Lecture Notes in Differential Equations - Bruce E. Shapiro

277
∑
∞ ∞∑
∞∑
0 = t 2 a k k(k − 1)t k−2 + αt ka k t k−1 + β a k t k (29.7)
k=0
k=0
k=0
k=0
∞∑
∞∑
∞∑
= a k k(k − 1)t k + αka k t k + βa k t k (29.8)
=
k=0
k=0
∞∑
a k t k [k(k − 1) + αk + β] (29.9)
k=0
Since this must hold for all values of t, by linear independence we require
that
a k [k(k − 1) + αk + β] = 0 (29.10)
for all k = 0, 1, ...
Hence either
or
a k = 0 (29.11)
k(k − 1) + αk + β = 0 (29.12)
for all values of k. Since α and β are given real numbers, it is impossible
for (29.12) to hold for all values of k because it is quadratic k. It will hold
for at most two values of k, which are most likely not integers.
This leads us to the following conclusion: the only time when (29.4) has a
series solution is when
r 1,2 = 1 − α ± √ (1 − α) 2 − 4β
2
(29.13)
are both non-negative integers, say p and q, in which case the series solution
is really just two terms
y = a p t p + a q t q (29.14)
When there is not a non-negative integer solution to (29.13) then (29.4)
does not have a seris solution.
The result we found in example 29.3 does suggest to us one way to solve
the Cauchy-Euler method. What if we relax the restriction on equation
29.10 that k be an integer. To see how this might come about, we consider
a solution of the form
y = x r (29.15)