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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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5. P = 2
7. p = 3
^ (-1)nx2n+1
9. > ----------------, p = oo
n—0 (2n + 1)!
11. 1 + (x — 1), p = ζ
13. ?(-1)n+1 ,
n
n= 1
ζ
15^ xn, p = 1
n=0
p = 1
2. p = 2
4. p = 2
6. p = 1
8. p = e ζ xn
10. ? —, p = ζ
n=0
n!
12. 1 — 2(x + 1) + (x + 1)2, p = ζ
ζ
14. ^(-1)nxn, p = 1
n=0
ζ
16. (-1)n+1(x - 2)n, p = 1
n=0
17. σ = 1 + 22x-
+ (n + 1)2 xn + •
σ" = 22 + 32 • 2x + 42 • 3x2 + 52 • 4X3 + ••• + (n + 2)2(n + 1)xn + ••• 18,19,23,25,28 18. / = a1 + 2a2x + 3a3x2 + 4a4x3 +-+ (n + 1)an+1 xn +-
ζζ
= E nanxn—1 ^ (n + 1)an+1 xn
n= 1
n=0
Σ = 2a2 + 6a3 x + 12a4 x2 + 20a5 x3 + ••• + (n + 2)(n + 1)an+2 xn + •••
ζζ
= XI n(n - 1)anxn-2 = X(n + 2)(n + 1)an+2xn
21. Y(n + 2)(n + 1)an+2 xn 22
n=0
23. ζ 24
J2(n + 1)anxn
n=0
25. ζ 26
J2 [(n + 2)(n + 1)an+2 + nan]xn
n=0
27. ζ 28
J2[(n + 1)nan+1 + an]xn
an-U
n= 1
28. an = (—2)na0/n!, n = 1, 2,...; a0e 2x
n=0
a
2
696
Answers to Problems
See SSM for detailed solutions to 2
3
Section 5.2, page 247
1. an+2 — an/(n + 2)(n + 1) x2 x4 x6
y1 (x) — 1 + — + — + — + '
2!
x3
4!
x5
6!
—
x2n
= coshx
y2(x) — x +--------1---------1----+ •
J2 3! 5! 7!
2. an+2 — an/(n + 2)
x2 x4 x6
σ (x) — 1 +--------+---------+----------
λ 2 2 • 4 2 • 4 • 6
x3 x5 x7
y2(x) — x +--------I---------I----------
J2 3 3-5 3-5-7
n—0 (2n)!
ζ x2 n+1
— V--------------- — sinh x
n—0 (2n+1)!
+ •

—
n—0
ζ
•—
n—0
x2 n
2"ni
2nn!x2 n|1 (2n + 1)!
3. (n + 2)an+2 — an+1 — an — 0
y1(x) — 1 + 2 (x — 1)2 + 6 (x — 1)3 + 1 (x — 1)4 + ••• y2(x) — (x — 1) + 2 (x — 1)2 + 2 (x — 1)3 + 1 (x — 1)4 +
4. an+4 — -k2an/(n + 4)(n + 3);
k2x4 k4x8
y1 (x) — 1 — — +
a2 — a3 — 0 k6 x12
3 • 4 3 • 4 • 7 • 8
ζ ΄ λ\Ψ-
— 1 + σ---------^
3478
3 4 7 8 11 12
+ •
(— 1)m+1(k2 x4)Ψ|1
Ψ—0
3 • 4 • 7 • 8 • • • (4m + 3)(4m + 4)
k2 x5
y2(x) — x — +
k4 x9
k6 x13
4.5 4-5-8-9 4-5-8-9-12-13
+ •
(—1)m|1(k2 x4)m|1
m—0
4 • 5 • 8 • 9 • • • (4m + 4)(4m + 5)
Hint: Let n — 4m in the recurrence relation, m — 1, 2, 3,... .
5. (n + 2)(n + 1)an+2 — n(n + 1)anl1 + an — 0, n > 1;
n+1
aΞ ------- ξ at
y1(x) — 1 — 2x2 — 1X3 — 24x4 + ••• > y2(x) — x — 6x3 — 12x4 — 24x5 + ¦
2 — 2 °0
¦1 x4 12
6. an+2 — -(n2 — 2n + 4)an/[2(n + 1)(n + 2)], n > 2;
2
0
y1(x) — 1 — x2 + 1 x4 — η! x6 + ••• ,
7. an+2 — —an/(n + 1), n — 0, 1, 2,...
y2(x) — x — 4x3 + ^x5 — msx7 + ¦
Σ³ (x) — 1 — — +-------—---------
λ 1 1 • 3 1 • 3 • 5
x3 x5 x7
y\ (x) — x — — +---------—-------
J2 2 2 ¦ 4 ^^6
+ •
= ^ (—1)nx2 n
f-T 1 • 3 • 5 ••• (2n — 1)
n—1
(—1)nx2 n|1
+-----— x +> —----------------------
n—12 • 4 • 6 •••(2n)
8. an+2 — -[(n + 1)2an|1 + an + an—1]/(n + 1)(n + 2), n — 1, 2,...
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