# Frobenius series without recurrence relation

1. Oct 27, 2013

### um0123

1. The problem statement, all variables and given/known data

Consider

$x^2y''-xy'+n^2y=0$

where n is a constant.

a) find two linearly independent solutions in the form of a Frobenius series, initially keeping at least the first 3 terms. Can you find the solution to all orders?

b) for n=1 you shouild find only one linearly independent solution. If y_1(x) and y_2(x) are two independent solution for n=/= 1, us the fact that

$\tilde{y} = [1/(n-1)][y_1(x) - y_2(x)]$

is a solution and take the $n \rightarrow 1$ of it to find another solution.

2. Relevant equations

3. The attempt at a solution

I did the normal method of frobenius by taking the maclaurin expansion of y(x) and plugging it back in and got

$y(z) = \sum_{k = 0}^{ \infty} a_kx^{(k+s)}[n^2-(k+s)+(k+s)(k+s-1)] = 0$

and by asserting that a_0 does not equal 0 i solved for s and got

$s = 1 \pm \sqrt{1-n^2}$

But all the examples i have looked at have had another summation that starts at a different value of k in order to get a recurrence relationship between the coefficients. I only have one summation so i don't know what to do.

Last edited: Oct 28, 2013