Mathematical Tools for Physics

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!

−...

]