Q:

Solve $9x(1-x)y^{\prime\prime} - 12y^\prime + 4y = 0$ using power series

Let $y(x) = \sum_{n=0}^\infty c_nx^n \Rightarrow y^\prime(x) = \sum_{n=0}^\infty nc_nx^{n-1}, y^{\prime\prime}(x) = \sum_{n=0}^\infty n(n-1)c_nx^{n-2}$. Now:
$$9x(1-x)\sum_{n=0}^\infty n(n-1)c_nx^{n-2} - 12\sum_{n=0}^\infty nc_nx^{n-1} + 4\sum_{n=0}^\infty c_nx^n = 0$$
$$9\sum_{n=0}^\infty n(n-1)c_nx^{n-1}- 9\sum_{n=0}^\infty n(n-1)c_nx^{n} - 12\sum_{n=0}^\infty nc_nx^{n-1} + 4\sum_{n=0}^\infty c_nx^n = 0,$$
$$9\sum_{n=1}^\infty n(n-1)c_nx^{n-1}- 9\sum_{n=0}^\infty n(n-1)c_nx^{n} - 12\sum_{n=1}^\infty nc_nx^{n-1} + 4\sum_{n=0}^\infty c_nx^n = 0,$$
Make $t = n-1$,
$$9\sum_{t=0}^\infty (t+1)(t)c_{t+1}x^{t}- 9\sum_{n=0}^\infty n(n-1)c_nx^{n} - 12\sum_{t=0}^\infty (t+1)c_{t+1}x^{t} + 4\sum_{n=0}^\infty c_nx^n = 0,$$
Make $t = n$, 
$$9\sum_{n=0}^\infty (n+1)(n)c_{n+1}x^{n}- 9\sum_{n=0}^\infty n(n-1)c_nx^{n} - 12\sum_{n=0}^\infty (n+1)c_{n+1}x^{n} + 4\sum_{n=0}^\infty c_nx^n = 0,$$
So, $9(n+1)(n)c_{n+1} - 9n(n-1)c_n - 12(n+1)c_{n+1} + 4c_n = 0 \Rightarrow (n+1)c_{n+1}(9n - 12) = c_n(9n^2-9n -4) \Rightarrow c_{n+1} = c_n\frac{9n^2-9n -4}{(n+1)(9n - 12)} = \frac{3n+1}{3(n+1)}c_n$
But I'm having troubles with the numerator when I try to solve the recurrence relation.

A:

Hints:
First, manipulate the recurrence as: $$c_n =c_0\prod_{k=1}^n\dfrac{3k-2}{3k}=\dfrac{c_0}{n!}\prod_{k=1}^n\left(k-\frac 23\right) .$$
On the other hand, recall the series expansion of $(1+x)^\alpha:$
$$(1+x)^\alpha = \sum_{n=0}^\infty\binom{\alpha}{n}x^n,$$
where:
$$\binom{\alpha}{n} = \prod_{k=1}^n\dfrac{\alpha-k+1}{k}.$$
So now all you need to do is to recognize correct $\alpha$ and do some small dirty work. 

