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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?

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Consider the differential equation \begin{align} (x-1)(2x-1) y''(x) + 2x y'(x) - 2 y(x) = 0 \end{align} with solutions expanded around $1/2$. For this it is seen that \begin{align} y(x) = \sum_{n=0}^{\infty} a_{n} \left(x - \frac{1}{2} \right)^{n+\sigma} \end{align} which leads to \begin{align} 0 &= 2(x-1)\left(x-\frac{1}{2}\right) \sum_{n=0}^{\infty} a_{n}(n+\sigma)(n+\sigma-1) \left(x - \frac{1}{2} \right)^{n+\sigma-2} + 2x \sum_{n=0}^{\infty} a_{n}(n+\sigma) \left(x - \frac{1}{2} \right)^{n+\sigma-1} \\ & \hspace{10mm} - 2\sum_{n=0}^{\infty} a_{n} \left(x - \frac{1}{2} \right)^{n+\sigma}\\ &= 2(x-1) \sum_{n=0}^{\infty} a_{n}(n+\sigma)(n+\sigma-1) \left(x - \frac{1}{2} \right)^{n+\sigma-1} + 2 x \sum_{n=0}^{\infty} a_{n}(n+\sigma)^2 \left(x - \frac{1}{2} \right)^{n+\sigma-1} \\ & \hspace{10mm} - 2 \sum_{n=0}^{\infty} a_{n} \left(x - \frac{1}{2} \right)^{n+\sigma} \\ &= 2[(x-1/2) - 1/2] \sum_{n=0}^{\infty} a_{n}(n+\sigma)(n+\sigma-1) \left(x - \frac{1}{2} \right)^{n+\sigma-1} \\ & \hspace{5mm} +2[(x-1/2) +1/2] \sum_{n=0}^{\infty} a_{n}(n+\sigma)^2 \left(x - \frac{1}{2} \right)^{n+\sigma} + \sum_{n=0}^{\infty} a_{n} \left(x - \frac{1}{2} \right)^{n+\sigma} \\ &= -\sum_{n=0}^{\infty} a_{n}(n+\sigma)(n+\sigma-2) \left(x - \frac{1}{2} \right)^{n+\sigma-1} +2 \sum_{n=0}^{\infty} a_{n}[(n+\sigma)^2-1] \left(x - \frac{1}{2} \right)^{n+\sigma} \\ 0 &= -a_{0} (\sigma)(\sigma-2) + \sum_{n=0}^{\infty} (n+\sigma+1)(n+\sigma-1) [ (a_{n+1} - 2 a_{n}) \left(x - \frac{1}{2} \right)^{n+\sigma} \end{align} From this it is seen that \begin{align} 0 &= a_{0} (\sigma)(\sigma-2) \\ a_{n+1} &= 2 a_{n}. \end{align} From the first equation it is seen that $a_{0} \neq 0$ and $\sigma = 0,2$. The coefficients $a_{n}$ are given by $a_{n} = 2^{n} a_{0}$. The solutions follow from \begin{align} y_{\sigma}(x) &=a_{0} \sum_{n=0}^{\infty} 2^{n} \left(x - \frac{1}{2} \right)^{n+\sigma}\\ &= a_{0} \left(x - \frac{1}{2}\right)^{\sigma} \sum_{n=0}^{\infty} (2x-1)^{n} \\ &= \frac{a_{0}}{2(1-x)} \, \left(x - \frac{1}{2} \right)^{\sigma}. \end{align}

Since $\sigma = 0,2$ then the general solution follows the form \begin{align} y(x) &= \frac{a_{2}}{1-x} + a_{3} \, \frac{(1-2x)^{2}}{1-x} = \frac{a_{2}}{1-x} + a_{3} \left( \frac{1}{1-x} - 4 x \right) \\ &= \frac{A}{1-x} + B x. \end{align}

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