We use the method of Frobenius to obtain two linearly independent series solution. Where x=0 is a regular singular point of the differential equation. The general solution on
, , and
Substitute into DE:
We want to have all the powers of x match. In order to do this we will substitute k for the value of n:
Now, in order to add together the sums we need to have the indexes for each sum match. To do this we list the terms for the first and second sum at . In this example we need to have the indexes match the . In order to reduce the amount of work required we will add together the sums where and then factor out the term. At this point we can also factor out the term.
The first two terms are:
From the sum we have:
Finally we now have:
Now we are ready to solve the equation. We start by equating the terms on the left hand side with the terms on the right hand side (which are all equal to zero). Which gives us the following implications:
(1) the indicial roots are: and .
The recurrence relation is:
We can now substitute the values for that satisfy (1) into (2) to yield two different recurrence relations:
Now we can let on equations (4) and (5) for the first 3 terms:
And so on ...
For simplification purposes we let
so we get the general solution to the differential equation on :