 0.1 0.2 1.0 y 1.0 0.91 0.83805 0.73708 [ [.1 [ [.2 [ [.3 [ [.4 [ [.5 [ [.6 [ [.7 [ [.8 [ [.9 [ [1.0 ] ] ] .82 ] ] .758 ] ] .7122 ] ] .68098 ] ] .662882 ] ] .6565938 ] ] .66093442 ] ] .674840978 ] ] .6973568802] .9 [x [ [0 [ [.1 [ [.2 [ [.3 [ [.4 [ [.5 [ [.6 [ [.7 [ [.8 [ [.9 [ [1.0 ] ] 1 ] ] .9100000000] ] .8380500000] ] .7824352500] ] .7416039013] ] .7141515307] ] .6988071353] ] .6944204575] ] .6999505140] ] .7144552152] ] .7370819698] y 0 0.5 1.0 The modified Euler Method is more accurate. 588 Chapter 6 1 kt 2 1 ln b k Differential Equations ln b . k eu eu e e u u kt 5. Let u 1 e 2u tanh u e kt 1 2 e be 2u kt e ln be Finally, 1 L1 2 tanh 1 kt 2 ln b k L 1 2 L 21 1 tanh u 2 be L be kt kt . The graph of the logistics function is just a shift of the graph of the hyperbolic tangent. 1 12 32 x2 y 6 2 2 6. k g 36, 36 12h Equation of tank y 6 2 x2 Ah Ah 12h 12h 18h 3 2 36 5 2 h 5 h3 2 36h 5 When h 6, t h2 h2 dh dt dh dt dh dt 12y y2 h2 , Area of cross section y k 2gh 1 144 1 12 h 18 x + (y − 6) = 36 2 2 6 ft 64h h x 216h1 2 dh 144 h 3 2 720 0 and C t t dt C C 63 2 5 504 1481.45. 0⇒t 1481.45 seconds 24 minutes and 41 seconds. The tank is completely drained when h P roblem Solving for Chapter 6 dh dt dh dt dh 589 7. (a) A h r2 h 12 k 2gh k 64 h (b) t 3600 sec ⇒ 2 h ⇒h r 0.000865 3600 62 7.21 feet 8k dt r2 Ct C1 C1 at t Ct C dt, C 8k r2 18 ft h 2h 2 18 0, h 6 2. 18 Hence, 2 h At t 30 60 2 12 1800, h 1800 C C 12: 62 0.000865 6 2. 62 43 1800 Hence, 2 h h 0⇒t 0.000865 t 62 0.000865 9809.1 seconds, (2 hr, 43 min, 29 sec) dh dt dh dt 8. Ah 64 h k 2gh 8h 36 1 2 dh 1 dt 288 t 288 C t 288 4 5 288 C 45 45 2575.95 sec x x2 joins 1, 1 and x0, 0 . 2h h 20: 2 20 2h h 0⇒t 9. Let the radio receiver be located at x0, 0 . The tangent line to y y 2 Transmitter (−1, 1) 1 y = x − x2 −1 Radio x1 x0 x — CONTINUED — 590 Chapter 6 Differential Equations 9. —CONTINUED— (a) If x, y is the point of tangency on y 1 x 2x2 x2 1 2x 2x 2x 2 x y Then 1 1 0 x0 3 1 1 6 x0 (c) 10 x 1 1 x2, then (b) Now let the transmitter be located at 1 x x2 2x2 2x 1 h 2x 2x 1 x y x x 0 2± 1 y x x2 h h 1 4 32 2h 2 2h x0 2h 4 h 32 h h 2 h 4 2 2 h h 1 x2 x x h x2 1, h . h 1 y x x 0 1 1 x2 x x 1 x2 2± 2 x x2 4 33 5 3 33 x0 6 8 5. 6 1 3 4h 1 33 11 x0 6 33 33 3 Then, h 1 0 x0 32 h 1 2h x0 h 1 32 4 4 h 33 43 6 6 33 1.155. h h 1. 0.25 −2 3 h2 h 4 32 There is a vertical asymptote at h height of the mountain. ds dt (a) 1 4, which is the 10. 3.5 0.019s ds 0.019s 0.019s 0.019s 0.019s 0.019s s t dt C1 C2 11. (a) dC C ln C C R dt V R t V Ke Rt V 3.5 1 ln 3.5 0.019 ln 3.5 3.5 K1 0.019t C3 e 3.5 0.019 t Since C C0 when t 0, it follows that K the function is C C0 e Rt V. (b) Finally, as t → , we have t→ C0 and C3 e 0.019t 0.019t t→ lim C lim C0 e Rt V 0. 184.21 Ce (b) 400 0 0 200 (c) As t → , Ce 0.019t → 0, and s → 184.21. P roblem Solving for Chapter 6 1 Q 1 ln Q R Q RC 1 dt V t V e K1 R tV K1 591 12. From Exercise 11, we have C (a) For V 2, R 0.5, and C0 C 0.6e 0.25t. 0.8 C0 e Rt V . 13. (a) dC 0.6, we have RC RC C 1 Q R 1 Q R e R tV K1 0 0 4 Ke Rt V (b) For V 2, R 1.5, and C0 C 0.6e 0.75t. 0.8 0.6, we have 0 when t 0, it follows that K Q we have C 1 e Rt V . R Since C (b) As t → , the limit of C is Q R. Q and 0 0 4 ...
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