Lecture Notes in Differential Equations - Bruce E. Shapiro

298 LESSON 30. THE METHOD OF FROBENIUS
Since
we can write
exp
{
e −u = 1 +
−
∞∑
k=1
∞∑
(−1) k uk
k!
k=1
p k
k tk }
= 1 +
(30.142)
∞∑
a k t k (30.143)
for some sequence of numbers a 1 , a 2 , .... We do not actually need to know
these numbers, only that they exist. Hence
}
∞∑
W (t) = |t|
{1 −p0 + a k t k (30.144)
By the method of reduction of order, a second solution is given by
∫ ∫ {
}
−p W (t)
|t|
0 ∞∑
v(t) = u(t)
u 2 (t) dt = u(t) u 2 1 + a k t k dt (30.145)
(t)
From equation (30.139), since u is analytic, so is 1/u, except possibly at its
zeroes, so that 1/u can also be expanded in a Taylor series. Thus
{ ∞
} −2 {
}
1
u 2 (t) = ∑
∞∑
t−2α c k t k = t −2α c −2
0 1 + b k t k (30.146)
k=0
k=1
for some sequence b 1 , b 2 , ... Letting K = 1/c 2 0,
∫
v(t) = Ku(t)
for some sequence d 1 , d 2 , ...
k=1
k=1
k=1
k=1
} }
∞∑
∞∑
|t|
{1 −p0−2α + b k t
{1 k + a k t k dt (30.147)
By the quadratic formula
α 1 = 1 (1 − p 0 + √ )
(1 − p 0 )
2
− 4q 0 = 1 2 (1 − p 0 + ∆) (30.148)
α 2 = 1 (1 − p 0 − √ )
(1 − p 0 )
2
− 4q 0 = 1 2 (1 − p 0 − ∆) (30.149)
and therefore
k=1
2α + p 0 = 1 + ∆ (30.150)
since we have chosen α = max(α 1 , α 2 ) = α 1 . Therefore
∫
v(t) = Ku(t)
|t| −(1+∆) {1 +
}
∞∑
d k t k dt (30.151)
k=1