Is there any help to solve this equation even I know it's inexact but I couldn't find the integrating factor
$$ {(}{y}^{5}{-}{2}{xy}{)}{dx}{+}{(}{3}{x}^{2}{+}{xy}^{4}{)}{dy}{=}{0} $$ If there any explanation to calculate the integrating factor I will be so thankful
On
Considering the differential equation $${(}{y}^{5}{-}{2}{xy}{)}\,x'{+}{(}{3}{x}^{2}{+}{xy}^{4}{)}{=}{0}\tag1$$ let first $x=\frac z y$ to get $$5z^2+y(y^5-2z)z'=0 \tag 2$$ Now, let $z=t+\frac {y^5}2$ to get $$-2 y t t'+5 t^2+\frac{5 y^{10}}{4}=0 \tag 3$$ Now, let $u=t^2$ to get$$-yu'+5u +\frac{5 y^{10}}{4}=0 \tag 4$$ which is separable leading to $$u=c_1 y^5+\frac{y^{10}}{4}\tag 5$$ which, finally seems to give $$x=\frac{y^5\pm\sqrt{y^5 \left(c_1+y^5\right)}}{2 y}=\frac 12\left( y^4\pm \sqrt{y^3 \left(c_1+y^5\right)}\right)\tag 5$$
If you use squaring, you would end with an quartic equation in $y$ $$ x y^4+c y^3- x^2=0\tag 6$$ yielding to the four (nightmares !) solutions shown by Wolfram Alpha if you start with ${(}{y}^{5}{-}{2}{xy}{)}{+}{(}{3}{x}^{2}{+}{xy}^{4}{)}\,y'{=}{0}$.
$$(y^5-2xy)dx+(3x^2+xy^4)dy=0$$ $$(y^5dx+xy^4dy)+(-2xydx+3x^2dy)=0$$ $$y^4d(xy)+(-2xydx+3x^2dy)=0$$ $$d(xy)+(-2xy^{-3}dx+3x^2y^{-4}dy)=0$$ $$d(xy)+[-y^{-3}d(x^2)-x^2d(y^{-3})]=0$$ $$d(xy)-d(x^2y^{-3})=0$$ $$xy-x^2y^{-3}=C$$