Lecture Notes in Differential Equations - Bruce E. Shapiro

289
Thus a Frobenius solution is
y 1 = √ ∞∑
t a k t k (30.58)
k=0
Returning to the second solution
= √ t(a 0 + a 1 t + a 2 t 2 + · · · ) (30.59)
√
= a 0 t
(1 − 1 3! t2 + 1 5! t4 − 1 )
7! t6 + · · · (30.60)
(
)
a
√ 0
t − t3
t 3! + t5
5! − t7
7! + · · · (30.61)
= a 0
√ sin t (30.62)
t
y 2 = b 0
√
t
+ b 1
√
t + b2 t 3/2 + b 3 t 5/2 + · · · (30.63)
This series cannot even converge at the origin unless b 0 = 0. This leads to
y 2 = b 1
√
t + b2 t 3/2 + b 3 t 5/2 + · · · (30.64)
= √ t(b 1 + b 2 t + b 3 t 2 + · · · ) (30.65)
which is the same as the first solution. So the Frobenius procedure, in this
case, only gives us the one solution.
Theorem 30.1. (Method of Frobenius) Let
p(t) =
q(t) =
∞∑
p k (t − t 0 ) k and (30.66)
k=0
∞∑
q k (t − t 0 ) k (30.67)
k=0
be analytic functions at t = t 0 , with radii of convergence r, and let α 1 , α 2
be the roots of the indicial equation
Then
α(α − 1) + αp(t 0 ) + q(t 0 ) = 0. (30.68)
1. If α 1 , α 2 are both real and α = max{α 1 , α 2 }, there exists some set of
constants {a 1 , a 2 , ...} such that
y = (t − t 0 ) α
∞ ∑
k=0
a k (t − t 0 ) k (30.69)