Lecture Notes in Differential Equations - Bruce E. Shapiro

258 LESSON 28. SERIES SOLUTIONS
by assuming a k = a 1 /(k − 1)! as an inductive hypothesis. Hence
y = a 1 t + a 2 t 2 + a 3 t 3 + · · · (28.22)
= a 1
(t + t 2 + 1 2 t3 + 1 3! t4 + 1 )
4! t5 + · · · (28.23)
(
= a 1 t 1 + t + 1 2 t2 + 1 3! t3 + 1 )
4! t4 + · · · (28.24)
= a 1 te t (28.25)
which is the same solution we found by separation of variables. We have
not applied the initial condition, but we note in passing the only possible
initial condition that this solution satisfies at t 0 = 0 is y(0) = 0.
Example 28.2. Find a solution to the initial value problem
⎫
y ′′ − ty ′ − y = 0⎪⎬
y(0) = 1
⎪⎭
y ′ (0) = 1
(28.26)
Letting y = ∑ ∞
k=0 c kt k and differentiating twice we find that
y ′ =
y ′′ =
∞∑
kc k t k−1 (28.27)
k=0
∞∑
k(k − 1)c k t k−2 (28.28)
k=0
The initial conditions tell us that
Substituting into the differential equation.
k=0
c 0 = 1, (28.29)
c 1 = 1 (28.30)
∞∑
∞∑
∞∑
0 = k(k − 1)c k t k−2 − t kc k t k−1 − c k t k (28.31)
k=0
Let j = k − 2 in the first term, and combining the last two terms into a
single sum,
0 =
j=−2
k=0
∞∑
∞∑
(j + 2)(j + 1)c j+2 t j − (k + 1)c k t k (28.32)
k=0