power series method to find differential solution use power series method to find solution of $(x^2+2x)y''+(x^2-2)y'-(2x+2)y=0$ i have tried $y=\sum_{0}^{\infty}{am*x^{m+r}}$ normalize the power to $m+r-1$ ,for m=0, $(2r^2-4r)a_0=0$, $r=0,2$ or a0=0 , m=1 , $(r^2-r+2)a_0=0$
m≥2
$2(m+r)(m+r-2)a_m+[(m+r-1)(m+r-2)+2]a_m-1+(m+r-4)a_m-2=0$
then i don't know the next step to find the regular pattern, can anyone help me?

 A: The approach suggested to you, only implemented more rigorously, is the following. Assume that $y(x)=\sum\limits_na_nx^{n+r}$ solves the ODE, then 
$$y'(x)=\sum\limits_n(n+r+1)a_{n+1}x^{n+r},\quad y''(x)=\sum\limits_n(n+r+2)(n+r+1)a_{n+2}x^{n+r},
$$
and, for every integer $i$, 
$$
x^iy(x)=\sum\limits_na_{n-i}x^{n+r},\quad x^iy'(x)=\sum\limits_n(n+r+1-i)a_{n+1-i}x^{n+r},
$$
and 
$$
x^iy''(x)=\sum\limits_n(n+r+2-i)(n+r+1-i)a_{n+2-i}x^{n+r}.
$$
Hence, the first term of the ODE becomes
$$
(x^2+2x)y''(x)=\sum\limits_n(n+r)(n+r-1)a_{n}+2(n+r+1)(n+r)a_{n+1}
x^{n+r},
$$
and similarly for the other terms. Putting everything together yields a relation between $a_n$, $a_{n+1}$ and $a_{n-1}$, to be solved. The initial condition comes from the terms of lowest degree. These are the initial term of $2xy''-2y'$, thus
$$
2r(r-1)a_0-2ra_0=0,
$$
which yields $r=0$ or $r=2$, hence everything is in place to solve the ODE.

Another approach, more direct, is to note that, considering $a(x)=x^2+2x$, the ODE becomes
$$
a(x)y''(x)+(a(x)-a'(x))y'(x)-a'(x)y(x)=0,
$$
that is, considering $z(x)=y'(x)+y(x)$,
$$
a(x)z'(x)-a'(x)z(x)=0.
$$
Thus, there exists some constant $\lambda$ such that
$$z(x)=\lambda a(x).
$$
Now, $a(x)=b(x)+b'(x)$ where $b:x\mapsto x^2$ hence there exists some $\lambda$ such that
$$
y'(x)+y(x)=\lambda (b(x)+b'(x)),
$$
that is,
$$
(y-\lambda b)'(x)=-(y-\lambda b)(x),
$$
hence there exists some $\mu$ such that
$$
(y-\lambda b)(x)=\mu\mathrm e^{-x},
$$
that is,
$$
y(x)=\lambda x^2+\mu\mathrm e^{-x}.
$$
Note that the case $r=0$ in the first approach corresponds to $\mu\ne0$ while the case $r=2$ corresponds to $\mu=0$ and $\lambda\ne0$.
