CHAPTER 6 Differential Equations

Transcription

1 CHAPTER 6 Differential Equations Section 6. Slope Fields and Euler s Method Section 6. Differential Equations: Growth and Deca Section 6. Separation of Variables and the Logistic Equation Section 6. First-Order Linear Differential Equations Review Eercises Problem Solving

2 CHAPTER 6 Differential Equations Section 6. Slope Fields and Euler s Method. Differential equation: Solution: Ce Check: Ce. Differential Equation: e e e Check: e e e e e. Differential equation: Solution: Check: C C C C. Differential Equation: Solution: Check: ln d 5. Differential equation: 0 Solution: Check: C cos C sin C sin C cos C cos C sin C cos C sin C cos C sin 0 6. Differential equation: 0 Solution: C e cos C e sin Check: C C e sin C C e cos C e sin C e cos C e sin C e cos C C e sin C C e cos C e cos C e sin C C C C e sin C C C C e cos 0 55

3 56 Chapter 6 Differential Equations 7. Differential Equation: cos ln sec tan cos cos sec tan sec tan sec sin ln sec tan sin ln sec tan sin sin sec cos ln sec tan Substituting, sec tan sec tan sec cos ln sec tan sin sec cos ln sec tan cos ln sec tan tan. tan sec tan sec tan sec sin ln sec tan 8. e e e e e e Substituting, e e e e e. 9. sin cos cos sin cos cos sin cos sin Differential Equation: sin cos cos cos sin sin cos sin sin Initial condition: sin cos cos 0 0. cos. sin Differential Equation: sin Initial condition: 0 0 cos0 6e 6e e Differential Equation: e 6e Initial condition: 0 6e 0 6e 0 6 6

4 Section 6. Slope Fields and Euler s Method 57. e cos e cos sin sin e cos Differential Equation: sin e cos sin sin Initial condition: ecos e 0 In Eercises 8, the differential equation is cos. cos cos 8 cos 6 5 cos 0, 6 8 cos 8 cos 0, No Yes 5. e 6. 5 ln 6e 6 6e 6e 0, Yes ln 0, No 7. C e C e C sin C cos 6C e 6C e 6C sin 6C cos 6 0, Yes 8. e sin 8e 6 sin 6 8e 6 sin 6e sin 0, Yes In 9, the differential equation is e. 9., 0. e, e e e 0 e, e e e, No Yes. e, e e e e e e, Yes. sin, cos cos sin e, No

5 58 Chapter 6 Differential Equations. ln,. e 5, e e 0 ln e, e e 0 e 5 e, Yes No 5. Ce passes through 0,. 6. C passes through 0,. Ce 0 C C 0 C C Particular solution: e Particular solution: 7. C passes through,. 6 C6 C Particular solution: or 8. C passes through,. 9 6 C C Particular solution: 9. Differential equation: General solution: Particular solutions: 00. Differential equation: 0 C C 0, Two intersecting lines C ±, C ±, Hperbolas General solution: Particular solutions: C 0, Point C, C, Circles C. Differential equation: General solution: Ce General solution: C Ce Ce 0 0. Differential equation: Initial condition: 0, Ce 0 C 0 Particular solution: e 0 Initial condition: : 8 C Particular solution:. Differential equation: General solution: C sin C cos C cos C sin, 9C sin 9C cos 9 9C sin 9C cos 9 0Initial conditions: 9C sin C cos 0 C sin C cos 6 C cos C C cos C sin C sin C C, 6 Particular solution: sin cos

6 Section 6. Slope Fields and Euler s Method 59. Differential equation: General solution: C, C C 0Initial conditions: C C ln C 0 0 C C ln C 0, C C, C ln Particular solution: ln ln ln 5. Differential equation: 0 General solution: C C C C, 6C 6C C C C C 0 Initial conditions: 0, 0 C 8C C C C C C C 0 C C C, C Particular solution: 6. Differential equation: 9 0 General solution: e C C e C C C e e C C C e C C C e C e C C C 9 9 Initial conditions: 0, 0 0 e C C C 0 C e C C C e C C C e C C 0 0 e C C Particular solution: e 7. d 8. d d C d C

7 550 Chapter 6 Differential Equations 9. d 0. d ln C u, du d d e e e e d ln e C. d d ln C ln C. cos d cos d sin C u, du d. sin d sin d cos C. d tan sec sec d tan C u, du d 5. d Let u, then u and d u du. d u uu du u u du u 5 5 u C 5 5 C Let u 5, u 5, d u du. d 5 d 5 u uu du 0u u du 0u u5 5 C C 7. d e 8. 5e d e d e C u, du d 5e d 5e d 0e C

8 Section 6. Slope Fields and Euler s Method d Undef. 0 d d d cos 5. d For,. d Matches. d sin For 0, 0. d Matches (c). 55. d e 56. As, d 0. Matches (d). d For 0, d is undefined (vertical tangent). Matches (a). 57. (a), 5 (, ) 58. (a), 6 5 (c) As, As, (c) As, As, 59. (a), (, ) 60. (a), 8 (c) As, As, (c) As, As,

9 55 Chapter 6 Differential Equations 6. (a), 06. (a), 0 (, 0) 6 (0, ) As, Note: The solution is ln., (, ) 6 As, As,, (, ) As, , d, 0 d , d 0., 0 9 d Slope field for with solution passing e8 sin 67. Slope field for with solution passing through 0,. through 0,

10 Section 6. Slope Fields and Euler s Method , 0, n 0, h 0. 0 hf 0, hf, , etc. n n n , 0, n 0, h hf 0, hf, , etc. The table shows the values for n 0,,,..., 0. n n n , 0, n 0, h hf 0, hf, , etc. n n n , 0, n 5, h 0. 0 hf 0, hf, , etc. n n n e, 0, n 0, h 0. 0 hf 0, 0 0.e 0. hf,. 0.e 0...6, etc. n n n

11 55 Chapter 6 Differential Equations 7. cos sin, 0 5, n 0, h 0. 0 hf 0, cos 0 sin hf, cos0. sin , etc. n n n d, e, (eact) h 0. h d,, (eact) h 0. h d cos, sin cos e, As h increases (from 0. to 0.), the error increases (eact) h 0. h , h 0. dt 7, (a) t 0 Euler , h 0.05 dt 7, (a) t 0 Euler e t eact 7 68e t eact t 0 Eact t 0 Eact The approimations are better using h 0.05.

