James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 17.4 Series Solutions 1179
We solve this recursion relation by putting n − 0, 1, 2, 3, . . . successively in Equation 7:
Put n − 0: c 2 − 21
1 2 c 0
Put n − 1: c 3 − 1
2 3 c 1
Put n − 2: c 4 − 3
3 4 c 3
2 − 2
1 2 3 4 c 0 − 2 3 4! c 0
Put n − 3: c 5 − 5
4 5 c 3 −
1 5
2 3 4 5 c 1 − 1 5 c 1
5!
Put n − 4: c 6 − 7
5 6 c 4 − 2 3 7
4! 5 6 c 0 − 2 3 7 c 0
6!
Put n − 5: c 7 − 9
6 7 c 5 − 1 5 9
5! 6 7 c 1 − 1 5 9 c 1
7!
Put n − 6: c 8 − 11
7 8 c 6 − 2 3 7 11 c 0
8!
Put n − 7: c 9 − 13
8 9 c 7 − 1 5 9 13 c 1
9!
In general, the even coefficients are given by
and the odd coefficients are given by
3 7 11 ∙ ∙ ∙ s4n 2 5d
c 2n − 2 c 0
s2nd!
The solution is
c 2n11 −
1 5 9 ∙ ∙ ∙ s4n 2 3d
s2n 1 1d!
c 1
or
y − c 0 1 c 1 x 1 c 2 x 2 1 c 3 x 3 1 c 4 x 4 1 ∙ ∙ ∙
− c 0S1 2 1 2! x 2 2 3 4! x 4 2 3 7
6!
1 c 1Sx 1 1 3! x 3 1 1 5
5!
x 6 2 3 7 11
8!
x 5 1 1 5 9
7!
x 8 2 ∙ ∙ ∙D
x 7 1 1 5 9 13
9!
2nD
8 y − c 0S1 2 1 2! x 3 7 ∙ ∙ ∙ s4n 2 5d
2 2 ò
x
n−2 s2nd!
2n11D
1 5 9 ∙ ∙ ∙ s4n 2 3d
1 c 1Sx 1 ò
x
n−1 s2n 1 1d!
x 9 1 ∙ ∙ ∙D
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