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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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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
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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