12 Section 6. Slope Fields and Euler s Method A general solution of order n has n arbitrar constants while in a particular solution initial conditions are given in order to solve for all these constants. F, 8. A slope field for the differential equation consists of small line segments at various points, in the plane. The line segment equals the slope of the solution at the point,. F, F,, Consider Begin with a point 0, 0 that satisfies the initial condition, 0 0. Then using a step size of h, find the point, 0 h, 0 hf 0, 0. Continue generating the sequence of points n, n n h, n hf n, n. 8. Ce k 85. False. Consider Eample. is a solution to Ckek d but is not a solution. Since d 0.07, we have Cke k 0.07Ce k. Thus, k C cannot be determined. 0, 86. True 87. True 88. False. The slope field could represent man different differential equations, such as. 89., 0, e solution d (a) If h is halved, then the error is approimatel halved r 0.5. (c) When h 0.05, the errors will again be approimatel halved. e e r , 0, e d solution (a) If h is halved, then the error is halved r 0.5. (c) When h 0.05, the error will again be approimatel halved. e e r

13 556 Chapter 6 Differential Equations 9. (a) L di RI Et 9. dt di I dt di dt I 6 I I e kt ke kt k e kt 6 0 k e kt 6e kt 0 k 6 0 k ± since e kt 0 t As t, I. That is, lim It. In fact, I t is a solution to the differential equation. 9. A sin t9. A cos t A sin t 6 0 A sin t 6A sin t 0 A sin t6 0 ± If A 0, then radians/sec. f f gf, g 0 ff ff g f d d f f g f For < 0, g f 0 For > 0, g f 0 Thus, f f is increasing for < 0 and decreasing for > 0. f f has a maimum at 0. Thus, it is bounded b its value at 0, f0 f0. Thus, f and f is bounded. 95. Let the vertical line k cut the graph of the solution f at k, t. The tangent line at k, t is Since t fk k p q, we have t qk pkt k For an value of t, this line passes through the point To see this, note that qk pk t? qk pkt k pk k? qkk pktk qk t kqk pkkt pk qk t. pk k pk, qk pk.

14 Section 6. Differential Equations: Growth and Deca 557 Section 6. Differential Equations: Growth and Deca. d. d d C d C. d d. d d d d ln C ln C e C Ce e C Ce Ce Ce d 5 d 5 d 6. d d C 7. d d d 5 C 9 C ln C 5 C e C e C e Ce d d d d d ln C d e C e C e Ce ln ln C ln ln ln C ln ln C C

15 558 Chapter 6 Differential Equations d d 00 d. dq dt k t dq dt dt k t dt dq k C t Q k t C ln00 C ln00 C 00 e C e C e Ce. dp k0 t dt. dn k50 s ds dp dt dt k0 t dt dn ds ds k50 s ds dp k 0 t C dn k 50 s C P k 0 t C N k 50 s C. kl d k L d L d d k d k L C lnl k C L e k C L e C e k L Ce k 5. (a) 5 9 (0, 0) 6, 0, 0 d d 6 ln 6 C 6 e C C e 6 C e 5 0, 0: 0 6 C C 6 6 6e 7 6 6

16 Section 6. Differential Equations: Growth and Deca (a) 7. dt t, 0, 0 (0, ) t dt t C d, 0, ln d C 0 0 C C 0 t 0 6 (0, 0) e C C e 0, : C e0 C e 8. dt t, 0, 0 9. dt, 0, 0 t dt dt t C ln t C 0 0 C C 0 e tc e C e t Ce t 0 Ce 0 C 0 t 0 0e t 5 6 (0, 0) (0, 0) dt, 0, 0. k d dt Ce k (Theorem 5.6) (0, 0) 5 5 0, : Ce 0 C ln t C 5, 0: 0 e k k ln 5 e tc e C e t Ce t 0 Ce 0 C 0 0e t When 6, e ln56 e ln5 5 5.

17 560 Chapter 6 Differential Equations. dn dt kn. dv kv dt N Ce kt (Theorem 5.6) V Ce kt (Theorem 5.6) 0, 50: C 50 0, 0,000: C 0,000, 00: 00 50e k k ln ln 8 5,,500:,500 0,000e k k ln 5 8 When t, N 50e ln85 50e ln85 When t 6, V 0,000e ln586 0,000e ln , dp kp dt 5. Ce kt, 0,, 5, 5 P Ce kt (Theorem 5.6) 0, 5000: C 5000 C, 750: e k k ln 9 0 ekt When t 5, P 5000e ln905 5 e5k k ln 0 5 eln 05t 0t5 or e0.605t Ce kt,,, 5, 5 Ce kt, 0,, 5, 8. C e kt e5k k ln8 5 e 0.59t 0.59 Ce k 5 Ce 5k 5Ce k Ce 5k 5e k e 5k 5 e k k ln 5 Ce 0.0t Ce 0.0 C e 0.0t C 5 Ce kt,,,, 5 Cek 5 Ce k Ce k 5 Cek 0e k e k 0 e k k ln 0.06 Ce.06t 5 Ce.06 C e.06t

18 Section 6. Differential Equations: Growth and Deca In the model Ce kt, C represents the initial value of when t 0. k is the proportionalit constant. 0. k dt. d when > 0. Quadrants I and III. d > 0. d when > 0. Quadrants I and II. d > 0. Since the initial quantit is 0 grams, 0e kt. Since the half-life is 599 ears, 5 0e k599 k 599 ln. Thus, 0e ln599t. When t 000, 0e ln g. When t 0,000, 0. g.. Since the half-life is 599 ears, ek599 k 599 ln. Since there are.5 g after 000 ears,.5 Ce ln C.. Hence, the initial quantit is approimatel. g. When t 0,000,.e ln5990, g. 5. Since the half-life is 599 ears, ek599 k 599 ln. Since there are 0.5 gram after 0,000 ears, 0.5 Ce ln5990,000 C Hence, the initial quantit is approimatel 8.58 g. When t 000, 8.58e ln g. 6. Since the half-life is 575 ears, ek575 k 575 ln. Since there are grams after 0,000 ears, Ce ln5750,000 C Hence, the initial quantit is approimatel 6.76 g. When t 000, 6.76e ln g. 7. Since the initial quantit is 5 grams, C 5. Since the half-life is 575 ears,.5 5e k575 k 575 ln. When t 000 ears, 5e ln g. When t 0,000 ears, 5e ln5750,000.9 g. 8. Since the half-life is 575 ears, ek575 k 575 ln. Since there are. grams when t 000 ears,. Ce ln C.6. Thus, the initial quantit is approimatel.6 g. When t 0,000,.6e ln5750, g.

