Solving the differential equation $9x(1-x)y''-12y'+4y=0$ Solve in series the following ODE:
$$9x(1-x)y''-12y'+4y=0$$
expanding $y(x)$ about $x_0=0$.
My guess: I think it is by Frobenius series since it is not an ordinary point.
 A: You can check that $x=0$ and $x=1$ are singular regular points of the equation. Thus, you can obtain the solution in $x \in (0,1)$ if you expand using Fröbenius about, for example, at $x=0$. If you do so, you can expand the solution as follows:

$$ y(x) = \sum^\infty_{n=0}a_n \, x^{n+s}, \quad s\in \mathbb{C}, \quad a_0 \ne 0.$$

Plug this into the equation, provided uniform convergence of $y, y'$ and $y''$, to have (double check please):

$$\small{ \sum_{n=-1}^\infty a_{n+1} \big{(} (n+s+1)(9(n+s)-12)  \big{)} x^{n+s} + \sum^\infty_{n=0} a_n \big{(}4-9(n+s)(n+s-1)\big{)} = 0 }\tag{1}$$

Since the first series has one term more than second, we can get the term corresponding for $n=-1$ out of the series to come up with the known as indicial equation for $s$ (if I did my maths well):

$$a_0 \, s \, (9(s-1)-12) = 0 \implies \boxed{s=s_1=0} \text{ or } \boxed{s= s_2 = 1+12/9}. $$

Since $|s_1-s_2|$ is not a integer, we can conclude that there's no patology and Fröbenius will provide us the two parts of the solutions$^1$, $y_1(x) = \sum_{n=0}^\infty a_n(s=s_1)x^{n+s_1}$ and $y_2(x) = \sum_{n=0}^\infty a_n(s=s_2)x^{n+s_2}$. The general term of $a_n$ can be obtained from eq. (1) which for $n\geq 0$ reads:

$$ \frac{a_{n+1}}{a_n} = \frac{4-9(n+s)(n+s-1)}{(n+s+1)\big{(}9(n+s)-12\big{)} }. $$

It's up to you now to determine the value of $a_n$ and check for convergence.
Cheers!

$^1$, if this hadn't been the case, see here to see how you can obtain not only the particular solution (which in your case is $0$) but the second part of the solution provided one of them.
