We want a solution in the form
[tex]y=\displaystyle\sum_{n\ge0}a_nx^n[/tex]
with derivatives
[tex]y'=\displaystyle\sum_{n\ge0}(n+1)a_{n+1}x^n[/tex]
[tex]y''=\displaystyle\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n[/tex]
Substituting [tex]y[/tex] and its derivatives into the ODE,
[tex]y''-2xy'+2y=0[/tex]
gives
[tex]\displaystyle\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n-2\sum_{n\ge0}(n+1)a_{n+1}x^{n+1}+2\sum_{n\ge0}a_nx^n=0[/tex]
Shift the index on the second sum to have it start at [tex]n=1[/tex]:
[tex]\displaystyle\sum_{n\ge0}(n+1)a_{n+1}x^{n+1}=\sum_{n\ge1}na_nx^n[/tex]
and take the first term out of the other two sums. Then we can consolidate the sums into one that starts at [tex]n=1[/tex]:
[tex]\displaystyle(2a_2+2a_0)+\sum_{n\ge1}\bigg[(n+2)(n+1)a_{n+2}+(2-2n)a_n\bigg]x^n=0[/tex]
and so the coefficients in the series solution are given by the recurrence,
[tex]\begin{cases}a_0=y(0)\\a_1=y'(0)\\(n+2)(n+1)a_{n+2}=2(n-1)a_n&\text{for }n\ge0\end{cases}[/tex]
or more simply, for [tex]n\ge2[/tex],
[tex]a_n=\dfrac{2(n-3)}{n(n-1)}a_{n-2}[/tex]
Note the dependency between every other coefficient. Consider the two cases,
[tex]k=0\implies n=0\implies a_0=a_0[/tex]
[tex]k=1\implies n=2\implies a_2=-a_0=2^1\dfrac{(-1)}{2!}a_0[/tex]
[tex]k=2\implies n=4\implies a_4=\dfrac{2\cdot1}{4\cdot3}a_2=2^2\dfrac{1\cdot(-1)}{4!}a_0[/tex]
[tex]k=3\implies n=6\implies a_6=\dfrac{2\cdot3}{6\cdot5}a_4=2^3\dfrac{3\cdot1\cdot(-1)}{6!}a_0[/tex]
[tex]k=4\implies n=8\implies a_8=\dfrac{2\cdot5}{8\cdot7}a_6=2^4\dfrac{5\cdot3\cdot1\cdot(-1)}{8!}a_0[/tex]
and so on, with the general pattern
[tex]a_{2k}=\dfrac{2^ka_0}{(2k)!}\displaystyle\prod_{i=1}^k(2i-3)[/tex]
[tex]k=0\implies n=1\implies a_1=a_1[/tex]
[tex]k=1\implies n=3\implies a_3=\dfrac{2\cdot0}{3\cdot2}a_1=0[/tex]
and we would see that [tex]a_{2k+1}=0[/tex] for all [tex]k\ge1[/tex].
So we have
[tex]y(x)=\displaystyle\sum_{k\ge0}\bigg[a_{2k}x^{2k}+a_{2k+1}x^{2k+1}\bigg][/tex]
so that one solution is
[tex]\boxed{y_1(x)=\displaystyle a_0\sum_{k\ge0}\frac{2^k\prod\limits_{i=1}^k(2i-3)}{(2k)!}x^{2k}}[/tex]
and the other is
[tex]\boxed{y_2(x)=a_1x}[/tex]
I've attached a plot of the exact and series solutions below with [tex]a_0=y(0)=1[/tex], [tex]a_1=y'(0)=1[/tex], and [tex]0\le k\le5[/tex] to demonstrate that the series solution converges to the exact one.
