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Given an equation:

$$xdy-2ydx+xy^2(2xdy+ydx)=0$$

I've tried a lot to group the dx and dy multipliers, but haven't proceed to at least go further in my solution, so I need help.

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  • $\begingroup$ are $x$ and $y$ independent variables, or how they are related? $\endgroup$
    – Masacroso
    Oct 29 '18 at 14:19
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Try to use $$ 2xdy+ydx = \frac1yd(xy^2) $$ and $$ xdy-2ydx = x^3\,d(x^{-2}y) $$ and express the remaining coefficients in terms of $u=xy^2$ and $v=x^{-2}y$.

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Hint: One way is finding integrating factor, so with $$(-2y+xy^3)dx+(x+2x^2y^2)dy=0$$ $M=-2y+xy^3$ and $N=x+2x^2y^2$, then $$p(z)=\frac{{M}_{y}-{N}_{x}}{Ny-Mx}=\frac{-(3+xy^2)}{xy(3+xy^2)}=\dfrac{-1}{xy}=\dfrac{-1}{z}$$ therfore $$I=e^{\int p(z)\ dz}=\dfrac{1}{z}=\color{blue}{\dfrac{1}{xy}}$$ is integrating factor.

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