Solving $(x-2)y'=xy$ by power series centered at nonzero point Given the equation $(x-2)y'=xy$, I was wondering if centering the power series at $2$ rather than $0$ would be "better".
Written in equations, I could start with $y=\sum_{n=0}^\infty a_nx^n$, but instead I wanted to try $$y=\sum_{n=0}^\infty a_n(x-2)^n$$ Then $xy=\sum_{n=0}^\infty a_nx^{n+1}$ and $(x-2)y'=\sum_{n=1}^\infty na_n(x-2)^{n-1}$.
How would I simplify this? It's hard for me to equate terms since, for example, I sometimes have a term like $x(x-2)$ and another term like $(x-2)^2$, both of which should be equated in some way because they have degree $2$.
I'm also wondering if there is a standard way to write the answer, or if you write it simply as a function (for example, in $1/(1-x)=\sum_{n=1}^\infty x^{n}$, the LHS would be preferred) if you can? Or are all forms generally accepted? Obviously it might depend on the professor, but I'm not taking a class and was just wondering what the common practice is.
 A: Yes you can do this you have to be careful however. If we set  $y=\sum_{n=0}^\infty a_n(x-2)^n$ we get
\begin{align}
xy&=x\sum_{n=0}^\infty a_n(x-2)^{n}=(2+x-2)\sum_{n=0}^\infty a_n(x-2)^{n}=\sum_{n=0}^\infty 2a_n(x-2)^{n}+\sum_{n=0}^\infty a_n(x-2)^{n+1}\\
(x-2)y'&=(x-2)\sum_{n=0}^\infty na_n(x-2)^{n-1}=\sum_{n=0}^\infty na_n(x-2)^{n}
\end{align}
In order to compare the terms we have to to a index shift in the sum with the $(x-2)^{n+1}$ terms.
$$
xy=\sum_{n=0}^\infty 2a_n(x-2)^{n}+\sum_{n=1}^\infty a_{n-1}(x-2)^{n}\
$$
Now we can compare the coefficients and get $2a_n+a_{n-1}=na_n \forall n\geq1$.

*

*$n=0$: $\Rightarrow a_0=0$

*$n=1$: $\Rightarrow a_1=0$

*$n=2$: $\Rightarrow a_1=0$ (No new information but also not contradicting!)

*$n\geq3$: $\Rightarrow a_n=\frac{a_{n-1}}{n-2}$
This implies that $a_n=\frac{a_{2}}{(n-2)!}$ and thus:
$$
y=a_{2}\sum_{n=2}^\infty \frac{(x-2)^{n}}{(n-2)!}
$$
By manipulating the solution you can rewrite it in terms of standard functions:
$$
y=a_{2}(x-2)^2\sum_{n=2}^\infty \frac{(x-2)^{n-2}}{(n-2)!}=a_{2}(x-2)^2\sum_{n=0}^\infty \frac{(x-2)^{n}}{n!}=a_{2}(x-2)^2e^{x-2}
$$