19 56 Chapter 6 Differential Equations 9. Since the half-life is,00 ears, ek,00 k,00 ln. Since there are. grams after 000 ears,. Ce ln,00000 C.6. Thus, the initial quantit is approimatel.6 g. When t 0,000,.6e ln,000,000.6 g. 0. Since the half-life is,00 ears, ek,00 k,00 ln. Since there are 0. grams after 0,000 ears, 0. Ce ln,000,000 C 0.5. Thus, the initial quantit is approimatel 0.5 g. When t 000, 0.5e ln, g.. Ce kt. C Cek599 k 599 ln When t 00, Ce ln C Therefore, 95.76% remains after 00 ears. C Cek C Ce ln575t ln0.5 Ce kt k 575 ln ln t 575 t 5,6.8 ears. Since A 000e 0.06t, the time to double is given b. Since A 0,000e 0.055t, the time to double is given b e 0.06t and we have 0,000 0,000e 0.055t and we have e 0.06t ln 0.06t t ln.55 ears Amount after 0 ears: A 000e $8. e 0.055t ln 0.055t t ln.6 ears Amount after 0 ears: A 0,000e $, Since A 750e rt and A 500 when t 7.75, we have 6. Since A 0,000e rt and A 0,000 when t 5, we the following. have the following e 7.75r r ln % 7.75 Amount after 0 ears: A 750e $8.67 0,000 0,000e 5r r ln % Amount after 0 ears: A 0,000e ln 50 $0,000

20 Section 6. Differential Equations: Growth and Deca Since A 500e rt and A 9.85 when t 0, we have the following e 0r r ln The time to double is given b e t t ln 7.0 ears % 8. Since A 000e rt and A when t 0, we have the following e 0r r ln The time to double is given b e 0.0t t ln 6.9 ears % ,000 P ,000 P P 500, P 500, $5,6.0 $, ,000 P ,000 P P 500, P 500, $0, $5,.9 5. (a) t (d) e 0.07t (c).07 t ln t ln.07 t ln 0. ears ln t t ln t ln 0.07 t t t ln 65t ln t ln 9.9 ears ln 0.07 ln 9.90 ears 65 ln e 0.07t ln 0.07t t ln 9.90 ears 0.07

21 56 Chapter 6 Differential Equations 5. (a) t 55. (a) t.06 t.085 t ln t ln.06 ln t ln.085 t ln.90 ears ln.06 t ln 8.50 ears ln t t 0.06 t t ln t ln 0.06 ln t ln t ln ln ears t ln ln ears (c) t (c) t t t ln 65t ln ln 65t ln t 65 ln ln ears t 65 ln ln ears 65 (d) e 0.06t (d) e 0.085t e 0.06t e 0.085t ln 0.06t ln 0.085t t ln.55 ears 0.06 t ln 8.5 ears (a) t (c).055 t ln t ln.055 t ln.95 ears ln t t ln t ln t ln ln ears (d) t t ln 65t ln t e 0.055t e 0.055t ln 0.055t ln ln ears 65 t ln.60 ears 0.055

22 Section 6. Differential Equations: Growth and Deca (a) P Ce kt Ce 0.009t 58. (a) P Ce kt Ce 0.08t P 7.7 Ce P.7 Ce 0.08 C C.7 P e 0.009t For t 5, P e million. (c) If k < 0, the population is decreasing. P.7e 0.08t For t 5, P.7e million. (c) For k > 0, the population is increasing. 59. (a) P Ce kt Ce 0.06t 60. (a) P Ce kt Ce 0.00t 5. P Ce 0.06 P.6 Ce 0.00 C C.607 P e 0.06t For t 5, P e million. (c) For k > 0, the population is increasing. P.607e 0.00t For t 5, P.607e million. (c) For k < 0, the population is decreasing. 6. (a) N t N 00 when t 6. hours (graphing utilit) Analticall, t.55 t t ln.55 ln.996 t ln hours. ln (a) Let Ce kt. At time : 5 Ce k C 5e k At time : 50 Ce k 50 5e k e k 5 C 5e k 5e ln Approimatel 5 bacteria at time eln5t.6e 0.58t (c) When t 8, k ln 5 k ln eln (d) 5, eln5t t.9 hours.6 ek

23 566 Chapter 6 Differential Equations 6. (a) 9 0 e 0k 6. (a) 0 0 e 0k 0e 0k 0e 0k 0 k ln k ln 0 ln N 0 e 0.050t N 0 e 0.066t 5 0 e 0.050t 5 0 e 0.066t e 0.050t 6 e 0.066t 6 t ln 6 6 das t ln 6 9 das (a) P Ce kt 8e kt 05 8e 0k k 0 ln (c) 00 P P P 8e 0.05t 8.05 t Using a graphing utilit, P t The model P fits the data better. (d) Using the model P, t 0.009t 8.8 t ln08.8 ln ears, or (a) R t (c) e 0.079t I 0.6t.988t 6.599t 0.8t (d) Pt I R 0 0 According to the model, Rt 68.6e 0.079t I 0 log 0 I I 0, I I 9 0 log 0 0 0log 6 0 I 6 (a) (c) (d) 0 0 log decibels log decibels log decibels log decibels log 0 I I I 80 0 log 0 0 0log 6 0 I 6 8 log 0 I I 0 8 Percentage decrease: %

