Find a first integral of an ODE system

Question

I have the system

$$ \left\{\begin{aligned} \dot x & = 2xy \\ \dot y & = x+y^2 \\ \end{aligned}\right. $$

and I need to find a first integral $H$ of the system. This is easy if the equation $$\left( x + y^2 \right) {\rm d} x - 2 x y \, {\rm d} y = 0$$ is exact, i.e., if the divergence of the field is $0$. In this case, the divergence is $4y$ so I need to find an integrating factor $\mu(x,y) $, but I don't know how to do it, because it doesn't work if I use that $\mu$ is just a function of $x$ or $y$. Any help is welcome!

Answer 1

Collect terms in the differential form and combine them into differentials of increasingly complex functions to reduce the number of terms, $$ 0=\frac12d(x^2)+y^2dx - xd(y^2)=\frac12d(x^2)-x^2d(\frac{y^2}x). $$ With $u=x^2$ and $v=\frac{y^2}{x}$ this reads as $$ 0=\frac12du-udv=-u\,d\!\left(-\frac12\ln u+v\right), $$ which gives $F=-\frac12\ln u+v=-\ln|x|+\frac{y^2}x$ as first integral. Test by differentiating along a solution $$ \frac{dF}{dt}=-\frac{\dot x}{x}-\frac{y^2\dot x}{x^2}+\frac{2y\dot y}{x} =-2y-\frac{2y^3}{x}+\frac{2y(x+y^2)}{x}=0 $$

Answer 2

$$\frac{dy}{dx}=\frac{x+y^2}{2xy}$$ $$2y\frac{dy}{dx}=1+\frac{y^2}{x}$$ Let $Y=y^2$ $$\frac{dY}{dx}=1+\frac{Y}{x}$$ This is a first order linear ODE easy to solve. $$Y=x\ln|x|+cx$$ $$y=\pm\sqrt{x\ln|x|+cx}$$