Consider the ODE
$$u' = \bar{u}^2, u(0) = u_0, \quad {\rm Im}(u^3) = c,$$ where $u: \mathbb{R} \to \mathbb{C}$ and $c$ is a real given constant. I'm looking for an explicit formula for the solution !
I tried the following: Using the equation, we can write
$$\bar u 'u - u'\bar u = u^3 - \bar u^3 = \frac 12 ic.$$
Hence, writing $u = re^{i\theta},$ we find
$$r(r'+ i \theta ' r)- r(r'-i\theta ' r) = \frac 12 ic,$$ which implies that $$\theta'r^2 = \frac 14 c.$$
On the other hand, form the given constraint, we get $$r^3 \sin(3\theta) = c.$$
Thank you for any further hint.
You already found $\quad r^3\sin(3\theta)=c$
HINT :
$$\sin(3\theta)=\frac{c}{r^3} \quad \text{and}\quad \cos(3\theta)=\left(1-\frac{c^2}{r^6}\right)^{1/2}$$ $$u^3=r^3\left(\cos(3\theta)+i\sin(3\theta)\right)=r^3\cos(3\theta)+ic$$ $$u^3=r^3\left(1-\frac{c^2}{r^6}\right)^{1/2}+ic $$ $$u^3=\left(r^6-c^2\right)^{1/2}+ic \tag 1$$ $$(u^3)'=\frac{3r^5\:r'}{\left(r^6-c^2\right)^{1/2}}$$ $u' = \bar{u}^2$
$ u^2u' = u^2\bar{u}^2 = r^4$
$\frac13 (u^3)' = r^2$
$$\frac13\frac{3r^5\:r'}{\left(r^6-c^2\right)^{1/2}}=r^4$$ $$r'=\frac{\left(r^6-c^2\right)^{1/2}}{r}$$ This is a separable ODE formally solvable in terms of elliptic functions.
So, you get $r$, then $u^3$ from Eq.$(1)$ from which $u$ can be formally derived.