24 Section 6. Differential Equations: Growth and Deca At Vte 0.0t 00,000e 0.8t e 0.0t 00,000e 0.8t0.0t da dt 00, t 0.0 e0.8t0.0t 0 when 6. The timber should be harvested in the ear 0, Note: You could also use a graphing utilit to graph At and find the maimum of At. Use the viewing rectangle 0 0 and 0 600, R ln I ln I 0 ln 0 ln I 0 ln 0, I er ln 0 0 R (a) 8. ln I ln I 0 ln 0 I ,56,.5 R ln I ln I 0 ln 0 R ln 0 I e R ln 0 e (c) R ln I ln I 0 ln 0 dr di l I ln 0 e R ln 0 0 R Increases b a factor of e R ln 0 or 0 R. 7. Since k 80 dt 7. k 0 dt 80 k dt ln 80 kt C. When t 0, 500. Thus, C ln 0. When t, 0. Thus, 0 Ce kt (See Eample 6) 60 0 Ce k0 C e k5 e5k 7 k ln 0 ln0 80 k ln 00 ln 0 ln 0. k 5 ln e5 ln7t Thus, 0e ln0t 80. When t 5, 79.. eln7ts 7 ts ln t 5 ln 7 t 5 ln 5 ln 0.5 minutes ln 7 ln 7 It will take minutes longer. 7. False. If Ce kt, Cke kt constant. 7. True 75. True 76. True

25 568 Chapter Differential Equations Section 6. Separation of Variables and the Logistic Equation. d. d d d C C C. dr 0.05r ds. dr 0.05s ds dr r 0.05 ds dr 0.05s ds ln r 0.05s C r 0.05s C r e 0.05sC Ce 0.05s 5. d ln ln ln C ln C C 6. d ln ln ln C ln C C 7. sin 8. d 6 cos sin d 6 cos d cos C cos C 6 sin C sin C d d 8 8 d 9 5 d 5 9 d 5 9 C C

26 Section 6. Separation of Variables and the Logistic Equation 569. ln 0 ln ln ln C d d u ln, du e ln C Ce ln. e d e d e C. e 0. e d e C e C Initial condition: 0, 6 C, C Particular solution: e 0 d C C Initial condition:, 8 9 C Particular solution: d ln Initial condition: Ce C, Ce, C e Particular solution: e e 6. ln 0 ln d ln d : C ln ln C d ln ln C ln ln ln C lnc C 0 : C C d 0 : d C 0 C C

27 570 Chapter 6 Differential Equations 9. du uv sin v dv 0. dr ers ds du u v sin v dv e r dr es ds lnu cos v C e r es C u Ce cos v Initial condition: u0, C e e Particular solution: u e cos v r0 0: C C e r es e r es r ln es ln es r ln e s. dp kp dt 0. dp kdt P dt kt 70 dt 0 dt T 70 k dt ln P kt C lnt 70 k t C P Ce kt T 70 Ce kt Initial condition: P0 P 0, P 0 Ce 0 C Particular solution: P P 0 e kt Initial condition: T0 0; Ce 0 C Particular solution: T 70 70e kt, T 70 e kt. 9 d d. d d 8 9 C Initial condition:, 8 9 C, C 5 ln ln ln C C Initial condition: 8, C8, C 8 Particular solution: Particular solution: 8, m d 0 6. m 0 d 0 d d ln C Ce ln ln C ln ln C ln C C

28 Section 6. Separation of Variables and the Logistic Equation f, 8. f t, t t tt t t Homogeneous of degree f, f t, t t t t Not homogeneous 9. f, 0. f, ft, t t t t t f t, t t t t t Homogeneous of degree t t t Homogeneous of degree. f, ln. f, tan ft, t lntt f t, t tant t tant lnt ln t ln Not homogeneous Not homogeneous. f, ln. f, tan ft, t ln t t ln f t, t tan t t tan Homogeneous degree 0 Homogeneous of degree 0 5. v dv v d dv v v v d dv v d ln v ln ln C lnc v C C C C, v 6. d v, dv v d v dv v d v d v dv v d d v d v dv d v dv d v ln C ln C ln C

29 57 Chapter 6 Differential Equations 7. v dv v d v v d dv v v d v v v dv d ln v v ln ln C ln C v v C C C, v 8. dv v v v v v d d v v dv d v v v d dv v d v v v dv d lnv ln ln C ln C v C C C, v 9. v d dv v dv d v v v v d dv v v dv v d, v 0. v ln v ln ln C ln C v ln C v ln C v v d Ce v v d v dv v v d dv d v dv v d ln v ln ln C ln C v C C, v C C. e d 0, v v d dv e v v d 0 e v dv d e v ln C e ln C ln e C ln Initial condition: 0, C Particular solution: e ln

30 Section 6. Separation of Variables and the Logistic Equation 57. d 0, v. v d vv d dv 0 v dv v d v ln v ln ln C ln C C ev v C v ln C v e Ce cos v dv d Initial condition: sec d 0, v sec v vd v d dv 0 sin v ln ln C Ce sin v Ce sin 0, Ce 0 C Particular solution: e sin sec v v d v d dv Initial condition:, Ce C e Particular solution: e. d 0 Let v, dv v d. 5. d v d v dv v d 0 v d v dv 0 v d v dv d v v dv ln ln v C ln ln v ln C d C C v C C C 0: C 0 C

31 57 Chapter 6 Differential Equations 6. d 7. d 8 d C C d ln C e C Ce d d 0.5 d d ln 8 C e C 8 Ce 8 Ce 8 9. (a) Euler s Method gives d 6 6 ln C Ce 0 5 C 5 5e (c) At, 5e Error: (a) Euler s Method gives 0.6. d 6 6 d C C C C 9 (c) At, Error:

32 Section 6. Separation of Variables and the Logistic Equation (a) Euler s Method gives.08. d d : (c) For, C C C Error: (a) Euler s Method gives d d arctan C arctan0 C C arctan tan (c) At.5, tan k, dt C 0 0 ek599 Ce kt k 599 ln Ce ln599t initial amount When t 5, 0.989C or 98.9%. k, Cekt dt Initial conditions: 0 0, 6 Particular solution: When 75% has been changed: 5 0e t ln5 et ln5 t ln 6. hr ln5 0 Ce 0 C 6 0e k k ln 5 0e t ln5 55. k d The direction field satisfies d 0 along ; but not along 0. Matches (a). 56. k d The direction field satisfies d 0 along : Matches. 57. d k 58. k d The direction field satisfies d 0 along 0 and The direction field satisfies d 0 along 0, and. Matches (c). grows more positive as increases. Matches (d).

