Solving $x^2y'' - 5xy' + 6y = 0$ Through Power Series

Question

I've been trying to solve the differential equation $$ x^2y'' - 5xy' + 6y = 0 $$ around the point $x_0 = 0$ through the Frobenius Method.

I've already gotten the roots of the indicial equation ($x = 3 + \sqrt{3}, x - \sqrt{3}$), but I'm struggling to get any form of recurrence, I'm currently stuck with the two cases

First, when $r = 3 + \sqrt{3}$, I end up with

$$ x^{3 + \sqrt{3}} \sum_{m=0}^{\infty} (m^2 + 2 \sqrt{3} m) a_m x^m = 0. $$

Second, when $r = 3 - \sqrt{3}$, I end up with

$$ x^{3 - \sqrt{3}} \sum_{m=0}^{\infty} (m^2 - 2 \sqrt{3} m) a_m x^m = 0. $$

So far, I don't have a base case for $a_0$, so I do not know what the recurrence relation should end up being.

Answer 1

Now you need to compare coefficients of equal degree. You should see that apart for $m=0$, every other coefficient $a_m$ has to be zero. This is not that astonishing as the given equation is Euler-Cauchy where the indicial equation is also the characteristic equation determining the degrees of the solution basis monomials.