# Can't solve a differential equation $xdy-2ydx+xy^2(2xdy+ydx)=0$.

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.

• are $x$ and $y$ independent variables, or how they are related? Oct 29 '18 at 14:19

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 $$v=x^{-2}y$$.
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.