I consider the followin Landau-Stuart equation, $$\frac{dA}{dt} = \omega A - \frac{\lambda}{2}\left|A\right|^2A,\quad\omega\in\mathbb{C},\,\lambda>0,$$ where $A$ is the complex amplitude and $\omega=i\omega_0+\Omega$. I would like to study all the possible solutions of this equation.
To do it, I use polar representation of $A$, $A=R\exp(i\phi)$ and write down $$\frac{d\phi}{dt}=\omega_0,\quad \frac{dR}{dt}=\left(\Omega-\frac{\lambda}{2}R^2\right)R,$$ where the first equation is not interesting. I am interested in the specific case $\Omega=\lambda/2$, so $$\frac{dR}{dt}=\frac{\lambda}{2}\left(1-R^2\right)R.$$ It is easy to solve this equation and write down the explicit solution $R=R(t)$ with a given initial condition. Indeed, let $R^2=Q$, so $$\frac{dQ}{(1-Q)Q}=\lambda dt,$$ $$\frac{dQ}{Q}-\frac{dQ}{1-Q}=\lambda dt,$$ $$\ln Q - \ln (1-Q)=\lambda t - \lambda t_0,$$ where $t_0$ can be determined from the initial conditions. Next, $$\ln Q - \ln (1-Q) = 2\text{arctanh}\,Q,$$ $$R(t)=\frac{1}{\sqrt{2}}\left(1+\tanh[\lambda(t-t_0)]\right)^{1/2}.$$ There is one more way (noticed by K.defaoite and also mentioned in wiki): the eq. becomes linear for $P=R^{-2}$, $$\frac{dP}{dt}=-\frac{2}{R^3}\frac{dR}{dt}\rightarrow \frac{dP}{dt}=\lambda-\lambda P,$$ which has the solution $$P(t) = 1+ \exp(-\lambda t)P(0)\rightarrow R(t)=\left(1+e^{-\lambda t}(R(0)^{-2}-1)\right)^{-1/2}$$
However, I have a vague feeling that there are more different solutions.
I have searched the literature but almost all papers are devoted to consideration of the more complicated Ginzburg-Landau equation (with spatial derivatives). Can anyone recommend the relevant references?