Lecture Notes in Differential Equations - Bruce E. Shapiro

290 LESSON 30. THE METHOD OF FROBENIUS
is a solution of
(t − t 0 ) 2 y ′′ + (t − t 0 )p(t)y ′ + q(t)y = 0 (30.70)
2. If α 1 , α 2 are both real and distinct, such that ∆ = α 1 − α 2 is not an
integer, then (30.69) gives a second solution with α = min{α 1 , α 2 }and
a (different) set of coefficients {a 1 , a 2 , ...}.
3. If α 1 , α 2 are a complex conjugate pair, then there exists some sets of
constants {a 1 , a 2 , ...} and {b 1 , b 2 , ...} such that
y 1 = (t − t 0 ) α1
y 2 = (t − t 0 ) α2
∞ ∑
k=0
∑
∞
k=0
form a fundamental set of solutions to (30.70).
a k (t − t 0 ) k (30.71)
b k (t − t 0 ) k (30.72)
Example 30.5. Find the form of the Frobenius solutions for
2t 2 y ′′ + 7ty ′ + 3y = 0 (30.73)
The differential equation can be put into the standard form
t 2 y ′′ + tp(t)y ′ + q(t)y = 0 (30.74)
by setting p(t) = 7/2 and q(t) = 3/2. The indicial equation is
0 = α 2 + (p 0 − 1)α + q 0 (30.75)
= α 2 + (5/2)α + 3/2 (30.76)
= (α + 3/2)(α + 1) (30.77)
Hence the roots are α 1 = −1, α 2 = −3/2. Since ∆ = α 1 − α 2 = 1/2 /∈ Z,
each root leads to a solution:
y 1 = 1 t
∞∑
y 2 = t −3/2
k=0
∑
∞
a k t k = a 0
t + a 1 + a 2 t + a 3 t 2 + · · · (30.78)
k=0
b k t k = b 0
t 3/2 + b 1
t 1/2 + b 2t 1/2 + b 3 t 3/2 + · · · (30.79)