Lecture Notes in Differential Equations - Bruce E. Shapiro

292 LESSON 30. THE METHOD OF FROBENIUS
Proof. (of theorem 30.1). Suppose that
is a solution of (30.70), where
y = (t − t 0 ) α S(x) (30.91)
S =
∞∑
c n (t − t 0 ) n . (30.92)
n=0
We will derive formulas for the c k .
For n = 1, we start by differentiating equation (30.38)
0=2(t − t 0 )S ′′ + (t − t 0 ) 2 S ′′′
+ [p(t) + 2α] S ′ + (t − t 0 )p ′ (t)S ′ + (t − t 0 ) [p(t) + 2α] S ′′′
+ [q ′ (t) + αp ′ (t)] S + [q(t) + α(α − 1) + αp(t)] S ′ (30.93)
Everything on the right hand side of this equation is analytic, and hence
continuous. Taking the limit as t → t 0 , we have,
0= [p(t 0 ) + 2α] S ′ (t 0 )
+ [q ′ (t 0 ) + αp ′ (t 0 )] S + [q(t 0 ) + α(α − 1) + αp(t 0 )] S ′ (t 0 ) (30.94)
By Taylor’s Theorem c n = S (n) (t 0 )/n! so that c 0 = S(t 0 ) and c 1 = S ′ (t 0 );
similarly, p(t 0 ) = p 0 , p ′ (t 0 ) = p 1 , q(t 0 ) = q 0 , and q ′ (t 0 ) = q 1 in (30.66), and
therefore
0 = (p 0 + 2α)c 1 + [q 1 + αp 1 ]c 0 + [q 0 + α(α − 1) + αp 0 ]c 1 (30.95)
The third term is zero (this follows from (30.68)) and hence
c 1 = − q 1 + αp 1
p 0 + 2α c 0. (30.96)
For n > 1, we will need to differentiate equation (30.38) n times, to obtain
a recursion relationship between the remaining coefficients, starting with
0 = D (n) [ (t − t 0 ) 2 S ′′] + D (n) {[2α + p(t)] (t − t 0 )S ′ }
We then apply the identity
+ D (n) {[α(α − 1) + αp(t) + q(t)] S} (30.97)
D n (uv) =
n∑
k=0
( n
k)
(D k u)(D n−k v) (30.98)