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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.
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(a) et 7e- t t2
e
2
\ 8et 0 --- e2t
(c)
6 -8 -1Ï
9 15 6
-5 -1 51
11.
13.
15.
12 1
2 4
1 3 1 3
\~3
-
0
0
17. Singular /
19.
6 5
0 -2
1 0
3 ³
1
1 1
- 3 3
0 1
3/
1
4 8
³ ³
1 1
2 - 4
0 \)
13 8
5 5
11 6
5 5
³ ³
1 5
- 5
4 4
5 5
(b)
(c)
' 2e2t - 2 + 3e^
4elt - 1 - Çå31 ,-2e2i - 3 + 6e3t -1 + 6e-2t - 2et -3e3t + 3et - 2e4
1 + 4e-2t - et
2 + 2e-2t + et
3e3t + 2et - å4Ë 6e3t + ^ + e4t
- 5 /
t
—2e-
e
2et -e-t
\ — et —3e-t
2e2 -2e2 4e2
(d) (e - 1)
j
2e-
e-
3e~
2(e+ 1)\ - 2(e+ 1)
e + 1 /
Section 7.3, page 366
1
2, 3 6, 8
14
15
1 x - -1
i. x1 — 3,
3
4
2 — 3 ’ ë3 — 3
X —-c, x2 — c + 1, x3 — c, where c is arbitrary x2 — - c, x3 — -c, where c is arbitrary
1
x1 — c,
x3 — 0
5. x1 — 0, x2 — 0,
7. x11) - 5x(2) + 2x(3) — 0 9. Linearly independent
12. 3x(1)(f) - 6x(2) (t) + x x(1^ t1
-1"—G1,
15. x1 — 2, x(1) —
(3)(t) — 0 Ë9 — 4, x(2) —
16. Ë1 — 1 + 2i, x(1) —
17. Ë1 — -3, x(1) —
6. Linearly independent 8. 2x(1) - 3x(2) + 4x(3) - x(4) — 0 10. x(1) + x(2) - x(4) — 0 13. Linearly independent
1
1 + i
k7 — 1 - 2i, x(2) —
k2 — -1, x(2) —
709
See SSM for detailed solutions to 18, 21
24
27
1, 2a
2bc, 6abcd
1
5
6, 7, 9
18. k1 = 0, x(1) = I1. I ; k2 = 2, x(2) = ^-1.
; k2 = —2, x(2) =
-V3
20. k = -1/2, x(1) = (130
k2 = -3/2, x(2) =
21. k = 1, x(1) = -3
k2 = 1 + 2i, x(2) = | 1
k3 = 1 - 2i, x(3) = 1
22. k1 = 1, x(1) =
k 2, x(2)
k3 = 3, x(3) =
23. k1 = 1, x(1) = -2 ; k2 = 2, x(2) = Ml ; k3 = -1, x(3) =
24. k1 = -1, x(1) = I -4
k2 = -1, x(2) =
k3 = 8, x(3) = 1
(d) x'=l 0,-2 0,-1 )x
Section 7.4, page 371
2. (c) W (t) = ñ exp J [ p11 (t) + p22(t)] dt
6. (a) W(t) = t2
(b) x(1) and x(2) are linearly independent at each point except t = 0; they are linearly
independent on every interval.
(c) At least one coefficient must be discontinuous at t = 0.
0 1
y—2t-2 2t-
7. (a) W(t) = t(t - 2)e'
(b) x(1) and x(2) are linearly independent at each point except t = 0 and t = 2; they are linearly independent on every interval.
(c) There must be at least one discontinuous coefficient at t = 0 and t = 2.
/0 1 \
(d) x' = I 2 — 2t t2 — 2 I x
Kt2 - 2t t2 - 2t)
Section 7.5, page 381
L x = ñ1 Q e-t + c2^j e2t
3. x = c1 (Ó) ª e—
2. x = c1{1) e— + ^(T) e-2t
c1 14
4. x = cl : I e-3t + c2 ( 1 ) e2t
5. x = c1 (—1) e 3t + c^1) e (
7. x = c1 (3) + c2 fye~2‘
9. x = cl Ë + c2( M e2t
6. x = c1 (-1) eU1 +c2 (j) e2t
8. x = c1 (-2) + c2 (-1) e 10.x=+c^_0e-it
2
2
2
710
See SSM for detailed solutions to 16
20, 25
31c
1
11
1
4
5 -7
1
4
1
11. x = c1 I 1 I e + c2 I - 2 I e + c3 | 0 I e
12. x = c1 I -4 I e + c2 ² 0 I e + c3 I 1 I e
13. x = c1 I -5 I e 2t + c2 I -4 I e t + c3 | 1 I e
14. x = c1 I -4 I et + c2 I -1 I e 2t + c3 I 2 I eit
1
Ä
3
4 -2
1 1 1
16. x
1 Ë\ „-t , 1 Ë\ e3t
21
e- +
25
17. x = I -2 j et + 2 È | e2t
18. x = 6² 2 j et + 3² -2 j e-t -
1 e4
20. x = c1 11 t + c2 13 t-1
21. x = c1 (3) f2 + c2
4
22.x=c^y+<öót-2
29. (a) = x2, x2 = —(c/a)x1 - (b/a)x2
30. (a)(2) e-»20 + 29 (-2) e--/4
(c) T = 74.39
31. (a) x = cY-^ e(-2+V2)t/2 + ^^ e(-2-V2)t/2.
23. x = c^2 t-1 + cJ1 t2
+ c2 1 e
r1 2 = (-2 ± v)2)/2; node
(b) x = c^-1) º(-1+^ + c2(e(-1-^'. r12 = _1 ±^2; saddle point
(c) r1 2 = -1 ± V®; à = 1
32. (a) (0=c^3)e-2t+^(0e-t 33.(a) (cR - L
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