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. 
 A: Setting 
$$
y(x)=\sum_{n=0}^\infty a_nx^n,
$$
we have
\begin{eqnarray}
(x-1)y''(x)+y'(x)&=&(x-1)\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum_{n=1}^\infty na_nx^{n-1}\\
&=&\sum_{n=2}^\infty n(n-1)a_nx^{n-1}-\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum_{n=1}^\infty na_nx^{n-1}\\
&=&\sum_{n=1}^\infty n^2a_nx^{n-1}-\sum_{n=2}^\infty n(n-1)a_nx^{n-2}\\
&=&\sum_{n=0}^\infty(n+1)^2a_{n+1}x^n-\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n\\
&=&\sum_{n=0}^\infty(n+1)[(n+2)a_{n+2}-(n+1)a_{n+1}]x^n.
\end{eqnarray}
Therefore
$$
(x-1)y''(x)+y'(x)=0 \iff a_{n+2}=\frac{n+1}{n+2}a_{n+1} \quad \forall n\ge 0,
$$
i.e.
$$
a_n=\frac{n-1}{n}a_{n-1} \quad \forall n\ge 2.
$$
Thus
$$
a_n=\frac{(n-1)\cdot(n-2)\ldots(2-1)}{n\cdot(n-1)\ldots(2-1)}a_1=\frac{1}{n}a_1 \quad \forall n \ge 2.
$$
The solution of the DE is then given by
$$
y(x)=a_0+a_1x+a_1\sum_{n=2}\frac{1}{n}x^n=a_0+a_1\sum_{n=1}^\infty\frac{1}{n}x^n\equiv a_0-a_1\ln(1-x),
$$
where $a_0,a_1$ are real constants.
A: 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
$$
a_1 - 2a_2 + \sum_{m=1}^{\infty} \Bigl[ m(m+1)a_{m+1} - (m+1)(m+2)a_{m+2} + (m+1)a_{m+1} \Bigr] x^m = 0.
$$
A: Note that, it is easier to go this way: 
i) make the substitution $z=y'$ to reduce the order of the ode and deal with the new ode 

$$ (x-1)z'+z=0. $$

ii) use power series technique to solve the above ode by assuming 

$$z(x)=\sum_{n=0}^{\infty}a_n x^n .$$ 

iii) Use $z=y'$ to get $y(x)$.   
