First note that [tex]x=0[/tex] is a regular singular point; in particular [tex]x=0[/tex] is a pole of order 1 for [tex]\dfrac2x[/tex].
We seek a solution of the form
[tex]y=\displaystyle\sum_{n\ge0}a_nx^{n+r}[/tex]
where [tex]r[/tex] is to be determined. Differentiating, we have
[tex]y'=\displaystyle\sum_{n\ge0}(n+r)a_nx^{n+r-1}[/tex]
[tex]y''=\displaystyle\sum_{n\ge0}(n+r)(n+r-1)a_nx^{n+r-2}[/tex]
and substituting into the ODE gives
[tex]\displaystyle x\sum_{n\ge0}(n+r)(n+r-1)a_nx^{n+r-2}+2\sum_{n\ge0}(n+r)a_nx^{n+r-1}-x\sum_{n\ge0}a_nx^{n+r}=0[/tex]
[tex]\displaystyle \sum_{n\ge0}(n+r)(n+r-1)a_nx^{n+r-1}+2\sum_{n\ge0}(n+r)a_nx^{n+r-1}-\sum_{n\ge0}a_nx^{n+r+1}=0[/tex]
[tex]\displaystyle \sum_{n\ge0}(n+r)(n+r+1)a_nx^{n+r-1}-\sum_{n\ge0}a_nx^{n+r+1}=0[/tex]
[tex]\displaystyle r(r+1)a_0x^{r-1}+(r+1)(r+2)a_1x^r+\sum_{n\ge2}(n+r)(n+r+1)a_nx^{n+r-1}-\sum_{n\ge0}a_nx^{n+r+1}=0[/tex]
[tex]\displaystyle r(r+1)a_0x^{r-1}+(r+1)(r+2)a_1x^r+\sum_{n\ge2}(n+r)(n+r+1)a_nx^{n+r-1}-\sum_{n\ge2}a_{n-2}x^{n+r-1}=0[/tex]
[tex]\displaystyle r(r+1)a_0x^{r-1}+(r+1)(r+2)a_1x^r+\sum_{n\ge2}\bigg((n+r)(n+r+1)a_n-a_{n-2}\bigg)x^{n+r-1}=0[/tex]
The indicial polynomial, [tex]r(r+1)[/tex], has roots at [tex]r=0[/tex] and [tex]r=-1[/tex]. Because these roots are separated by an integer, we have to be a bit more careful, but we'll get back to this later.
When [tex]r=0[/tex], we have the recurrence
[tex]a_n=\dfrac{a_{n-2}}{(n+1)(n)}[/tex]
valid for [tex]n\ge2[/tex]. When [tex]n=2k[/tex], with [tex]k\in\{0,1,2,3,\ldots\}[/tex], we find
[tex]a_0=a_0[/tex]
[tex]a_2=\dfrac{a_0}{3\cdot2}=\dfrac{a_0}{3!}[/tex]
[tex]a_4=\dfrac{a_2}{5\cdot4}=\dfrac{a_0}{5!}[/tex]
[tex]a_6=\dfrac{a_4}{7\cdot6}=\dfrac{a_0}{7!}[/tex]
and so on, with a general pattern of
[tex]a_{n=2k}=\dfrac{a_0}{(2k+1)!}[/tex]
Similarly, when [tex]n=2k+1[/tex] for [tex]k\in\{0,1,2,3,\ldots\}[/tex], we find
[tex]a_1=a_1[/tex]
[tex]a_3=\dfrac{a_1}{4\cdot3}=\dfrac{2a_1}{4!}[/tex]
[tex]a_5=\dfrac{a_3}{6\cdot5}=\dfrac{2a_1}{6!}[/tex]
[tex]a_7=\dfrac{a_5}{8\cdot7}=\dfrac{2a_1}{8!}[/tex]
and so on, with the general pattern
[tex]a_{n=2k+1}=\dfrac{2a_1}{(2k+2)!}[/tex]
So the first indicial root admits the solution
[tex]y=\displaystyle a_0\sum_{k\ge0}\frac{x^{2k}}{(2k+1)!}+a_1\sum_{k\ge0}\frac{x^{2k+1}}{(2k+2)!}[/tex]
[tex]y=\displaystyle \frac{a_0}x\sum_{k\ge0}\frac{x^{2k+1}}{(2k+1)!}+\frac{a_1}x\sum_{k\ge0}\frac{x^{2k+2}}{(2k+2)!}[/tex]
[tex]y=\displaystyle \frac{a_0}x\sum_{k\ge0}\frac{x^{2k+1}}{(2k+1)!}+\frac{a_1}x\sum_{k\ge0}\frac{x^{2k+2}}{(2k+2)!}[/tex]
which you can recognize as the power series for [tex]\dfrac{\sinh x}x[/tex] and [tex]\dfrac{\cosh x}x[/tex].
To be more precise, the second series actually converges to [tex]\dfrac{\cosh x-1}x[/tex], which doesn't satisfy the ODE. However, remember that the indicial equation had two roots that differed by a constant. When [tex]r=-1[/tex], we may seek a second solution of the form
[tex]y=cy_1\ln x+x^{-1}\displaystyle\sum_{n\ge0}b_nx^n[/tex]
where [tex]y_1=\dfrac{\sinh x+\cosh x-1}x[/tex]. Substituting this into the ODE, you'll find that [tex]c=0[/tex], and so we're left with
[tex]y=x^{-1}\displaystyle\sum_{n\ge0}b_nx^n[/tex]
[tex]y=\dfrac{b_0}x+b_1+b_2x+b_3x^2+\cdots[/tex]
Expanding [tex]y_1[/tex], you'll see that all the terms [tex]x^n[/tex] with [tex]n\ge0[/tex] in the expansion of this new solutions are already accounted for, so this new solution really only adds one fundamental solution of the form [tex]y_2=\dfrac1x[/tex]. Adding this to [tex]y_1[/tex], we end up with just [tex]\dfrac{\sinh x+\cosh x}x[/tex].
This means the general solution for the ODE is
[tex]y=C_1\dfrac{\sinh x}x+C_2\dfrac{\cosh x}x[/tex]
