Lecture Notes in Differential Equations - Bruce E. Shapiro

264 LESSON 28. SERIES SOLUTIONS
Since p(t) and q(t) are analytic then they also have power series expansions
which we will assume are given by
∞∑
p(t) = p j t j (28.76)
q(t) =
j=0
∞∑
q j t j (28.77)
j=0
with some radius of convergence R. Substituting (28.71), (28.74) and (28.76)
into (28.69)
⎛ ⎞ (
∞∑
∞∑
∞
)
∑
0 = (k + 2)(k + 1)c k+2 t k + ⎝ p j t j ⎠ (k + 1)c k+1 t k
k=0
⎛ ⎞ (
∞∑
∞
)
∑
+ ⎝ q j t j ⎠ c k t k
j=0
k=0
j=0
k=0
(28.78)
Since the power series converge we can multiply them out term by term:
∞∑
∞∑ ∞∑
0 = (k + 2)(k + 1)c k+2 t k + p j t j (k + 1)c k+1 t k
k=0
+
∞∑
j=0 k=0
We next use the two identities,
and
k=0 j=0
j=0 k=0
∞∑
q j t j c k t k (28.79)
∞∑ ∞∑
∞∑
(k + 1)c k+1 p j t j+k =
k=0 j=0
k=0 j=0
∞∑ ∞∑
∞∑
q j c k t k+j =
k∑
(j + 1)c j+1 p k−j t k (28.80)
k=0 j=0
k∑
q k−j c j t k (28.81)
Substituting (28.80) and (28.81) into the previous result
∞∑
0 = (k + 2)(k + 1)c k+2 t k (28.82)
k=0
Since the t k are linearly independent,
(k + 2)(k + 1)c k+2 = −
k∑
[(j + 1)c j+1 p k−j + c j q k−j ] (28.83)
j=0