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
$$y=\sum_{n=0}^{\infty}c_{n}(x-2)^{n}\\y'=\sum_{n=1}^{\infty}c_{n}n(x-2)^{n-1}\\y''=\sum_{n=2}^{\infty}c_{n}n(n-1)(x-2)^{n-2}$$
 A: Hint.
Note that $\frac 1x$ is a solution for the DE so try something with
$$
y = \frac 12\sum_{k=0}^{\infty}(-1)^k\frac{(x-2)^k}{2^k}+\sum_{k=0}^{\infty}a_k (x-2)^k
$$
note also that
$$
x = (x-2) + 2\\
x^2 = (x-2)^2+4(x-2)+4
$$
A: 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:
$$ y(x) = ax + \frac{b}{x} $$
A: Define the operators $D$ and $X$ by $(D\,h)(x):=h'(x)$ and $(X\,h)(x):=x\,h(x)$.  For any function $\phi$, we also define the operator $\phi(X)$ to be $\big(\phi(X)\,h\big)(x):=\phi(x)\,h(x)$. Note that
$$D^2+\frac{1}{X}\,D-\frac{1}{X^2}=\left(D+\frac{2}{X}\right)\,\left(D-\frac{1}{X}\right)\,.$$
First, suppose $z:=\left(D-\dfrac{1}{X}\right)y$.  Then, $\left(D+\dfrac{2}{X}\right)\,z=0$, or equivalently, $D\,\left(X^2\,z\right)=0$.  Consequently,
$$z(x)=-\frac{2a}{x^2}\text{ for some constant }a\,.$$
Now, $\left(D-\dfrac{1}{X}\right)\,y=z$ implies that $D\,\left(\dfrac{1}{X}\,y\right)=\dfrac{1}{X}\,z$.  Therefore, for some constant $b$, we have
$$y(x)=x\,\int\,\frac{1}{x}\,\left(-\frac{2a}{x^2}\right)\,\text{d}x=x\,\Biggl(a\left(\frac{1}{x^2}\right)+b\Biggr)=a\,\left(\frac{1}{x}\right)+b\,x\,.$$
Thus, the only power-series solution to this differential equation is $y(x)=b\,x$ for some constant $b$.

I should have realized that there is an easier decomposition
$$D^2+\frac{1}{X}\,D-\frac{1}{X^2}=D\,\left(D+\frac{1}{X}\right)\,.$$
This leads to a very simple solution.  Note that 
$\left(D^2+\dfrac{1}{X}\,D-\dfrac{1}{X^2}\right)\,y=0$ implies
$$D\,\Biggl(\frac{1}{X}\,D\,(X\,y)\Biggr)=0\,.$$
Thus, $\big(D\,(X\,y)\big)(x)=2A\,x$ for some constant $A$, and
$$y(x)=\frac{1}{x}\,\left(A\,x^2+B\right)=A\,x+B\,\left(\frac1x\right)$$
for some constant $B$.
