Lecture Notes in Differential Equations - Bruce E. Shapiro

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Example 30.6. Find the form of the Frobenius solutions to
t 2 y ′′ − ty + 2y = 0 (30.80)
This equation can be written in the form t 2 y ′′ + tp(t)y ′ + q(t)y = 0 where
p(t) = −1 and q(t) = 2. Hence the indicial equation is
0 = α 2 + (p 0 − 1)α + q 0 = α 2 − 2α + 2 (30.81)
The roots of the indicial equation are α = 1 ± i, a complex conjugate pair.
Since ∆α = (1 + i) − (1 − i) = 2i ∉ Z, each root gives a Frobenius solution:
y 1 = t 1+i
∞ ∑
k=0
a k t k (30.82)
= (cos ln t + i sin ln t) ( a 0 t + a 1 t 2 + a 2 t 3 + · · · )
y 2 = t 1−i
∞ ∑
k=0
(30.83)
b k t k (30.84)
= (cos ln t − i sin ln t) ( b 0 t + b 1 t 2 + b 2 t 3 + · · · ) . (30.85)
Example 30.7. Find the form of the Frobenius solution for the Bessel
equation of order -3,
t 2 y ′′ + ty ′ + (t 2 + 9) = 0 (30.86)
This has p(t) = 1 and q(t) = t 2 + 9, so that p 0 = 1 and q 0 = 9. The indicial
equation is
α 2 + 9 = 0 (30.87)
The roots are the complex conjugate pair α 1 = 3i, α 2 = −3i. For a
complex conjugate pair, each root gives a Frobenius solution, and hence we
have two solutions
y 1 = t 3i
y 2 = t −3i
∞ ∑
k=0
∑
∞
k=0
where we have used the identity
a k t k = [cos(3 ln t) + i sin(3 ln t)]
b k t k = [cos(3 ln t) − i sin(3 ln t)]
∞∑
a k t k (30.88)
k=0
∞∑
b k t k (30.89)
k=0
t ix = e ix ln t = cos(x ln t) + i sin(x ln t) (30.90)
in the second form of each expression.