Frobenius series without recurrence relation

[um0123]

Homework Statement

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.

Homework Equations

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.

 

[Nino MMP]

I am also stuck on the second part. I know that \tilde{y} = [1/(n-1)][y_1(x) - y_2(x)]is a solution when n=/= 1, but i don't know how to take the limit of it as n \rightarrow 1.