Find the orbits of the system: $\dot{x} = y +x^2y$, $\dot{y} = 3x + xy^2$

Question

This is my answer. So the orbits is $y = \sqrt{c(1 + x^2) - 3}$?

Answer 1

A method more or less equivalent to yours, but motivated by groups $(1+x^2)$ and $(3+y^2)$.

$\begin{cases} x = \sinh(u)\\ y = \sqrt{3}\sinh(v)\end{cases}\iff \begin{cases} 1+x^2=\cosh(u)^2\\ 3+y^2=3\cosh(v)^2 \end{cases}$

$\begin{cases}\dot x=\cosh(u)\dot u=y\cosh(u)^2\\ \dot y=\sqrt{3}\cosh(v)\dot v=3x\cosh(v)^2 \end{cases}\implies\begin{cases}\dot u = \sqrt{3}\sinh(v)\cosh(u)\\\dot v = \sqrt{3}\sinh(u)\cosh(v)\end{cases}$

$\implies\tanh(v)\dot v=\tanh(u)\dot u$

$\implies \ln(\cosh(v))=\ln(\cosh(u))+cst$

$\implies\cosh(v)=A\cosh(u)$ with $A>0$

So with $c=3A^2$ we get $3+y^2=c(1+x^2)$ too.