I came across the following riccati equation:
$$\frac{d}{dx}y(x)=Ae^{-i \lambda x}\cos(\lambda x)y(x)^2+Ae^{i \lambda x}\cos(\lambda x)$$ with $$y(0)=0$$
I am not very experienced with these kind of problems and couldn't find any hints in Handbooks on ODEs, but the numerical solution suggests that for: $$\lambda \gg A$$
the function is well approximated by:
$$e^{\frac{i}{2}\lambda x}\tan(\frac{2A \sin(\frac{1}{2}\lambda x)}{\lambda})$$
I was wondering if anyone came across similar problems and if there is any hope of obtaining analytic solutions?
$$\frac{d}{dx}y(x)=Ae^{-i \lambda x}\cos(\lambda x)y(x)^2+Ae^{i \lambda x}\cos(\lambda x)$$ In order to simplify the writing, let $g(x)= Ae^{-i \lambda x}\cos(\lambda x)$ and $h(x)= Ae^{i \lambda x}\cos(\lambda x) $. $$\frac{d}{dx}y(x)=g(x)y(x)^2+h(x)$$ Change of function : $\quad y(x)= -\frac{u'}{g\:u}$ $$y'=-\frac{u''}{g\:u}+\frac{(u')^2}{g\:u^2}+\frac{u'g'}{g^2\:u}=g\:\left(-\frac{u'}{g\:u} \right)^2+h$$ $$u'' -\frac{g'}{g}u'+h\:g\:u=0$$ This is a linear second order ODE.
Generally, it is easier to solve a linear second order ODE than a non-linear first order ODE because the solution is a linear combination of particular solutions, which is not the case for the non-linear ODEs.
$g'= -\lambda Ae^{-i \lambda x}\left(i\cos(\lambda x)+\sin(\lambda x) \right)$ $$u'' -\frac{-\lambda Ae^{-i \lambda x}\left(i\cos(\lambda x)+\sin(\lambda x) \right)}{Ae^{-i \lambda x}\cos(\lambda x)}u'+\left(Ae^{i \lambda x}\cos(\lambda x) \right)\left(Ae^{-i \lambda x}\cos(\lambda x) \right)u=0$$ $$\cos(\lambda x)u'' + \left(i\cos(\lambda x)+\sin(\lambda x)\right)u'+A^2e^{-2i \lambda x}\cos^3(\lambda x)u=0$$ $$\cos(\lambda x)u'' + ie^{-i\lambda x}u'+A^2\cos^3(\lambda x)u=0$$ Unfortunately, this second order linear ODE seems not easy to solve.