33 576 Chapter 6 Differential Equations 59. dw k00 w dt dw 00 w k dt ln 00 w kt C 00 w e ktc Ce kt w 00 Ce kt w C C w 00 0e kt (a) k 0.8: t. ears k 0.9: t.6 ears k.0: t.05 ears (c) Maimum weight: 00 pounds lim w 00 t 60. From Eercise 0: w 00 Ce kt, k w 00 Ce t w0 w 0 00 C C 00 w 0 w w 0 e t 6. Given famil (circles): Orthogonal trajector (lines): C 0 d ln ln K ln K 6. Given famil (hperbolas): Orthogonal trajector: C 0 d ln ln ln k k k 6 6

34 Section 6. Separation of Variables and the Logistic Equation Given famil (parabolas): Orthogonal trajector (ellipses): C C C d K K 6. Given famil (parabolas): Orthogonal trajector (ellipse): C C C d K K Given famil: C C C Orthogonal trajector (ellipses): 66. Given famil (eponential functions): Orthogonal trajector (parabolas): Ce Ce d d K K K K e 68. Since 0 6, it matches (c) or (d). Since (d) approaches its horizontal asmptote slower than (c), it matches (d). Since e 0, it matches (a) e Since 0 8, it matches. e Since 0 6, it matches (c) or (d). Since approaches L faster for (c), it matches (c).

35 578 Chapter 6 Differential Equations 7. Pt (a) (c) (d) 500 e 0.75t 7. k 0.75 L 500 P e 0.75t e 0.75t e 0.75t 0.75t ln ln t ln dp (e) P0 60 dt 0.75P 500 P, Pt (a) (c) (d) e 0.t k 0. L 5000 P e 0.t e 0.t 9 0.t ln ln 9 9 t e 0.t ln dp (e) P0 5 dt 0.P 5000 P, 7. dp dt P 00 P (d) d P P dt P 00 P 00 P (a) (c) k L 00 P t P 9P 9P P P P P 00 P P P P 00 P P 00 d P for P 50, and b the first Derivative Test, this dt 0 is a maimum. Note: P 50 L dp dt 0.P 0.000P 0.P 0.00P 75. (a) (c) k 0. 0 L P 50 P dt 0, k, 0 8: L 0 Solution: 8 0 b b 0 e t 0 8 L 0 bekt be t 0 00 (d) P (Same argument as in Eercise 7)

36 Section 6. Separation of Variables and the Logistic Equation k., L be kt 8 be.t 0 5: L b b 8 5 b 5 8 Solution: 5 e.t 0 5 dt. 8, 77. dt , k 0.8, 5 L be 0 kt be 0.8t 0 8: Solution: L b 0 e 0.8t b dt ; k 0, L be 0 kt be 0t 0 5: L 0 Solution: 5 0 b 0 5e 0t b (a) 7e k 00 9 L be 5 00 b b 7 9 e k 9 k ln 9 ln e 0.60t L 00, 00 7e k For t 5, 70 panthers. 0 5 kt, (c) 00 7e 0.6t 0.6t ln e 0.6t t 7.7 ears (d) 0 5 dt k L , Using Euler s Method, 0.5 when t 6. (e) is increasing most rapidl where 00 00, corresponds to t 7.7 ears.

37 580 Chapter 6 Differential Equations 80. (a) 0 b b 9 L be kt, 0 9e k 9e k 5 L 0, 0 8. A differential equation can be solved b separation of variables if it can be written in the form M N 0. d To solve a separable equation, rewrite as, M d N and integrate both sides. e k 9 0 9e 0.057t Note: k ln 9 ln 9 ln 0 9e t ln 0 9 t For t 5,.58 grams. (c) (d) 8 0 9e t ln 7et ln dt ln t 6 t 8.8 hours For t 5, Euler s Method gives.09 grams. (e) The weight is increasing most rapidl when 0 5, corresponding to t 5. hours t t 0 5 Eact Euler M, d N, 0, where M and N are homogeneous functions of the same degree. See Eample 7a. 8. Two families of curves are mutuall orthogonal if each curve in the first famil intersects each curve in the second famil at right angles.

38 Section 6. Separation of Variables and the Logistic Equation (a) dv kw v dt dv W v k dt ln W v kt C Initial conditions: v W Ce kt 85. False. is separable, but 0 is not a solution. d W 0, v 0 when t 0, and v 5 when t. C 0, k ln Particular solution: v 0 e lnt 0 t or v 0 e 0.877t s 0 e 0.877t dt 0t.76e 0.877t C Since s0 0, C 69.5 and we have s 0t 69.5e 0.877t. 86. True d 87. False f t, t t t t f, 88. True C d C K K C C K C K K d 89. be kt be kt bkekt k be kt be kt be kt k be kt bekt be kt k be kt kt be k

39 58 Chapter 6 Differential Equations 90. Product Rule fg gf fg f fg gf 0 g f g 0 f f Need f f e e e 0, so avoid. g f g f f e e lng ln C g Ce Hence there eists g and interval a, b, as long as a, b. Section 6. First-Order Linear Differential Equations. e. e Linear ln 0 Linear ln 0 ln. cos. Not linear, because of the -term. Linear 5. d Integrating factor: e d e ln d C 6. d Integrating factor: e d e ln d C C C 7. 0 Integrating factor: e d e 8. Integrating factor: e d e e e 0e e 0e d 0e C 0 Ce e e d e C Ce

40 Section 6. First-Order Linear Differential Equations cos d 0. sin d 0 cos cos cos cos cos sin sin Integrating factor: e sin d e cos Integrating factor: e cos d e sin e sin cos e sin cos e sin e sin cos e sin d e cos sine cos d e cos C Ce cos e sin C Ce sin. Integrating factor: e d e ln. e Integrating factor: e d e e e e d e 6 d 6 e6 C d C 6 e Ce C. e Integrating factor: e d e. cos Integrating factor: e d e e e e d d C e e cos d Ce e cos sin C sin cos Ce 5. (a),(c) 5 d e e d Integrating factor: e d e e e e e e d 6 e e C C C e e e e e e

