Let $a, b \in \mathbb{R}$ and consider the system of ODEs $$\left\{ \begin{matrix} x' = ax+ by^2, & x(0) = x_0\\ y' = ay +bx^2, & y(0) = y_0 \end{matrix} \right. $$
where $x, y :\mathbb{R} \to \mathbb{C}.$ I wonder whether there is, in general, $F$ such that $F(x,y) =0.$ For example, if $a =0,$ and $b \neq 0,$ we can write $$\frac{dx}{dy} = \frac{y^2}{x^2},$$ which implies that $$x^3- y^3 + c = 0$$ for some constant $c.$
Thank you for any hint.
$$\left\{ \begin{matrix} x' = ax+ by^2, & x(0) = x_0\\ y' = ay +bx^2, & y(0) = y_0 \end{matrix} \right.$$ Multiply first equation by $x^2$ and second equation by $y^2$ $$\left\{ \begin{matrix} x^2x' = ax^3+ by^2x^2 & . \\ y'y^2 = ay^3 +bx^2y^2 & . \end{matrix} \right.$$ $$x'x^2-ax^3=y'y^2-ay^3$$ $$\frac 13 (x^3)'-ax^3=\frac 13 (y^3)'-ay^3$$ $$\frac 13( x^3-y^3)'=a(x^3-y^3)$$ You can integrate, substitute $z=x^3-y^3$ $$z'=3az$$ $$ \implies x^3(t)=y^3(t)+Ke^{3at}$$
You can try to integrate the differential equation with exactness $$\frac {dx}{dy}=\frac {ax+by^2}{ay+bx^2}$$ $$({ay+bx^2})dx-({ax+by^2})dy=0$$ $$Pdx+Qdy=0 \implies \partial_y P=a, \partial_x Q=-a $$ $$\implies \partial_y P -\partial_x Q=2a$$ Try an integrating factor $z=x^3-y^3$
And the formula for the integratinf factor $z=z(x,y)$ is $$\boxed {\frac {d\mu}{\mu}=\frac {\partial_y P- \partial_x Q}{Q \partial_x z -P \partial_y z}}$$ $$\frac {d\mu}{\mu}=\frac {2a dz}{Q \partial_x z -P \partial_y z}$$ $$\frac {d\mu}{\mu}=-\frac {2 dz}{3z}$$ $$\mu =\frac 1 {\sqrt[3]{z^2}}$$ $$\mu (x,y) =\frac 1 {\sqrt[3]{(x^3-y^3)^2}}$$