# 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 $r=-1$

$$c_n=-\frac{(n-3)c_{n-2}}{n^2-4n},n\neq4$$