Elementary Differential Equations and Boundary Value Problems  Boyce W.E.
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7
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detailed solutions to 11. 7 = 0 and 7 = 4 are equilibrium solutions; 7 ^ 4 if initial value is positive; 7 diverges
2, 4, 7, 11, 13, from 0 if initial value is negative.
15ab. 12. 7 = 0 and 7 = 5 are equilibrium solutions; 7 diverges from 5 if initial value is greater than
5; 7 ^ 0 if initial value is less than 5.
13. 7 = 0 is equilibrium solution; 7 ^ 0 if initial value is negative; 7 diverges from 0 if initial
value is positive.
14. 7 = 0 and 7 = 2 are equilibrium solutions; 7 diverges from 0 if initial value is negative;
7 ^ 2 if initial value is between 0 and 2; 7 diverges from 2 if initial value is greater than
15. 2.
(a) dq/dt = 300(102  q 106) (b) q ^ 104 g; no
16,21,22 16. dV/dt = kV2/3 for some k > 0.
17. (a) dq/dt = 500  0.4q (b) q ^ 1250 mg
18. (a) mv' = mg  kv2 (b) v ^ Jmg/ k (c) k = 2/49
19. 7 is asymptotic to t  3as t ^ æ 20. 7 ^ 0as t ^ æ
21. 7 ^ æ, 0, or æ depending on the initial value of 7
22. 7 ^ æ or æ depending on the initial value of 7
23. 7 ^ æ or æ or 7 oscillates depending on the initial value of 7
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Answers to Problems
See SSM for 24. y — —æ or is asymptotic to V2t — 1 depending on the initial value of y
detailed solution 25. y — 0 and then fails to exist after some tf > 0
for 24 26. y — æ or —æ depending on the initial value of y
Section 1.2, page 14
1b 1. (a) y = 5 + (y0 — 5)e—t (b) y = (5/2) + [yQ — (5/2)]e—2t
(c) y = 5 + (y, — 5)e—2t
Equilibrium solution is y = 5 in (a) and (c), y = 5/2 in (b); solution approaches equilibrium faster in (b) and (c) than in (a).
2. (a) y = 5 + (y, — 5)e( (b) y = (5/2) + [y, — (5/2)]e2t
(c) y = 5 + (Óî — 5)e2t
Equilibrium solution is y = 5 in (a) and (c), y = 5/2 in (b); solution diverges from equi
librium faster in (b) and (c) than in (a).
3. (a) y = ce—at + (b/a)
(c) (i) Equilibrium is lower and is approached more rapidly. (ii) Equilibrium is higher.
(iii) Equilibrium remains the same and is approached more rapidly.
4ab, 6b, 4. (a) Ó³ (t) = ceat (b) Ó = ceat + (b/a)
8ab,10a, 13abc 5. y =?e a + (b/a)
6. (a) T = 2ln18 = 5.78 months (b) T = 2ln[900/(900 — p0)] months
(c) p0 = 900(1 — e—6) = 897.8
7. (a) r = (ln2)/30 day—1 (b) r = (ln2)/N day—1
8. (a) T = 5ln50 = 19.56 sec (b) 718.34 m