I'm trying to solve the ODE: $$ y''(x) + \frac{2x}{(x-1)(2x-1)} y'(x) - \frac{2}{(x-1)(2x-1)} y(x) = 0 $$
I'm trying to find a solution by the Frobenius method, expanding a power series of the solution around $x = \frac 12$, that is in a series of terms $(x - 1/2)^{n}$. The indicial equation has two roots, $\alpha = 0$ and $\alpha=2$. For $\alpha= 2$ the solution will be $$ y(x) = \sum_{k=0}^{\infty} a_k \left( x - \frac 12 \right)^{k+2}$$
If I say that $$ p(x) =\frac{1}{x-\frac12} \frac{x}{x-1} =\frac{1}{x-\frac12} \left(-1 + \sum_{i = 0}^{\infty} -2^{i+1} \left( x -\frac 12\right)^i \right) $$ and $$ q(x) = -\frac{1}{\left( x -\frac12 \right)^2} \frac{x-1/2}{x-1} =\frac{1}{\left( x -\frac12 \right)^2} \left( -1 + \sum_{j=0}^{\infty} 2^{j+1} \left( x - \frac 12\right)^{j+1} \right) $$
and plug that and the power series expansion for $y, y', y''$ in the ODE, I get:
$$\sum_{k=0}^{\infty} (k+2)(k+1) a_k \left( x- \frac 12\right)^k + \frac{1}{x-\frac12} \left(-1 + \sum_{i = 0}^{\infty} -2^{i+1} \left( x -\frac 12\right)^i \right) \left(\sum_{k=0}^{\infty} (k+2) a_k \left( x- \frac 12\right)^{k+1}\right) + \frac{1}{\left( x -\frac12 \right)^2} \left( -1 + \sum_{j=0}^{\infty} 2^{j+1} \left( x - \frac 12\right)^{j+1} \right) \left(\sum_{k=0}^{\infty} a_k \left( x - \frac 12 \right)^{k+2} \right)$$
I go through the math and get
$$\sum_{k=0}^{\infty} (k+2)(k+1) a_k \left( x- \frac 12\right)^k + \sum_{k=0}^{\infty} \left( \sum_{i=0}^{k} -a_i (i+2) 2^{k-i+1} \right) \left( x- \frac 12\right)^{k} + \sum_{k=0}^{\infty} (k+2) a_k \left( x - \frac 12 \right)^k + \sum_{k=0}^{\infty} \left( \sum_{j=0}^{k} 2^{k-j} a_j \right) \left(x - \frac 12 \right)^k - \sum_{k=0}^{\infty} a_k \left( x - \frac 12 \right)^k = 0 $$
Now I equate the coefficients ofequal powers to 0: $$k = 0 , 2 a_0 - 4 a_ + 2 a_0 + a_0 - a_0 = 0 <=> 0 a_0 = 0 $$ $$k = 1 , 6 a_1 - 8 a_0 - 6 a_1 + 3 a_1 + 2 a_0 + a_1 - a_1 = 0 <=> a1 = (6/3) a_0$$
Am I now getting this right?