Lecture Notes in Differential Equations - Bruce E. Shapiro

284 LESSON 30. THE METHOD OF FROBENIUS
Example 30.1. Attempt to find a series solution for equation (30.5).
Following our usual procedure we let y = ∑ ∞
0 c kt k , differentiate twice, and
substitute the results into the differential equation, leading to
∑
∞ ∞∑
∞∑
0 = t 2 c k k(k − 1)t k−2 + t c k kt k−1 + c k t k (30.7)
k=0
k=0
k=0
k=0
k=0
∞∑
∞∑
∞∑
= c k k(k − 1)t k + c k kt k + c k t k (30.8)
=
=
k=0
∞∑
c k (k(k − 1) + k + 1)t k (30.9)
k=0
∞∑
c k (k 2 + 1)t k (30.10)
k=0
By linear independence,
c k (k 2 + 1) = 0 (30.11)
for all k. But since k 2 + 1 is a positive integer, this means c k = 0 for all
values of k. Hence the only series solution is the trivial one, y = 0.
Before one is tempted to conclude that there is no non-trivial solution to
equation (30.5), consider the following example.
Example 30.2. Use the substitution x = ln t to find a solution to (30.5).
By the chain rule
y ′ = dy
dx
Differentiating a second time,
y ′′ = d ( 1
dt t
)
dy
dx
= − 1 t 2 dy
dx + 1 t
=− 1 t 2 dy
dx + 1 t
dx
dt = 1 dy
t dx
( )
d dy
dt dx
( dy
dx
d
dx
) dx
dt
Substituting (30.12) and (30.16) into (30.5) gives
(30.12)
(30.13)
(30.14)
(30.15)
=− 1 t 2 dy
dx + 1 t 2 d 2 y
dx 2 (30.16)
d 2 y
dx 2 + y = 0 (30.17)