I am solving this second order ODE:
$$-gdf''+(-w^2 + iw^2(mx+b))f(x)=0$$
I used power series expansion about $x=0$. Assuming $f(x) = \sum_{n=0}^{\infty} a_n x^n$, and $f''(x) = \sum_{n=2}^{\infty} n(n-1)a_nx^{n-2}$ and substituting these to the ODE, I arrived at the following:
$$\sum_{n=2}^{\infty} -gd n(n-1)a_n x^{n-2} - \sum_{n=0}^{\infty} w^2 a_n x^n + \sum_{n=0}^{\infty} iw^2 ma_nx^{n+1} + \sum_{n=0}^{\infty} iw^2ba_nx^n = 0 $$
Now, expressing everything in terms of $x^n$ I got:
$$\sum_{n=0}^{\infty} -gd(n+2)(n+1)a_{n+2}x^n - \sum_{n=0}^{\infty} w^2 a_n x^n + \sum_{n=1}^{\infty} iw^2 ma_{n-1}x^n + \sum_{n=0}^{\infty} iw^2ba_nx^n = 0$$
Changing the lower index to $n=1$:
$$-2gda_2 + \sum_{n=1}^{\infty} -gd(n+2)(n+1)a_{n+2}x^n - w^2a_0 - \sum_{n=1}^{\infty} w^2 a_n x^n + \sum_{n=1}^{\infty} iw^2ma_{n-1}x^n + iw^2ba_0 + \sum_{n=1}^{\infty} iw^2ba_n x^n = 0$$
Equating all coefficients to $0$ we have:
$$-2gda_2 - gd(n+2)(n+1)a_{n+2} - w^2a_0 - w^2a_n + iw^2ma_{n-1} + iw^2ba_0 + iw^2ba_n = 0$$
Since $-2gd, -w^2, iw^2 \neq 0$, then $a_2, a_0 = 0$ (is this correct?), so the equation above reduces to:
$$ -gd(n+2)(n+1)a_{n+2} + (-w^2 + iw^2b)a_n + iw^2ma_{n-1} = 0$$
which is now my recursion formula but involves 3 unknowns $(a_{n+2}, a_n,$ and $a_{n-1})$. Any help or tips to proceed?
First think that $a_k\in \mathbb{C}$. Reserve $a_0, a_1$ for four real initial conditions and then proceed
$$ a_{k+2} = \frac{(ib-1)w^2 a_k+im w^2 a_{k-1}}{gd(k+2)(k+1)} $$