Method of Frobenius to solve $x^2 Y''(x) + x Y'(x) = c^2 Y(x)$

Question

I'm trying to $x^2 Y''(x) + x Y'(x) = c^2 Y(x)$ using the method of Frobenius. I assume the solution is of the form $Y = x^k \sum_{n=0}^\infty a_n s^n $ and then put that into my ODE and found $0 = \sum_{n=0}^\infty a_n[(n+k)(n+k-1)+(n+k)-c^2]x^{k+n} $. Then I found the indicial polynomial to be $a_0[k^2 - c^2]x^k = 0$ which means $k = c$. This is where I'm stuck. I'm not sure how to solve the equation after knowing this part. I was trying to follow http://mathworld.wolfram.com/FrobeniusMethod.html.

Answer 1

Actually, $k= \pm c$. This equation is really an Euler equation, so using Frobenius is overkill. You should get that $a_0$ and $a_1$ are arbitrary constants and that all the other $a_n$'s are zero.