Solve the differential equation $x^2\frac{d^2y}{dx^2}+(x^3-x)\frac{dy}{dx}-3y=0$ $$x^2\frac{d^2y}{dx^2}+(x^3-x)\frac{dy}{dx}-3y=0$$
$$\sum_{n=0}^{\infty}(n+r)(n+r-2)c_nx^{n+r}+\sum_{n=2}^{\infty}(n+r-2)c_{n-2}x^{n+r}-3\sum_{n=0}^{\infty}c_nx^{n+r}=0$$
$$r(r-2)c_0x^r+(r^2-1)c_1x^{1+r}-3c_0x^r-3c_1x^{r+1}+\sum_{n=2}^{\infty}[(n+r)(n+r-2)c_n+(n+r-2)c_{n-2}-3c_n]x^{n+r}=0$$
The indicial equation is
$$(r-3)(r+1)=0$$
We have two possibility
Deal with the larger value $3$ first,
$$c_n=-\frac{(n+1)c_{n-2}}{n^2+4n}$$
Taking values for $n\geq2$
$$c_2=-\frac{c_0}{4}$$
$$c_3=0$$
$$c_4=-\frac{5c_2}{32}$$
$$c_5=0$$
$$y_1(x)=C_1x^3(1-\frac{x^2}{4}+\frac{5x^4}{128}+...)$$
The second possibility $r=-1$
$$c_n=-\frac{(n-3)c_{n-2}}{n^2-4n},n\neq4$$
we know that $c_4 is an arbitrary constant
We know that $c_4$ will make $c_2=0$ from the above equation
$$c_2=-\frac{c_0}{4}$$
We see that it contradicts our initial assumption that $c_0\neq0$
$$c_3=0$$
$$c_5=-\frac{2c_3}{5}=0$$
$$c_6=-\frac{c_4}{4}$$
$$c_7=0$$
$$c_8=-\frac{5c_6}{32}$$
$$y_2=(x^{-1})c_4(x^4-\frac{x^6}{4}+\frac{5x^8}{128}+...)$$
Obviously they are two sides of the same coin.
How to find another linearly independent solution $y_2$. Is there a closed form for the above series?
The thing seems very messy.

Wolfram Alpha gives the solutions as
$$
y_1(x) = x e^{-x^2/4}~I_1(x^2/4) = \frac{x^3}{8} \left( 1 - \frac{x^2}{4} + \frac{5 x^4}{128} - \frac{7 x^6}{1536} + \dots\right)
$$
which is the one you found in the expansion. The other solution diverges at $x=0$ and is
$$
y_2(x) = x e^{-x^2/4}~K_1(x^2/4)  
$$
$I_1$ and $K_1$ are modified Besselfunctions.
