Solving a nonlinear ODE $$
 y''-i(\sin(x)y)'-i\omega y-\lvert y \rvert^2y'=0 , \quad y\rvert_{x=0}=0 \quad y'\rvert_{x=0}=0
$$
Hello, im looking for advice on how to solve this equation, im intrested in knowing what possible values of $\omega$ could be. Ive tried to get a solution by using the WKBJ method by assuming that $y$ is
\begin{equation}
y=\psi(x)\exp(iS(z))
\end{equation}
And then subbing this solution into the above equation and then assuming that both $\psi$ and $S$ are real then having seperating that equation into real and imaginary parts but then I have no idea what to do with the nonlinear $y$ term. Is there any methods or ways anyone could suggest on how to approach this problem?
I also tried to find a solution to this equation by assuming that $y$ is real and splitting the problem into real and imaginary part. The imaginary part of that equation is separable and gives a solution but how then do i use the real part?
Thank you.
Update:
So i tried another way solving this equation i haven't tried implementing the boundary conditions yet but does this seem like a reasonable approach?
$$
 y''-i(\sin(x)y)'-i\omega y-\lvert y \rvert^2y'=0 , \quad y\rvert_{x=0}=0 \quad y'\rvert_{x=0}=0
$$
I then split this equation in to real and imaginary parts:
$$
\textrm{real: } y'' - \lvert y\rvert^2 y' = 0\\
\textrm{Imaginary: } \omega y - (sin(x)y)' = 0 
$$
The imaginary part can be solved to get
$$
y' = \dfrac{\omega - cos(x)}{sin(x)}y \\
$$
which directly integrates to
$$
\lvert y \rvert = \exp\left(\int\dfrac{\omega - cos(x)}{sinx(x)} \right)
$$
I can rearrange the real part and integrate it to find that
$$
\lvert y' \rvert = \exp\left(\int \lvert y\rvert^2 dx\right)
$$
This integral
$$
\int \lvert y\rvert^2 dx = \dfrac{2sin(x)}{\omega - cos(x)}\exp\left(2\int\dfrac{\omega - cos(x)}{sin(x)}dx\right)
$$
combining all these together i get that
$$
\lvert y \rvert = \left| \dfrac{sinx(x)}{\omega - cos(x)}\right|\exp\left(-\dfrac{2sin(x)}{\omega - cos(x)}\exp\left(2\int \dfrac{\omega-cos(x)}{sin(x)} dx\right)\right) 
$$
The solution i found looks really messy so i cant help but feeling like ive done some bad maths somehow does this seem reasonable?
Thanks to Gribouillis for the suggestion, this equation originally had a $dy/dt$ term but im looking for a steady state solution, numerical solutions found from timestepping were oscilatory so thats where the $\omega$ was coming from, i was hoping there was some sort of expression i could get from looking at the steady state.
Thanks again to anyone who can help.
 A: This is a partial solution.
Let $y(x)=r(x)\exp\left(i \theta(x)\right)$ where $r(x)$ and $\theta(x)$ are real functions. For simplicity, we omit the $x$ argument and replace $y$ with $r e^{i\theta}$ in the main equation, and then divide the result by $y$ to get:
$$r''/r+2i \theta' r'/r-(\sin x) r'/r-r^2+i \theta''-(\theta')^2 -i\theta' \sin x-\cos x -i\omega=0 $$
with initial conditions
$$\cases{r(0)=0 \\ r'(0)+i\theta'(0)=0}$$
Now both the real and imaginary parts must be zero:
$$\cases{r''-r'\sin x-r \cos x-(\theta')^2 r-r^3=0\\
\theta''+\theta'\left(2r'/r  -\sin x\right)-\omega=0}\tag{*}\label{*}$$
with initial conditions
$$\cases{r(0)=r'(0)=0\\ \theta'(0)=0}$$
By letting $\rho(x)=\theta'(x)$ the second equation becomes a simple first-order ODE:
$$\rho'+\rho\left((2\ln r)'-\sin x\right)=\omega,\qquad \rho(0)=0$$
To solve this, let
$$h(x)=e^{\int (2\ln r)'-\sin u\;du}=r^{2}e^{\cos x}$$
then
$$\theta'(x)=\rho(x)=r^{-2}e^{-\cos x}\left(\omega\int_0^x h(u)du+c\right)\tag{**}\label{**}$$
According to the initial conditions of $\eqref{*}$ we have $c=0$ (I am not sure about this). Anyway, put the above into the first ODE of $\eqref{*}$ to get a highly complex and likely unsolvable equation.
