I want to solve the above ODE just using power series centered at $x_0=-3$(not using the Frobenius method).I give some answers below, but the recursion formula looks strange, because there are three coefficients. I don't know how to proceed. Your help would be highly appreciated!
$$y=\sum_{n=0}^{\infty}a_{n}(x+3)^{n}$$ $$y'=\sum_{n=1}^{\infty}a_{n}n(x+3)^{n-1}$$ $$y''=\sum_{n=2}^{\infty}a_{n}n(n-1)(x+3)^{n-2}$$ $$\sum_{k=2}a_{n}n(n-1)(x+3)^{n-2}+\frac{1}{x}\sum_{k=1}a_{n}n(x+3)^{n}+\sum_{n=0}a_{n}(x+3)^{n}=0$$ Compare the coefficient, $$a_{n+2}(n+2)(n+1)x^{n}+a_{n+1}(n+1)x^{n}+a_{n}x^{n}=0$$ $$\implies a_{n+2}(n+2)(n+1)+a_{n+1}(n+1)+a_{n}=0$$
We have this recursion: $$ (n+2)!a_{n+2}+(n+1)!a_{n+1}+n!a_{n}=0 $$
Take $b_n=n!a_n$. Then we get $b_{n+2}+b_{n+1}+b_n=0$.
You can use the characteristic equation: $$ x^2+x+1=0 $$ Solve $x_1=\frac{1}{2}(-1+i\sqrt{3})$ and $x_2=\frac{1}{2}(-1-i\sqrt{3})$.
So $$ b_n=Ax_1^{n-2}+Bx_2^{n-2} $$