<s>[INST] Find a power series solution centered at 0 (Differential equations Here's the problem: $$(x-1)y''+y'=0$$  This is the work that I've already done: $$y=\sum_{n=0}^{\infty}a_{n}x^n$$
$$y'=\sum_{n=0}^{\infty}(a_{n+1})(n+1)x^n$$
$$y''=\sum_{n=0}^{\infty}(a_{n+2})(n+2)(n+1)x^n$$
I then plug those into the original equation:
$$(x-1)\sum_{n=0}^{\infty}(a_{n+2})(n+2)(n+1)x^n+\sum_{n=0}^{\infty}(a_{n+1})(n+1)x^n=0$$
$$\sum_{n=0}^{\infty}(a_{n+2})(n+2)(n+1)x^{n+1}-\sum_{n=0}^{\infty}(a_{n+2})(n+2)(n+1)x^n+\sum_{n=0}^{\infty}(a_{n+1})(n+1)x^n=0$$
And this is where I'm stuck and don't know what to do from here. If someone could help me out that would be great. 
 [/INST] You have
$$
\sum_{n=0}^{\infty}(a_{n+2})(n+2)(n+1)x^{n+1}-\sum_{n=0}^{\infty}(a_{n+2})(n+2)(n+1)x^n+\sum_{n=0}^{\infty}(a_{n+1})(n+1)x^n = 0.
$$
Rewrite the first sum as
$$
\sum_{n=0}^{\infty}(a_{n+2})(n+2)(n+1)x^{n+1} = \sum_{m=1}^{\infty}(a_{m+1})(m+1)(m)x^m.
$$
Rewrite the second sum as
$$
\begin{align}
\sum_{n=0}^{\infty}(a_{n+2})(n+2)(n+1)x^n &= 2a_2 + \sum_{n=1}^{\infty}(a_{n+2})(n+2)(n+1)x^n \\
&= 2a_2 + \sum_{m=1}^{\infty}(a_{m+2})(m+2)(m+1)x^m.
\end{align}
$$
Rewrite the third sum as
$$
\begin{align}
\sum_{n=0}^{\infty}(a_{n+1})(n+1)x^n &= a_1 + \sum_{n=1}^{\infty}(a_{n+1})(n+1)x^n \\
&= a_1 + \sum_{m=1}^{\infty}(a_{m+1})(m+1)x^m.
\end{align}
$$
The first expression thus becomes
$$
\sum_{m=1}^{\infty}(a_{m+1})(m+1)(m)x^m - 2a_2 - \sum_{m=1}^{\infty}(a_{m+2})(m+2)(m+1)x^m + a_1 + \sum_{m=1}^{\infty}(a_{m+1})(m+1)x^m = 0.
$$
Now combine the sums to get
$$
a_1 - 2a_2 + \sum_{m=1}^{\infty} \Bigl[ m(m+1)a_{m+1} - (m+1)(m+2)a_{m+2} + (m+1)a_{m+1} \Bigr] x^m = 0.
$$
</s>