<s>[INST] Solve by a series of powers around the given ordinary point Solve by a series of powers around the given ordinary point, $$ x^{2}y'' + xy'-y = 0 \hspace{5mm}x_{0} = 2 $$
I have problems to continue with the development, I start with the assumption
$$y=\sum_{n=0}^{\infty}c_{n}(x-2)^{n}\\y'=\sum_{n=1}^{\infty}c_{n}n(x-2)^{n-1}\\y''=\sum_{n=2}^{\infty}c_{n}n(n-1)(x-2)^{n-2}$$
 [/INST] Let $t=x-2$, then the equation becomes
$$ (t+2)^2y'' + (t+2)y' - y = 0 $$
Simplifying each term at a time:
\begin{align} 
y &= \sum_{n=0}^\infty c_nt^n \\ \\ 
(t+2)y' &=  (t+2)\sum_{n=1}^\infty c_nnt^{n-1} \\
&= \sum_{n=1}^\infty nc_nt^n + \sum_{n=1}^\infty 2nc_nt^{n-1} \\
&= \sum_{n=1}^\infty nc_nt^n + \sum_{n=0}^\infty 2(n+1)c_{n+1}t^n \\ \\
(t+2)^2y'' &= (t^2+4t+4)\sum_{n=2}^\infty n(n-1)c_nt^{n-2} \\
&= \sum_{n=2}^\infty n(n-1)c_nt^n + \sum_{n=2}^\infty 4n(n-1)c_nt^{n-1} + \sum_{n=2}^\infty 4n(n-1)c_nt^{n-2} \\
&= \sum_{n=2}^\infty n(n-1)c_nt^n + \sum_{n=1}^\infty 4(n+1)nc_{n+1}t^n + \sum_{n=0}^\infty 4(n+2)(n+1)c_{n+2}t^n
\end{align}
Combining terms:
$$ (8c_2+2c_1-c_0) + (24c_3 + 12c_2)t + \sum_{n=2}^\infty \big[4(n+2)(n+1)c_{n+2} + 2(n+1)(2n+1)c_{n+1} + (n^2-1)c_nt^n \big] $$
Then we have the recurring relations
\begin{align} 
c_2 &= \frac18(c_0-2c_1) \\
c_3 &= -\frac12c_2 \\
c_{n+2} &= \frac{2n+1}{2(n+2)}c_{n+1} + \frac{n-1}{4(n+2)}c_n
\end{align}
Finding a closed form (in terms of $c_0$ and $c_1$) is a bit more involved. 
Using the Cauchy-Euler method, you can obtain the "official" solution:
$$ y(x) = ax + \frac{b}{x} $$
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