Lecture Notes in Differential Equations - Bruce E. Shapiro

302 LESSON 30. THE METHOD OF FROBENIUS
Example 30.9. Find the form of the Frobenius solutions to Bessel’s equation
of order 3,
t 2 y ′′ + ty + (t 2 − 9)y = 0 (30.176)
Equation (30.176) has the form t 2 y ′′ + tp(t)y ′ + q(t)y = 0, where p(t) = 1
and q(t) = t 2 − 9 are both analytic functions at t = 0. Hence p 0 = 1,
q 0 = −9, and the indicial equation α 2 + (p 0 − 1)α + q 0 = 0 is
α 2 − 9 = 0 (30.177)
Therefore α 1 = 3 and α 2 = −3. The first Frobenius solution is
y 1 = t 3
∞ ∑
k=0
Since ∆ = α 1 − α 2 = 6 ∈ Z, the second solution is
y 2 = a (ln t) t 3
a k t k = a 0 t 3 + a 1 t 4 + a 2 t 5 + · · · (30.178)
∞ ∑
k=0
a k t k + t −3
∞ ∑
k=0
b k t k (30.179)
The coefficients a, a k and b k are found by substituting the expressions
(30.178) and (30.179) into the differential equation.