Lecture Notes in Differential Equations - Bruce E. Shapiro

261
Hence
∞∑
∞∑
∞∑
0 = c k+2 (k + 1)(k + 2)t k + (−2)kc k t k + (6 − k 2 + k)c k t k
k=0
k=0
k=0
(28.54)
∞∑
∞∑
= c k+2 (k + 1)(k + 2)t k + (k 2 + k − 6)c k t k (28.55)
=
k=0
k=0
∞∑
[c k+2 (k + 1)(k + 2) − (k + 3)(k − 2)c k ] t k (28.56)
k=1
By linear independence
c k+2 =
(k + 3)(k − 2)
(k + 2)(k + 1) c k, k = 0, 1, ... (28.57)
or
c k =
(k + 1)(k − 4)
c k−2 , k = 2, 3, ... (28.58)
k(k − 1)
From the initial conditions we know that c 0 = 1 and c 1 = 0. Since c 3 , c 5 ,
... are all proportional to c 1 , we conclude that the odd-indexed coefficients
are all zero.
Starting with k = 2 the even-indexed coefficients are
c 2 =
⎫
(2 + 1)(2 − 4)
c 0 = −3c 0 = −3
2(2 − 1)
⎪⎬
(4 + 1)(4 − 4)
c 4 = c 2 = 0
4(4 − 1)
⎪⎭
c 6 = c 8 = c 10 = · · · = 0
(28.59)
Hence
y = c 0 + c 1 t + c 2 t 2 + c 3 t 3 + · · · = 1 − 3t 2 (28.60)
which demonstrates that sometimes the infinite series terminates after a
finite number of terms.