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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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2. (b, c) (0, 0); u' = u,v = — o v; r = 1, — 4; saddle point, unstable
(1, 2); u' = —4v, v' = u; r = ±1 i; center or spiral point, indeterminate
3. (b, c) (0, 0); u' = u,v = — 4 v; r = 1, — 1; saddle point, unstable
(2, 0); u' = — u — v,v'= 3 v; r =— 1, f; saddle point, unstable
(2, f); u' = — 4u — 1 v, v = 3 u; r = (—1 ± ė/Či)/8; spiral point,
asymptotically stable
4. (b, c) (0, 0); u' = 9 u,v' = —v; r = |, —1; saddle point, unstable
(8, 0); u' = — 8u — 16v,v' = 1 v; r = —|, 8; saddle point, unstable
(1, 4); u' = —u — 2v, v = 4u; r = (—1 ± V0.5)/2; node, asymptotically stable
Answers to Problems
723
See SSM for detailed solutions to 7abc, 11,13
1, 2, 4, 7, 8a
8b, 9, 11, 13, 16a
5. (b, c) (0, 0); u! = — u, v = — 2 v; r = —1, — 3; node, asymptotically stable
(1, 0); u! = 4 u — Žv, v = —v; r =— 1, 3; saddle point, unstable
(2, 0); il = — 3u — 5v,v'= 2v; r =— 3, 2; saddle point, unstable
(3, 3); vl = — 3 u — Žv, v1 = 5 u; r = (—3 ± \/39 i)/8; spiral point,
asymptotically stable
6. t = 0, T, 2T,... : His a max., dP/dt is a max.
t = T/4, 5T/4,... : dH/dt is a min., P is a max.
t = T/2, 3T/2,... : H is a min., dP/dt is a min.
t = 3T/4, 7T/4,... : dH/dt is a max., P is a min.
7. (a) Jcą/Ja y (b) Jb
(d) The ratio of prey amplitude to predator amplitude increases very slowly as the initial point moves away from the equilibrium point.
8. (a) 4n/\/3 = 7.2552
(c) The period increases slowly as the initial point moves away from the equilibrium point.
9. (a) T = 6.5 (b) T = 3.7, T = 11.5 (c) T = 3.8, T = 11.1
11. Trap foxes in half-cycle when dP/dt > 0, trap rabbits in half-cycle when dH/dt > 0, trap rabbits and foxes in quarter-cycle when dP/dt > 0 and dH/dt > 0, trap neither in quarter-cycle when dP/dt < 0 and dH/dt < 0.
12. dH/dt = aH — ą HP — ā H, dP/dt = —cP + y HP — & P, where ā and & are constants of proportionality. New center is H = (c + &)/y > c/y and P = (a — ā)/ą < a/ą, so equilibrium value of prey is increased and equilibrium value of predator is decreased!
13. Let A = a/ą — c/y > 0. The critical points are (0, 0), (a/a, 0), and (c/y, a A/ą), where (0, 0) is a saddle point, (a/a, 0) is a saddle point, and (c/y, a A/ą) is an asymptotically stable node if (ca/Y)2 — 4ca A > 0 or an asymptotically stable spiral point if (ca/Y)2 — 4ca A < 0. (H, P) ^ (c/y, a A/ą) as t ^ ę.
Section 9.7, page 530
1. r = 1,9 = t + t0, stable limit cycle
2. r = 1,9 = —t + t0, semistable limit cycle
3. r = 1,9 = t + t0, stable limit cycle; r = 3, 9 = t + t0, unstable periodic solution
4. r = 1, 9 = —t + t0, unstable periodic solution; r = 2, 9 = —t + t0, stable limit cycle
5. r = 2n — 1,9 = t + t0, n = 1, 2, 3,..., stable limit cycle;
r = 2n, 9 = t + t0, n = 1, 2, 3,..., unstable periodic solution
6. r = 2, 9 = —t + t0, semistable limit cycle;
r = 3, 9 = —t + t0, unstable periodic solution
8. (a) Counterclockwise
(b) r = 1, 9 = t + f0, stable limit cycle; r = 2, 9 = t + f0, semistable limit cycle; r = 3, 9 = t + t0, unstable periodic solution
9. r = v/0, 9 = — t + t0, unstable periodic solution
14. (a) /ė = 0.2, T = 6.29; ä = 1, T = 6.66; ä = 5, T = 11.60
15. (a) x ^ y = —x + äó - ėó3/3
(b) 0 < ä < 2, unstable spiral point; ä > 2, unstable node
(c) A = 2.16, T = 6.65
(d) ä = 0.2, A = 1.99, T = 6.31; ä = 0.5, A = 2.03, T = 6.39;
ä = 2, A = 2.60, T = 7.65; ä = 5, A = 4.36, T = 11.60
16. (a) k = 0, (1.1994, -0.62426); k = 0.5, (0.80485, -0.13106)
(b) k0 = 0.3465 (c) T = 12.54 (d) k1 = 0.3369
Section 9.8, page 538
1ab
1. (b) X = ė1, ?(1) = (0, 0, 1)T; X = A.2, g(2) = (20, 9 - V81 + 40r, 0)T;
X = X3, ?(3) = (20, 9 + J81 + 40r, 0)T
724
See SSM for detailed solutions to 1c, 2abc, 3b
3c, 4, 5ab
CHAPTER 10
2, 3, 7, 11, 15
3, 5, 7, 10
13ab, 15ab
15a, 21 abed, 25
Answers to Problems
(c) X1 = -2.6667, ?(1) = (0, 0, 1)T; X2 = -22.8277, g(2) = (20, -25.6554, 0)T; X3 = 11.8277, g(3) = (20, 43.6554, 0)T
2. (c) X1 = -13.8546; X2,X3 = 0.0939556 ± 10.1945Ó
5. (a) dV/dt = -2a[rx2 + y2 + b(z - r)2 - br2]
Section 10.1, page 547
2. y = (cot \fln cosV2x + sin v/2x)/v/2
1. y =- sin x
3. y = 0 for all ³; y = c2 sin x if sin L = 0.
4. y = - tan L cos x + sin x if cos L = 0; no solution if cos L = 0.
5. No solution.
6. y = (-n sinV^x + xsinv/2n)/2 sinv/2n
7. No solution. 8. y = c2sin2x + 1 sin x
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