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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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24. (b) ำ = Jgg U(2k—i)n(t)e—[t—(2k—1)n]/2ฐsm{V399[t — (2k — 1)n]/20}
Section 6.6, page 335
3. sin t * sin t = 2 (sin t — t cos t) is negative when t = 2n, for example.
4. F (s) = 2/s2(s2 + 4) 5. F (s) = 1/(s + 1)(s2 + 1)
6. F (s) = 1/s2(s — 1) 7. F (s) = s/(s2 + l)2
8. f(t) = ff (t — r)3sint dr 9. f(t) = ? e—(t—T) cos2r dr
10. f( t) = f f (t — r)e—(t—r) sin2r dr 11. f( t) = f sin(t — r)g(r) dr
2 0 s s 0
1 - ft ด t (t )
12. ๓ =— sin mt +-------- sin m(t — r)g(r) d r 13. ๓ = e—(x) sin(t — r) sin ar dr
6) 6) -J0 -J0
14. ๓ = 8 f e-(t—x)/2 sin2(t — r)g(r) dr
15. ๓ = e—t/2cos t — 1 e—s/2sin t + ? e—(t—x)/2sin(t — r)[1 — un(r)] dr
16. ๓ = 2e—2t + te—2t + (t — r)e—2(t—T)g(r) dr
17. ๓ = 2e— — e—2t + j? [e
—x) — e—2(t—x)] cos ar dr 18. ๓ = ijj [sinh(t — r) — sin(f — r)]g(r) dr
19. ๓ = 4 cos f — - cos2t + - [2 sin(t — r) — sin2(t — r)]g(r) dr
F(s)
20. ฎ(s) = ------V—
1 + K (s)
21. (c) ๔(ด) = -(4sin2t — 2sin t)
(d) u(t) = f (2 sin t — sin2t)
CHAPTER 7 Section 7.1, page 344
2, 4, 5
-. x1 = X2,
x2 = —2x, — 0.5x2
2. xf = x2, x2 = —2x, — 0.5x2 + 3 sin t
3. xf = x2, x2 = —(1 — 0.25t—2)x, — t—1 x2
4. xf =
=
x3,
x3 =
x4 = x,
5. xf = X2, x2 = —q(t)x, — p(t)x2 + g(t); x,(0) = u0, x2(0) = u0
6. ำฏ = ำ2, ำ2 = -(k1 + k2) y,/ml + k2 ำ๚/ m 1 + Fl(t)/ml,
ำ = ำ4, ำ4 = k2ำณ /m2 — (k2 + k3)y3/m2 + F2(t)/m2
2
Answers to Problems
707
See SSM for detailed solutions to 8, 9
12, 14, 19
21abc
— t — 3t
x2 = ั?1 e - c2e
7. (a) x1 = c1e—t + c2e—
(b) c1 = 5/2, c2 = —1/2 in solution in (a)
(c) Graph approaches origin in the first quadrant tangent to the line x1 = x2.
8. x,
= 11 e2t 2 e— t
Graph is asymptotic to the line x1 = 2x2 in the first quadrant.
x - _3et/2 _ 1 e2t ห1 = 2 2 ,
Graph is asymptotic to the line x1 = x2 in the third quadrant.
10. x1 = - 7e-1 + 6e-2t
x2 = - 7e t + 9e 2t
Graph approaches the origin in the third quadrant tangent to the line x1 = x2.
11. x1 = 3cos2f + 4sin2f, x2 = - 3sin2f + 4cos2f
Graph is a circle, center at origin, radius 5, traversed clockwise.
12. x1 = -2e-t/2cos2t + 2e-t/2sin2f, x2 = 2e- t/2cos2t + 2e-t/2sin2f Graph is a clockwise spiral, approaching the origin.
13. LRCI" + LI'+ RI = 0
21. (a) Q1 = ง - 1 Q1 + 40 Q2, Q1 (0) = 25
Q2 = 3 + 110 Q1 - 5 Q2, Q2(0) = 15
(b) QE = 42, Q2 = 36
(c) x1 = -110x1 + 40x2’ x1(0) = -17
x2 = 110 x1 - 1 x2 > x2(0) = -21
22. (a) Q1 = 3q1 - 11! Q1 + 110 Q2, Q1 (0) = Q0
Q2 = q2 + 30 Q1 - 100 Q2, Q2(0) = Q2
(b) QE = 6(9q1 + q2), Q2e = 20(3q1 + 2q2)
(c) No
(d) 10 < Q22/ Qf < 20
Section 7.2, page 355
1ac
/6 -6 3>
1. (a) 5 9 -2
\2 3 8/
-15 6 -12
(b) 7 -18 -1
26 3 5
(c) 4
12
89
(d) 14
, , /-3 + 5i 7 + 5 A
(cH 2 + i 7 + 2ij
(d)
1ร
12 5
5 -8 5/
3 + 4i 6i
11 + 6i 6 - 5i
8 + 7i 4 - 4i
2
1
1
3. (a4 1 0 -1 (b4 2 -1
23
1
31
4. (a)
3 2i
2i
10
0
4
๗1 + i -2 + 3i
6 -4\
4 10
4 6
(b)
6 4i
4
3 -2>
(c), (d) f 3
4
-1
4
3 + 2i
1i
2 + i -2 - 3i
(c)
'3 + 2i 2 + i
ห- i -2- 3i
708
Answers to Problems
7 -11 -3 5 0 -1
See SSM for 6. (a) 11 20 17 (b) 2 7 4
detailed solutions -4 3 -12 -1 1 4
to 6, 10 8. (a) 4i (b) 12 - 8i (c) 2 + 2i (d) 16
12
14
22, 25
10.
3 4 \
11 11 1
2 J_
11 11/
-3 2
-3 3 -1
2 1 0
14. Singular
16.
18.
/ 1 3 ฑ \
10 10 10
2 4 2
10 10 10
7 1 3
- 10 10 10/
/ 1 1 0 1 \
1 0 1 1
1 1 1 1
V 0 1 0 1 /
7 et 5e- t 10e2t
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