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$$\left(2x-3\right)y''-xy'+y=0$$

First I found the first to derivatives of the following power series:

$$y(x)=\sum_{n=0}^{\infty}a_nx^n$$

$$y'(x)=\sum_{n=1}^{\infty}na_nx^{n-1}$$

$$y''(x)=\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}$$

I used these equations to rewrite my initial equation:

$$\left(2x-3\right)\sum_{n=0}^{\infty}n(n-1)a_nx^{n-2}-x\sum_{n=1}^{\infty}na_nx^{n-1}+\sum_{n=0}^{\infty}a_nx^n=0$$

After making all the summations the same ($n=0$), I got the following:

$$\left(2x-3\right)\sum_{n=0}^{\infty}n(n-1)a_nx^{n-2}-x\sum_{n=0}^{\infty}(n+1)a_{n+1}x^{n}+\sum_{n=0}^{\infty}a_nx^n=0$$

I finally got to the following step, but I'm stuck here not knowing how I can factor out the $x$ so I can find the first non-zero terms:

$$\left(2x-3\right)\sum_{n=0}^{\infty}\left[n(n-1)a_nx^{n-2}-x(n+1)a_{n+1}x^n+a_nx^n\right]=0$$

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To solve the differential equation, you must write it in the form $$\sum_{n=0}^\infty b_nx^n=0$$ so you must put together the same powers of $x$.

For instance, you may want to use more convenient indexing, such as \begin{align}\left(\sum_{n=0}^\infty b_nx^n\right)'&=\sum_{n=0}^\infty (n+1)b_{n+1}x^n\\\left(\sum_{n=0}^\infty b_nx^n\right)''&=\sum_{n=0}^\infty (n+2)(n+1)b_{n+2}x^n\\ x^{\color{red}{k}}\cdot\sum_{n=0}^\infty b_nx^n&=\sum_{\color{red}{n=k}}^\infty b_{n\color{red}{-k}}x^n\end{align}

Using these should put you in the right direction.

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