41 58 Chapter 6 Differential Equations 6. (a),(c) sin, P, Q sin u e d e ln sin sin d cos C cos C 0 cos C C cos 7. cos 0 sec sec Integrating factor: e sec d e tan 8. e e e tan sec e tan d e tan C Ce tan Initial condition: 0 5, C Particular solution: e tan Integrating factor: e d e e d C e C Initial condition: e, C Particular solution: e 9. tan sec cos Integrating factor: e tan d e ln sec sec sec sec sec cos d tan C sin cos C cos Initial condition: 0, C Particular solution: sin cos 0. sec sec Integrating factor: e sec d e ln sec tan sec tan sec tan sec tan sec d sec tan C C sec tan Initial condition: 0, C 0, C Particular solution: cos sec tan sin

42 Section 6. First-Order Linear Differential Equations Integrating factor: e d e ln Separation of variables: d d. 0 Integrating factor: e d e e C Ce Separation of variables: d ln ln ln C ln ln C ln ln C C e C Ce Initial condition:, C Initial condition:, C Particular solution: Particular solution: e. d. d d u e d ln ln C ln C Linear d d 0 C C d 5 5 C C Linear u e d d 6 5 C C 7 5 d n, Q, P n, Q, P, e d e e d e d d e e d e e d e C Ce e e C Ce Ce

43 586 Chapter 6 Differential Equations n, Q, P e d e ln d C C C n, Q, P e d e ln d 5 5 C 5 C 5 5 C 5 9. e, n, Q e, P e d e e e e d e d e e C e Ce 0. e e n, Q e, P e d e e e e d e C e Ce. (a) 8 d 6 Integrating factor: e d e ln d C (c) 8 C 6, : 8 C C 8 (, 8): 8 8 C C

44 Section 6. First-Order Linear Differential Equations 587. (a) 5 Integrating factor: e d e e e e 5 e e d e C Ce (c) 5 0, 7 : 7 C C e 0, : C C e 5. (a) 5 cot Integrating factor: sin cos sin e cot d e lnsin sin sin sin d cos C cot C csc (c) (, : cot C csc C cot sin cos csc cot csc sin cos 5, : cot C csc C cot cos sin csc cot csc cos sin. (a) 6 5 Bernoulli equation, n letting z, ou obtain e d e and e d e. The solution is: e e C Ce Ce (c) 6 5 0, : C C C 6 Ce e 6 e 0, : C C C e

45 588 Chapter 6 Differential Equations 5. dp kp N, N constant dt dp dt kp N kp N dp dt k lnkp N t C lnkp N kt C kp N e ktc P C e kt N k 6. da ra P dt da dt ra P da ra P dt r lnra P t C lnra P rt C ra P e rtc A C e rt P r P Ce kt N k When t 0: P P 0 A Ce rt P r When t 0: A 0 P 0 C N k C P 0 N k 0 C P r C P r P P 0 N k ekt N k A P r ert 7. (a) A P r ert A 00, e ,098.0 A 50, e0.050,, ,000 75, e0.08t.85 e 0.08t t ln ears dq 9. (a) q kq, q constant dt Let Pt k, Qt q, then the integrating factor is ut e kt. (c) Q kq q Q e kt When t 0: Q Q 0 lim t Q q k qe kt dt e kt q k ekt C q k Q 0 q k C C Q 0 q k Q q k Q 0 q k ekt Cekt 0. (a) dn dt k0 N N kn 0k Integrating factor: e kt Ne kt 0ke kt dt 0e kt C N 0 Ce kt (c) For t, N 0: For t 0, N 9: 0 ln Ce k 0 Ce k 9 0 Ce 0k ek e0k e9k 9k 0 Ce k Ce 0k k 9 ln C 0e k N e 0.088t

46 Section 6. First-Order Linear Differential Equations 589. Let Q be the number of pounds of concentrate in the solution at an time t. Since the number of gallons of solution in the tank at an time t is v 0 r r t and since the tank loses gallons of solution per minute, it must lose concentrate at the rate Q v 0 r r t r. The solution gains concentrate at the rate r q. Therefore, the net rate of change is dq dt q r Q v 0 r r t r or r dq dt r Q v 0 r r t q r.. From Eercise, and using r r r, dq dt rq v 0 q r.. (a) Q r Q v 0 r r t q r Q0 q 0, q 0 5, q 0, v 0 00, r 0, r 0, Q 0 Q 0 dq Q 0 dt ln Q 0 t ln C Q Ce 0t Initial condition: Q0 5, C 5 Particular solution: Q 5e 0t (c) 5 5e 0t ln 5 0 t t 0 ln 0. min 5 lim t 5e0t 0. (a) Q r Q v 0 r r t q r Q0 q 0 5, q 0.0, v 0 00, r 0, r 0, Q Q 0. 0 Integrating factor: e 0t Qe 0t 0.e 0t dt 8e 0t C 5 8 7e 0t (c) 7 7e 0t ln t t 0 ln min lim Qt 8 lbs t Q 8 Ce t Q C C 7 Q 8 7e 0t

47 590 Chapter 6 Differential Equations 5. (a) The volume of the solution in the tank is given b v 0 r r t. Therefore, 00 5 t 00 or t 50 minutes. Q r Q v 0 r r t q r Q0 q 0, q 0 0, q 0.5, v 0 00, r 5, r, Q Integrating factor: Q50 t Initial condition: Particular solution: e 00t dt 50 t.550 t dt 50 t 5 C Q 50 t C50 t Q0 0, 0 50 C50, C 50 5 Q 50 t t Q.5 00 t Q lbs 6. (a) The volume of the solution is given b v 0 r r t 00 5 t 00 t 50 minutes. Q r Q v 0 r r t q r Q0 q 0 0, q, v 0 00, r 5, r Q Q 00 t 5 Integrating factor is 50 t, as in #. Q50 t 550 t dt 50 t 5 C Q 50 t C50 t Q0 0: 0 00 C50 C Q 50 t t When t 50, Q lbs (double the answer to #) 7. From Eample 6, dv dt kv m g v mg k ektm, Solution g, mg 8, v5 0, m 8 implies that g 0 8 k e5k. Using a graphing utilit, k , and v 59.7 e 0.007t. As t, v 59.7 ftsec. The graph of v is shown below st vt dt 59.7 e 0.007t dt 59.7t 79.57e 0.007t C s C C st 59.7t 79.57e 0.007t The graph of st is shown below st 0 when t 6. sec. 00

48 Section 6. First-Order Linear Differential Equations L di dt RI E 0, L 50. Integrating factor: e RL dt e RtL I e E RtL 0 L ertl dt E 0 R ertl C I E 0 CeRtL R I R L I E 0 I0 0, E 0 0 volts, R 600 ohms, L henrs I E 0 CeRtL R C C 5 I 5 5 e50t lim I t 5 amp e50t 0.9 e 50t e 50t 0. 50t ln0. t ln sec P Q d Standard form u ep d Integrating factor P Q n 5. Standard form Let z n n 0,. Multipling b n n produces n n np n nq z npz nq. Linear d d C ln C Matches c. Ce Matches d d ln C Matches a. Ce d ln C C e Ce Matches b.

