I'll assume the ODE is actually
[tex]y''+(x-2)y'+y=0[/tex]
Look for a series solution centered at [tex]x=2[/tex], with
[tex]y=\displaystyle\sum_{n\ge0}c_n(x-2)^n[/tex]
[tex]\implies y'=\displaystyle\sum_{n\ge0}(n+1)c_{n+1}(x-2)^n[/tex]
[tex]\implies y''=\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n[/tex]
with [tex]c_0=y(2)=2[/tex] and [tex]c_1=y'(2)=0[/tex].
Substituting the series into the ODE gives
[tex]\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge0}(n+1)c_{n+1}(x-2)^{n+1}+\sum_{n\ge0}c_n(x-2)^n=0[/tex]
[tex]\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge1}nc_n(x-2)^n+\sum_{n\ge0}c_n(x-2)^n=0[/tex]
[tex]\displaystyle2c_2+c_0+\sum_{n\ge1}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge1}nc_n(x-2)^n+\sum_{n\ge1}c_n(x-2)^n=0[/tex]
[tex]\displaystyle2c_2+c_0+\sum_{n\ge1}\bigg((n+2)(n+1)c_{n+2}+(n+1)c_n\bigg)(x-2)^n=0[/tex]
[tex]\implies\begin{cases}c_0=2\\c_1=0\\(n+2)c_{n+2}+c_n=0&\text{for }n>0\end{cases}[/tex]
[tex]k=0\implies n=0\implies c_0=c_0[/tex]
[tex]k=1\implies n=2\implies c_2=-\dfrac{c_0}2=(-1)^1\dfrac{c_0}{2^1(1)}[/tex]
[tex]k=2\implies n=4\implies c_4=-\dfrac{c_2}4=(-1)^2\dfrac{c_0}{2^2(2\cdot1)}[/tex]
[tex]k=3\implies n=6\implies c_6=-\dfrac{c_4}6=(-1)^3\dfrac{c_0}{2^3(3\cdot2\cdot1)}[/tex]
and so on, with
[tex]c_{2k}=(-1)^k\dfrac{c_0}{2^kk!}[/tex]
So we have
[tex]y(x)=\displaystyle\sum_{k\ge0}c_{2k}(x-2)^{2k}=\sum_{k\ge0}(-1)^k\frac{(x-2)^{2k}}{2^{k-1}k!}[/tex]
Recall that
[tex]e^x=\displaystyle\sum_{n\ge0}\frac{x^k}{k!}[/tex]
The solution we found can then be written as
[tex]y(x)=\displaystyle2\sum_{k\ge0}\frac1{k!}\left(-\frac{(x-2)^2}2\right)^k[/tex]
[tex]\implies\boxed{y(x)=2e^{-(x-2)^2/2}}[/tex]
