Lecture Notes in Differential Equations - Bruce E. Shapiro

301
Example 30.8. In example 30.4 we found that one solution of Bessel’s
equation of order 1/2, given by
t 2 y ′′ + ty ′ + (t 2 − 1/4)y = 0 (30.167)
near the origin is
y 1 = sin t √
t
(30.168)
Find a second solution using the method of Frobenius.
In example 30.4 we found that the the roots of the indicial equation are
Since the difference between the two roots is
α 1 = 1/2 and α 2 = −1/2 (30.169)
∆ = α 1 − α 2 = 1 (30.170)
We find the result from theorem 30.2, case 2, which gives a second solution
y 2 (t) = ay 1 (t) ln |t| + t α2
∞
∑
k=0
a k t k (30.171)
The numbers a and a 0 , a 1 , ... are found by substituting (30.171) into the
original differential equation and using linear independence. In fact, since
we have a neat, closed form for the first solution that is not a power series,
it is easier to find the second solution directly by reduction of order.
In standard form the differential equation can be rewritten as
y ′′ + 1 t y′ + t2 − 1/4
t 2 y = 0 (30.172)
which has the form of y ′′ + py ′ + q = 0 with p(t) = 1/y. By Abel’s formula,
one expression for the Wronskian is
W = C exp
∫ −1
t dt = C t
(30.173)
According to the reduction of order formula, a second solution is given by
∫
y 2 = y 1 (t)
∫ ( √ ) 2
W (t) sin t 1 t
dt = √
y 1 (t)
2 t t · dt = sin t ∫
√
sin t t
csc 2 t (30.174)
= − sin t √
t
cot t = − cos t √
t
(30.175)