Lecture Notes in Differential Equations - Bruce E. Shapiro

260 LESSON 28. SERIES SOLUTIONS
where n is an integer (n = 2in this case). Expanding as usual we obtain
∞∑
y = c k t k (28.44)
k=0
∞∑
y ′ = kc k t k−1 (28.45)
k=0
∞∑
y ′′ = k(k − 1)c k t k−2 (28.46)
k=0
Substituting into the differential equation,
∞∑
∞∑
∞∑
0 = (1 − t 2 ) c k k(k − 1)t k−2 − 2t c k kt k−1 + 6 c k t k (28.47)
k=0
k=0
∞∑
∞∑
∞∑
∞∑
= c k k(k − 1)t k−2 − c k k(k − 1)t k − 2t c k kt k−1 + 6 c k kt k
k=0
k=0
k=0
k=0
(28.48)
∞∑
∞∑
∞∑
= c k k(k − 1)t k−2 − 2t c k kt k−1 + (6 − k 2 + k)c k t k (28.49)
k=0
k=0
k=0
k=0
The first term can be rewritten as
∞∑
∞∑
c k k(k − 1)t k−2 = c k k(k − 1)t k−2 (28.50)
k=0
k=2
∞∑
= c m+2 (m + 1)(m + 2)t m (28.51)
m=0
∞∑
= c k+2 (k + 1)(k + 2)t k (28.52)
k=0
and the middle term in (28.49) can be written as
∞∑
∞∑
−2t c k kt k−1 = (−2)kc k t k (28.53)
k=0
k=0