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Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Download (direct link): elementarydifferentialequations2001.pdf
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3
=
X
0.
2
-
7
=
X7
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(10-2 - q 10-6) (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 ^ 20. 7 ^ 0as t ^
21. 7 ^ , 0, or - depending on the initial value of 7
22. 7 ^ or - depending on the initial value of 7
23. 7 ^ or - or 7 oscillates depending on the initial value of 7
679
680
Answers to Problems
See SSM for 24. y 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 or depending on the initial value of y
Section 1.2, page 14
1b 1. (a) y = 5 + (y0 5)et (b) y = (5/2) + [yQ (5/2)]e2t
(c) y = 5 + (y, 5)e2t
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 + ( 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 = ceat + (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) ӳ (t) = ceat (b) = 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 e6) = 897.8
7. (a) r = (ln2)/30 day1 (b) r = (ln2)/N day1
8. (a) T = 5ln50 = 19.56 sec (b) 718.34 m
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