# Solution to a complex first order cubic nonlinear ODE

I'm attempting to solve the differential equation $$\left(i\omega+\sigma\right)\dot{y}+\mu\left|y\right|^2y=0$$ with the initial condition $y(0)=0$ I've tried to use separation of variables which means I have the following integral to solve $$t-t_0=-\frac{i\omega+\sigma}{\mu}\int_{y_0}^{y}\frac{dz}{\left|z\right|^2z}.$$ I guess the general solution to this is $$\left|y\right|^2=y_0^2\frac{i\omega+\sigma}{y_0^2\mu (t-t_0)+i\omega+\sigma},$$ but I know somewhere in there something isn't quite right (maybe because uniqueness isn't guaranteed). Does someone know of a different way to find the solution(s) to this equation for y instead of $|y|^2$?

For simplicity, we denote $a+ib=\mu/(\sigma+i\omega)$. Assume $y$ is not the trivial solution $y=0$. Letting $y(t)=\rho(t) e^{i\theta(t)}$ be its polar representation, your equation $$\dot y+(a+ib)|y|^2y=0$$ becomes $$e^{i\theta}(i\rho\dot \theta+\dot\rho+(a+ib)\rho^3)=0.$$ Equating real and imaginary parts to zero gives a system for $\rho(t)$ and $\theta(t)$: \begin{align} \dot\rho+a\rho^3&=0,\\ \dot\theta+b\rho^2&=0. \end{align} The first equation is decoupled, so the second equation is linear in $\theta$ once $\rho$ is found.