It looks like you're trying to solve the ODE via Frobenius' method. It's not immediately clear where the solution is to be centered, but I think we can safely assume it's to be taken about [tex]x=0[/tex]. Dividing through by [tex]x[/tex], we get
[tex]y''+\dfrac{y'}x-\dfrac yx=0[/tex]
and we observe that [tex]x=0[/tex] is a regular singular point.
Now, taking a solution of the form
[tex]y=\displaystyle\sum_{n\ge0}a_nx^{n+r}[/tex]
[tex]\implies y'=\displaystyle\sum_{n\ge0}(n+r)a_nx^{n+r-1}[/tex][tex]\implies y''=\displaystyle\sum_{n\ge0}(n+r)(n+r-1)a_nx^{n+r-2}[/tex]
and substituting into the ODE gives
[tex]\displaystyle\sum_{n\ge0}(n+r)(n+r-1)a_nx^{n+r-1}+\sum_{n\ge0}(n+r)a_nx^{n+r-1}-\sum_{n\ge0}a_nx^{n+r}=0[/tex]
[tex]\displaystyle\sum_{n\ge0}(n+r)^2a_nx^{n+r-1}-\sum_{n\ge0}a_nx^{n+r}=0[/tex]
[tex]\displaystyle r^2a_0x^{r-1}+\sum_{n\ge1}(n+r)^2a_nx^{n+r-1}-\sum_{n\ge0}a_nx^{n+r}=0[/tex]
[tex]\displaystyle r^2a_0x^{r-1}+\sum_{n\ge1}(n+r)^2a_nx^{n+r-1}-\sum_{n\ge1}a_{n-1}x^{n+r-1}=0[/tex]
[tex]\displaystyle r^2a_0x^{r-1}+\sum_{n\ge1}\bigg((n+r)^2a_n-a_{n-1}\bigg)x^{n+r-1}=0[/tex]
The indicial polynomial admits only one root (of multiplicity two) at [tex]r=0[/tex], so in fact a regular power series solution will exist. So in fact the ODE above reduces to
[tex]\displaystyle\sum_{n\ge1}\bigg(n^2a_n-a_{n-1}\bigg)x^{n-1}=\sum_{n\ge0}\bigg((n+1)^2a_{n+1}-a_n\bigg)x^n=0[/tex]
leaving us with the recurrence
[tex]a_{n+1}=\dfrac{a_n}{(n+1)^2}[/tex]
We can solve this by recursively substituting the right hand side:
[tex]a_n=\dfrac{a_{n-1}}{n^2}=\dfrac{a_{n-2}}{n^2(n-1)^2}=\cdots=\dfrac{a_0}{(n!)^2}[/tex]
and so one of the solutions will be
[tex]y_1(x)=\displaystyle a_0\sum_{n\ge0}\frac{x^n}{(n!)^2}[/tex]
Now the literature on the Frobenius method when the indicial polynomial has repeated roots suggests a second solution will take the form
[tex]y_2(x)=Cy_1\ln x+\displaystyle\sum_{n\ge0}b_nx^n[/tex]
but it's not clear to me *why* this works.
