I have to solve the following differential equation near ordinary point $x=0:$
$$\left(x^2-1\right)\frac{d^2y}{dx^2}+3x\,\frac{dy}{dx}+xy=0. $$
Proceeding with the method of power series I end up with
$$\sum a_{n-1}x^n+\sum(n(n-1)+3)a_nx^n-\sum(n+2)(n+1)a_{n+2}x^n=0. $$
I do not know how to proceed further.
Partial Solution
We have the DE $$\left(x^2-1\right)y''+3xy'+xy=0. $$ We let $y=\sum a_nx^n,$ so that \begin{align*} y&=\sum_{n=0}^\infty a_nx^n\\ y'&=\sum_{n=1}^\infty n a_n x^{n-1}\\ y''&=\sum_{n=2}^\infty n(n-1)a_nx^{n-2}. \end{align*} Plugging this into the DE yields $$ \underbrace{\sum_{n=2}^\infty n(n-1)a_nx^{n}}_{\text{1st term: } x^2}- \underbrace{\sum_{n=2}^\infty n(n-1)a_nx^{n-2}}_{\text{1st term: } x^0}+ \underbrace{\sum_{n=1}^\infty 3n a_n x^{n}}_{\text{1st term: } x^1}+ \underbrace{\sum_{n=0}^\infty a_nx^{n+1}}_{\text{1st term: } x^1}=0. $$ We break off terms in the second, third, and fourth sums so that the first term is $x^2,$ as in the first sum. Then we re-index. The result I get is as follows: $$-2a_2-6a_3x+3a_1x+a_0x+\sum_{j=0}^\infty\left[a_{j+1}+(j+2)(j+4)a_{j+2}-(j+4)(j+3)a_{j+4}\right]x^{j+2}=0. $$ Equating coefficients of like powers of $x,$ we obtain these three equations: \begin{align*} a_2&=0\\ a_0+3a_1-6a_3&=0\\ a_{j+1}+(j+2)(j+4)a_{j+2}-(j+4)(j+3)a_{j+4}&=0. \end{align*} Note that the second equation is consistent with the third, for $j=-1.$ Without more information, it would appear this is as far as we can go.