Using Frobenius' Theorem find two linearly independent solutions to the following equation. is a regular singular point.
, , and
We now want to have all the powers of x match. In order to do this we will substitute k for the value of n:
Notice that in the third term, the index of summation, the index for the constant, and exponent of all changed. 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 each sum at until each index is the same. In this example we need to have the indexes match the index so we list the terms at and . 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.
From the sum:
The first two terms are:
Which finally gives us:
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) hence and
We can now substitute the values for that satisfy (1) into (3) to yield two different recurrence relations:
And then using our values for we can find the series solutions for the differential equation.
And so on...
This gives us the terms for our infinite sum and from that we have the solutions to the differential equation.
For (4) where let so the sum is
For (5) where let so the sum is
Which means the general solution is: