Lecture Notes in Differential Equations - Bruce E. Shapiro

293
term by term to (30.97). Starting with the first term,
D n [ (t − t 0 ) 2 S ′′] n∑
( ) n [D
=
j (t − t
j
0 ) 2] [ D n−j S ′′] (30.99)
j=0
( ( n
= (t − t
0)
0 ) 2 D n S ′′ n
+ 2 (t − t
1)
0 )D n−1 S ′′
(
n
+ 2 D
2)
n−2 S ′′ (30.100)
Even for n > 2 there are only 3 terms because D n (t − t 0 ) 2 = 0 for n ≥ 3.
Hence
At t = t 0 ,
D n [ (t − t 0 ) 2 S ′′] = (t − t 0 ) 2 S (n+2) + 2(n − 1)(t − t 0 )S (n+1)
+ n(n − 1)S (n) (30.101)
D n [ (t − t 0 ) 2 S ′′]∣ ∣
t=t0
= n(n − 1)S (n) (t 0 ) = n(n − 1)c n n! (30.102)
Similarly, the second term in (30.97)
D n {(t − t 0 )[2α + p(t)]S ′ (t)}
n∑
(
n
= D
k)
k (t − t 0 )D n−k {S ′ (t)[2α + p(t)]} (30.103)
At t = t 0 ,
k=0
= (t − t 0 )D n {S ′ (t)[2α + p(t)]} + nD n−1 {S ′ (t)[2α + p(t)]} (30.104)
D n {(t − t 0 )[2α + p(t)]S ′ (t)} (t 0 )
= nD n−1 {S ′ (t)[2α + p(t)]} (t 0 ) (30.105)
n−1
∑
( ) n − 1 {
= n
[2α + p(t)] (k) (t
k
0 )}
S (n−k) (t 0 ) (30.106)
k=0
= n[2α + p(t 0 )]S (n) (t 0 )
n−1
∑
( ) { n − 1
+ n
[2α + p(t)] (k) (t
k
0 )}
S (n−k) (t 0 ) (30.107)
k=1
n−1
∑
( )
= n[2α + p(t 0 )]S (n) n − 1
(t 0 ) + n
p (k) (t
k
0 )S (n−k) (t 0 ) (30.108)
k=1