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How can one find the first integral of system $\begin{cases} \dot{x} = x^4+x^3y^3 \\ \dot{y} = -2x^3y -x^2y^4 \end{cases}$?

From this system we can get that $\displaystyle\frac{2ydx+xdy}{y^3} = \displaystyle\frac{ydx+xdy}{-x}$, or $(y^4+2xy)dx+(x^2+xy^3)dy=0$. Then I've tried to find $U(x,y)$, such that $dU = \mu(y^4+2xy)dx+\mu(x^2+xy^3)dy$, but I didn't succeed.

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$$(y^4+2xy)dx+(x^2+xy^3)dy=0$$ $$y^4dx+(ydx^2+x^2dy)+xy^3dy=0$$ $$y^4dx+d(x^2y)+xy^3dy=0$$ $$y^3d(xy)+d(x^2y)=0$$ $$y^3du+dv=0$$

Where $u=xy$ and $v=x^2y$. Note that: $$u^2=(xy)^2=x^2y\times y=vy$$ Deduce $y^3= \dfrac {u^6}{v^3}$. Then the DE is separable(exact): $$u^6du+v^3dv=0$$

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