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

1178 Chapter 17 Second-Order Differential Equations
Putting these values back into Equation 2, we write the solution as
y − c 0 1 c 1 x 1 c 2 x 2 1 c 3 x 3 1 c 4 x 4 1 c 5 x 5 1 ∙ ∙ ∙
− c 0S1 2 x 2
2! 1 x 4
4! 2 x 6
1 c 1Sx 2 x 3
− c 0 ò s21d n
n−0
6! 1 ∙ ∙ ∙ 1 s21dn
3! 1 x 5
5! 2 x 7
x 2n
s2nd! 1 c 1 ò s21d n
n−0
7! 1 ∙ ∙ ∙ 1 s21dn
x 2n11
s2n 1 1d!
x
D
2n
s2nd! 1 ∙ ∙ ∙
x
D
2n11
s2n 1 1d! 1 ∙ ∙ ∙
Notice that there are two arbitrary constants, c 0 and c 1 .
■
Note 1 We recognize the series obtained in Example 1 as being the Maclaurin series
for cos x and sin x. (See Equations 11.10.16 and 11.10.15.) Therefore we could write the
solution as
ysxd − c 0 cos x 1 c 1 sin x
But we are not usually able to express power series solutions of differential equations in
terms of known functions.
ExamplE 2 Solve y0 2 2xy9 1 y − 0.
SOLUtion We assume there is a solution of the form
y − ò c n x n
n−0
Then
y9 − ò nc n x n21
n−1
and
y0 − ò nsn 2 1dc n x n22 − ò sn 1 2dsn 1 1dc n12 x n
n−2
n−0
as in Example 1. Substituting in the differential equation, we get
ò sn 1 2dsn 1 1dc n12 x n 2 2x ò nc n x n21 1 ò c n x n − 0
n−0
n−1
n−0
ò sn 1 2dsn 1 1dc n12 x n 2 ò 2nc n x n 1 ò c n x n − 0
n−0
n−1
n−0
ò
n−1
2nc n x n − ò 2nc n x n
n−0
ò fsn 1 2dsn 1 1dc n12 2 s2n 2 1dc n gx n − 0
n−0
This equation is true if the coefficients of x n are 0:
7 c n12 −
sn 1 2dsn 1 1dc n12 2 s2n 2 1dc n − 0
2n 2 1
sn 1 1dsn 1 2d c n n − 0, 1, 2, 3, . . .
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