49 59 Chapter 6 Differential Equations 57. e d e Separation of variables: e e d e e e d e e e C e e C d 0 Separation of variables: d 6 6 C C 59. cos cos d Separation of variables: cos d Separation of variables: d sin ln ln C arcsin C ln sin ln C sin C Ce sin 6. d 0 Homogeneous: v, v d dv v v d v v d dv 0 5 d v dv 0 v v ln 5 ln v v ln C 5 v v C C 6. d 0 Linear: Integrating factor: d ln C ln C e d e ln 6. e d 0 6. Linear: e Integrating factor: e d e ln e d e C e C d 0 Homogeneous: v, v d dv v v d v d dv 0 v d dv 0 d dv v ln v C C ln

50 Review Eercises for Chapter d d 0 Bernoulli: n, Q, P, e d e ln Homogeneous: v, d v dv v dv v 0 v dv C d C ln v ln ln C v C C 67. d 68. d e 0 Integrating factor: d e d e ln Separation of variables: d e ln e C ln e C d 5 5 C 5 C e False. The equation contains. 70. True. is linear. Review Eercises for Chapter 6., 6 Not a solution. sin cos 8 sin 6 cos 8 6 cos 8 sin 0 Not a solution. d 5. d 5 d 5 C d C 5. cos 6. d sin d cos d sin C sin d cos C

51 59 Chapter 6 Differential Equations 7. 7 d 7 d 8. e d e d 9e C Let u 7, du d, u 7: u 7u du 5 u5 8 u C C 5 7 C 9. d d sin d Undef. 0 d ,, (, )., 0, 5.,. 5 0, (0, ),, 5. 0,, 6. (0, ), 0, 8

52 Review Eercises for Chapter d 6 d 6 C 6 d 6 d 9. d d 0. ln 6 C 6 e C Ce 6 Ce d d C C C C C C C C. 0. d d d ln ln C Ce Ce 0 d d ln ln C Ce. Ce kt. Ce kt 0, : 5, 5: C 0 5 ek5 e5k k 5 ln 0 eln 05t e0.79t, :, 5: Hence, Cek C ek 5 Ce k ek ek ek 0 ek k ln 0 C e ln Finall, 9 0 e ln 0t.

53 596 Chapter 6 Differential Equations 5. Ce kt 6. 0, 5: 5, 6 : C 5 5e5k 6 k 5 ln 0 5e t ln 05 5e 0.680t ln 0 5 Ce kt, 9: 6, : Hence, 9 Ce k C 9e k Ce 6k 9e k e 6k 9e 5k k 5 ln C 9e 5 ln Finall,.586e 0.008t. 7. dp kp, P Ce dh kt 5e kt.5 5e Ph k599 0e kh P8,000 0e 8,000k 5 k ln ln 8,000 8,000 Ph 0eh ln 8,000 P5,000 0e 5,000 ln 8, inches k 599 ln When t 600, 5e g. 9. S Ce kt (a) S 5 when t (c) 0 5 Ce k lim t Cekt C 0 5 0e k k ln S 0 6 0e.798t When t 5, S which is 0,965 units S 5 e kt (a) 5 e k e k 5 ek 5 k ln ,000 units lim S t 5 (c) When t 5, S.55 which is,55 units. (d) P Ce 0.05t C Ce 0.05t e 0.05t ln 0.05t t ln 6. ears 0.05

54 Review Eercises for Chapter (a) ds 0.0, s > 50 Speed(s) ds 0.0 ln s C Ce 0.0s When s 50, 8 Ce C 8e 0.6 8e s, s > 50. Miles per Gallon () d. d e e d e e d e e d ln C ln e C d d ln C e C Ce e sin 0 d e sin e sin d e cos C e cos C ln C C cos C ln cos C 7. (homogeneous differential equation) d d 0 Let v, dv v d. v d v dv v d 0 v v d v dv 0 v d v dv v d v dv d v v dv ln ln v C ln v ln C C v C C C or C

55 598 Chapter 6 Differential Equations 8. (homogeneous differential equation) d d 0 Let v, dv v d. v d dv v d 0 v d dv 0 v d dv d v dv ln ln v C ln v ln C C v C v C C C C C C 9. C C C C 6C 6C C C C C, 0: 0 C 8C C C, : C C 6C C 9C C C 0 C C 8C C, C 0. dv kv 9.8 dt (a) dv kv 9.8 dt k ln kv 9.8 t C ln kv 9.8 kt C kv 9.8 e ktc C e kt v k 9.8 C ekt (c) 9.8 lim vt t k st k 9.8 kv 0 9.8ekt dt k 9.8t k kv 0 9.8ekt C 9.8t k k kv 0 9.8ekt C At t 0, v 0 k 9.8 C C kv s0 k kv C C s 0 k kv v k 9.8 kv 0 9.8ekt. Note that k < 0 since the object is moving downward. st 9.8t k k kv 0 9.8ekt s 0 k kv t k k kv 0 9.8ekt s 0

56 Review Eercises for Chapter d d C. d d ln C C ellipses ln C C e C e Ce. Pt (a) (c) (d) 700 e 0.55t. k 0.55 L 700 P e 0.55t e 0.55t 700 e 0.55t (e) dp dt 0.55P 700 P t 0.55 ln 6.88 rs. Pt (a) (c) (d) 800 e 0.5t k 0.5 L 800 P e 0.5t 800 e 0.5t t 0.5 ln 7.59 rs. (e) dp dt 0.5P 800 P 5. (a) L 0,00, 0 00, ,00 be kt ,00 b 000 0,00 6e k 8 7,8 trout (c) 0,000 6e k 6 5 k ln 0 ln ,00 6e 0.55t b 6 0,00 t.9 rs. 0.55t 6e dt ,00, Use Euler s method with h t Eact 00 7,95 7,7 Euler ,869 6,70 Euler s method gives 8 6,70 trout.

