4—Differential Equations 75Let=kin the first sum. Let=k+ 2in the second. Explicitlyshow the limits of the index
on the sums, or you’re bound to get it wrong.
∑∞`=0a`
[
(`+s)^2 −n^2
]
x`+s+
∑∞
`=2a− 2 x+s= 0
The lowest power ofxin this equation comes from the`= 0term in the first sum. That coefficient of
xsmust vanish. (a 06 = 0)
a 0
[
s^2 −n^2
]
= 0 (4.20)
This is called theindicialequation. It determiness, or in this case, maybe twos’s. After this, set to
zero the coefficient ofx`+s.
a`
[
(`+s)^2 −n^2
]
+a`− 2 = 0 (4.21)
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) (4.22)
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. There are procedures that will let you find
the second solution to this sort of second order differential equation. See problem4.49for one such
method.
For the casen=^1 / 2 the calculations just above 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 + 2s
=−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