Lecture Notes in Differential Equations - Bruce E. Shapiro

263
has an analytic solution at t = t 0 , given by
∞∑
y = c k (t − t 0 ) k (28.67)
k=0
where
c k = y k
, k = 0, 1, ..., n − 1 (28.68)
k!
and the remaining c k may be found by the method of undetermined coefficients.
Specifically, if p(t) and q(t) are analytic functions at t = t 0 , then
the second order initial value problem
has an analytic solution
y ′′ + p(t)y ′ + q(t)y = 0
y = y 0 + y 1 (t − t 0 ) +
y(t 0 ) = y 0
y ′ (t 0 ) = y 1
⎫
⎪⎬
⎪ ⎭
(28.69)
∞∑
c k (t − t 0 ) k (28.70)
Proof. (for n = 2). Without loss of generality we will assume that t 0 = 0.
We want to show that a non-trivial set of {c j } exists such that
∞∑
y = c j t j (28.71)
j=0
To do this we will determine the conditions under which (28.71) is a solution
of (28.69). If this series converges, then we can differentiate term by term,
∞∑
y ′ = jc j t j−1 (28.72)
y ′′ =
j=0
k=2
∞∑
j(j − 1)c j t j−2 (28.73)
j=0
Observing that the first term of the y ′ series and the first two terms of the
y ′′ series are zero, we find, after substituting k = j − 1 in the first and
k = j − 2 in the second series, that
∞∑
y ′ = (k + 1)c k+1 t k (28.74)
y ′′ =
k=0
∞∑
(k + 2)(k + 1)c k+2 t k (28.75)
k=0