<s>[INST] differential equation power series solution I am trying to solve this equation using power series 
$$
  (1-x)y"-xy'+y=0  
$$
Knowing that $y(0)=-2$ and $y'(0)=6$.
Please I need someone's help, I get a relation between 
$c(n)$,$c(n+1)$, and $c(n+2)$.
 [/INST] taking
$$\begin{align}
y&=\sum_{n=0}^{+\infty}a_n x^n\\
y'&=\sum_{n=0}^{+\infty}na_n x^{n-1}\\
y''&=\sum_{n=0}^{+\infty}n(n-1)a_nx^{n-2}
\end{align}$$
we have
$$\begin{align}
(1-x)y''-xy'+y&=0\\
(1-x)\sum_{n=0}^{+\infty}n(n-1)a_nx^{n-2}-x\sum_{n=0}^{+\infty}na_n x^{n-1}+\sum_{n=0}^{+\infty}a_n x^n&=0\\
\sum_{n=0}^{+\infty}n(n-1)a_nx^{n-2}-x\sum_{n=0}^{+\infty}n(n-1)a_nx^{n-2}-\sum_{n=0}^{+\infty}na_n x^{n}+\sum_{n=0}^{+\infty}a_n x^n&=0\\
\sum_{n=0}^{+\infty}n(n-1)a_nx^{n-2}-\sum_{n=0}^{+\infty}n(n-1)a_nx^{n-1}-\sum_{n=0}^{+\infty}na_n x^{n}+\sum_{n=0}^{+\infty}a_n x^n&=0\\
\sum_{n=2}^{+\infty}n(n-1)a_nx^{n-2}-\sum_{n=1}^{+\infty}n(n-1)a_nx^{n-1}-\sum_{n=0}^{+\infty}na_n x^{n}+\sum_{n=0}^{+\infty}a_n x^n&=0\\
\sum_{n=0}^{+\infty}(n+2)(n+1)a_{n+2}x^{n}-\sum_{n=0}^{+\infty}(n+1)na_{n+1}x^{n}-\sum_{n=0}^{+\infty}na_n x^{n}+\sum_{n=0}^{+\infty}a_n x^n&=0\\
\sum_{n=0}^{+\infty}\left[(n+2)(n+1)a_{n+2}x^{n}-(n+1)na_{n+1}x^{n}-na_n x^{n}+a_n x^n\right]&=0\\
\sum_{n=0}^{+\infty}\left[(n+2)(n+1)a_{n+2}-(n+1)na_{n+1}-na_n+a_n \right]x^n&=0
\end{align}$$
then you get
$$\begin{align}
(n+2)(n+1)a_{n+2}-(n+1)na_{n+1}-na_n+a_n&=0\\
(n+2)(n+1)a_{n+2}-(n+1)na_{n+1}+(1-n)a_n&=0\\
a_{n+2}=\frac{(n+1)na_{n+1}+(n-1)a_n}{(n+2)(n+1)}
\end{align}$$
since you got $y(0)=-2$ and $y'(0)=6$ you got $a_0$ and $a_1$ because
$$\begin{align}
y(0)&=\sum_{n=0}^{+\infty}a_n x^n\\
&=a_0\\
y'(0)&=\sum_{n=0}^{+\infty}na_n x^{n-1}\\
&=a_1
\end{align}$$
then with the value of $a_0$ and $a_1$ you can get $a_2,a_3,\cdots$ and then get the solution (valid within the series convergence radius)
</s>