57 600 Chapter 6 Differential Equations P, 8 Ce Q 8 u e d e e 8e d e 8e C e e e P, Q e u e d e e e e d e e C e Ce 9. e e P, u e d e e e e d e C e Ce Q e d P 5, 5 Ce5 Q u e 5 d e 5 e 5 e5 d e 5 5 e5 C 5. d P, u e d e ln c Q d d P, u e d e ln d C Q C sin d 0 sin Integrating factor: e d e e e sin d e sin cos C sin cos Ce 5. tan e d tan e d 55. Integrating factor: e tan d e ln cos cos cos e cos d e cos sin C e tan C sec 5 e 5 Integrating factor: e 5 d e 5 e 5 e 0 d 0 e0 C 0 e5 Ce 5

58 Review Eercises for Chapter Bernoulli equation b n, let z, Integrating factor: e a d e a ln a a b a d b a a C Linear equation u e d e b Ca a z e e d e e e C a z z z. Ce Ce 59.Bernoulli equation 58. Bernoulli equation n, let z, u e d e z e e d e e C Ce z z C e Ce z. Linear equation n, let z, u e d e ln z C z z z. d C Linear equation 60. Bernoulli Equation n, let z, z z z. Linear equation u e d z d ln ln C C ln C 6. Answers will var. Sample Answer: d 0 Solution: Let v, dv v d. v d v dv v d 0 v d v dv 0 v d v dv d v v dv ln ln v C C v C C

59 60 Chapter 6 Differential Equations 6. Answers will var. 6. Answers will var. Sample Answer: 6. Answers will var. u e d d ln C Problem Solving for Chapter 6. (a).0 dt k dt.0 dt kt C t C t C 0.0 C 0.0t C 0.0t 00 Hence, kt C 0 C kt 0 0 C C 0 C 0 kt. 0 : C00 C For t 0 k,. Hence, 0.0t 00. For T 00, lim t T.. Since k 0, dt 0 k dt ln 0 kt C Ce kt 0. When t 0, 7. Therefore, C 5. When t, 8. Therefore, 8 5ek 0, and k ln 7 5 7,. Thus, 5e ln7t 0. When t 5, 5e 5 ln ek 8

60 Problem Solving for Chapter (a) (c) ds dt k SL S S ds dt L Cekt Cke kt L 00. Also, S 0 when t 0 C 9. And, S 0 when t k ln 9. Particular Solution: 5 L kt is a solution because Ce LC kekt Ce kt k k L L L C Le kt Ce kt Ce kt L Ce L L kt kt Ce k SL S, where k k L. s 00 9e ln9t 00 9e 0.809t (d) ds dt k S00 S d S dt k S ds ds 00 S dt dt Choosing S 50, we have: S k 00 S ds dt 0 when S 50 or ds 0. dt 50 ln9 ln9 t 9e ln9t t.7 months (This is the inflection point.) 00 9e ln9t t (e) Sales will decrease toward the line S L.. (a) dt k ln L k dt ln L ln ln ln L ln kt C ln L Cekt (d) e C e C 0 C ln L ecekt Le Cekt (c) As t, L, the carring capacit. dt k ln L d dt k ln L k dt Hence, k dt ln L k ln L ln L d dt 0 when L L dt ln L L e L e. L and t.7. e e The graph is concave upward on 0,.7 and downward on.7,.

61 60 Chapter 6 Differential Equations 5. Let tanh u e e u e u e u e u e u e kt ln bk e ln b e kt be kt Finall, u k ln b t k. L tanh k ln b t k L tanh u L be kt L be kt. The graph of the logistics function is just a shift of the graph of the hperbolic tangent, as shown in Section k g 6 6 cross section 8h 6h dh dt When 6 5 h5 h t C h 6h 70 t C 5 h 6, t 0 and Equation of tank 6 6 h h dh dt 6h h h dh dt 8 h Area of The tank is completel drained when h 0 t 8.5 seconds minutes and seconds. Ah h h Ah dh dt kgh 6 ft C (a) h Ct C 8 C at t 0, h 8 Hence, h Ct 6. At t , h : 800 C 6 6 C Hence, h t 6. h 0 t seconds ( hr, min, 9 sec) t 600 sec h ft Ah dh dt kgh r dh dt k6h h dh 8k dt C dt, r r h 7. feet h C 8k r + ( 6) = 6 h

62 Problem Solving for Chapter Ah dh dt kgh 6 dh dt 6 8h h dh 88 dt h t 88 C h 0: 0 C 5 h t 5 88 h 0 t sec 9. Let the radio receiver be located at 0, 0. The tangent line to joins, and 0, 0. Transmitter (, ) = Radio 0 (a) If, is the point of tangenc on, then (c) Then ± Now let the transmitter be located at, h. h 0 Then, 0 h h h h 0 0 h ± h h h h h h h h h h h h h h h h 0. h h There is a vertical asmptote at h, which is the height of the mountain.

63 606 Chapter 6 Differential Equations 0. ds s dt (a) ds s dt 0.09 ln s t C 00 ln s 0.09t C s C e 0.09t 0.09s.5 C e 0.09t s 8. Ce 0.09t. (a) dc C R V dt ln C R V t K C Ke RtV Since C C 0 when t 0, it follows that K C 0 and the function is C C 0 e RtV. Finall, as t, we have lim C lim C 0e RtV 0. t t (c) As t, Ce 0.09t 0, and s 8... From Eercise 9, we have C C 0 e RtV. (a) For V, R 0.5, and C 0 0.6, we have C 0.6e 0.5t (a) Q RC dc V dt R ln Q RC t V K Q RC e RtVK C R Q ertvk 0 0 For V, R.5, and C 0 0.6, we have C 0.6e 0.75t. 0.8 R Q KeRtV Since C 0 when t 0, it follows that K Q and we have C Q R ertv. As t, the limit of C is QR. 0 0