4—Differential Equations 94
This determinesa 2 in terms ofa 0 ; it determinesa 4 in terms ofa 2 etc.
a`=−a`− 2
1
(`+s)^2 −n^2
, `= 2, 4 , ...
For example, ifn= 0, the indicial equation sayss= 0.
a 2 =−a 0
1
22
, a 4 =−a 2
1
42
= +a 0
1
2242
, a 6 =−a 4
1
62
=−a 0
1
224262
a 2 k= (−1)ka 0
1
22 kk!^2
then y(x) =a 0
∑∞
k=0
(−1)k
(x/2)^2 k
(k!)^2
=a 0 J 0 (x) (17)
and in the last equation I rearranged the factors and used the standard notation for the Bessel function,Jn(x).
This is a second order differential equation. What about the other solution? This Frobenius series method
is guaranteed to find one solution near a regular singular point. Sometimes it gives both but not always, and in
this example it produces only one of the two solutions. There is a procedure that will let you find the second
solution to this sort of second order differential equation but I’ll leave that for elsewhere.
For the casen=^1 / 2 the calculations that I just worked out will produce two solutions. The indicial equation
givess=±^1 / 2. After that, the recursion relation for the coefficients give
a`=−a`− 2
1
(`+s)^2 −n^2
=−a`− 2
1
`^2 + 2`s
=−a`− 2
1
`(`+ 2s)
=−a`− 2
1
(
±1)
For thes= +^1 / 2 result
a 2 =−a 0
1
2. 3
a 4 =−a 2
1
4. 5
= +a 0
1
2. 3. 4. 5
a 2 k= (−1)ka 0
1
(2k+ 1)!
This solution is then
y(x) =a 0 x^1 /^2
[
1 −
x^2
3!
+
x^4
5!