Lecture Notes in Differential Equations - Bruce E. Shapiro

288 LESSON 30. THE METHOD OF FROBENIUS
Canceling the common factor of √ t,
0 =
=
∞∑
k=0
∞∑
k=0
k=0
(
a k k 2 − 1 )
t k +
4
k=0
∞∑
k=0
(
a k k + 1 )
t k
2
∞∑
+ a k t k+2 − 1 ∞∑
a k t k (30.50)
4
k=0
(
a k k 2 − 1 4 + k + 1 2 − 1 )
+
4
∞∑
a k t k+2 (30.51)
k=0
∞∑
∞∑
= a k (k 2 + k) + a k t k+2 (30.52)
k=0
Letting j = k + 2 in the second sum,
∞∑
∞∑
0 = a k (k 2 + k)t k + a j−2 t j (30.53)
k=0
j=2
Since the k = 0 term is zero and the k = 1 term is 2a 1 t in the first sum,
∞∑ [
0 = 2a 1 t + aj (j 2 ]
+ j) + a j−2 t
j
j=2
(30.54)
By linear independence,
a 1 = 0 (30.55)
a j = −a j−2
j(j + 1) , j ≥ 2 (30.56)
Since a 1 = 0, all subsequent odd-numbered coefficients are zero (this follows
from the second equation). Furthermore,
a 2 = −a 0
3 · 2 , a 4 = −a 2
5 · 4 = a 0
5! , a 6 = −a 4
7 · 6 = −a 0
7! , · · · (30.57)