Lecture Notes in Differential Equations - Bruce E. Shapiro

300 LESSON 30. THE METHOD OF FROBENIUS
Integrating term by term,
⎡
∫
v(t) = Ku(t) ⎣d ∆ t −1 dt +
⎡
= Ku(t) ⎣d ∆ ln |t| +
∞∑
k=0,k≠∆
∞∑
k=0,k≠∆
d k
∫
Substituting equation (30.139) in the second term,
v(t) = au(t) ln |t| + Kt α
∞ ∑
k=0
c k t k
⎤
t k−(1+∆) dt⎦ (30.159)
⎤
d k
k − ∆ tk−∆ ⎦ (30.160)
∞ ∑
k=0,k≠∆
d k
k − ∆ tk−∆ (30.161)
where a is a constant, and we have factored out the common t −∆ in the
second line.
Since the product of two power series is a power series, then there exists a
sequence of numbers {a 0 , a 1 , ...} such that
v(t) = au(t) ln |t| + t α2
∞ ∑
k=0
a k t k (30.162)
where we have used the fact that α 2 = α − ∆. This proves (30.136).
Case 3. ∆ ≠ 0 and ∆ is not an integer. Integrating (30.151) term by term,
[
]
v(t) = Ku(t) − t−∆
∞ ∆
+ ∑ d k
k − ∆ tk−∆ (30.163)
k=1
∑
∞
= u(t)t −∆ f k t k (30.164)
k=0
where f 0 = −K/∆ and f k = Kd k /(k − ∆), for k = 1, 2, .... Substitution
for u(t) gives
v(t) = t α−∆
∞
∑
k=0
c k t k
∞
∑
j=0
f j t j (30.165)
Since α 2 = α − ∆, and since the product of two power series is a power
series, there exists some sequence of constants {a 0 , a 1 , ...} such that
which proves (30.137).
v(t) = t α2
∞ ∑
k=0
a k t k (30.166)