<s>[INST] Solve $(y^2+2x^2y)dx+(2x^3-xy)dy=0$ Solve $(y^2+2x^2y)dx+(2x^3-xy)dy=0$
My attempt
$Mdx + Ndy$ form
$M = (y^2+2x^2y)$, $N =(2x^3-xy) , M_y=2y+2x^2, N_x=6x^2-y$
I am not getting in of standard forms here...
Another attempt:
$2x^2 d(xy)+y(ydx-xdy)=0$, 
$2d(xy)=yd(y/x)$ Any clue pls???
 [/INST] $$(y^2+2x^2y)dx+(2x^3-xy)dy=0....(i)\\
\implies y^2dx+2x^2ydx+2x^3dy-xydy=0\\
\implies y(ydx-xdy)+x^2(2ydx+2xdy)=0$$
Which is the form of :
$$x^ay^b(mydx+nxdy)+x^{a'}y^{b'}(m'ydx+n'xdy)=0$$
So, comparing both equations we get:
$a=0, b=1, m=1, n=-1, a'=2, b'=0, m'=2, n'=2$
Now,
$$\frac{a+h+1}{m}=\frac{b+k+1}{n}\\
\implies \frac{h+1}{1}=\frac{1+k+1}{-1}\\
\implies h+k+3=0.....(ii)$$
and $$\frac{a'+h+1}{m'}=\frac{b'+k+1}{n'}$$
$$\implies \frac{2+h+1}{2}=\frac{k+1}{2}\\
\implies h-k+2=0.....(iii)$$
Solving equation (ii) & (iii) we get, $h=-\frac{5}{2}, k=-\frac{1}{2}.$
So,I.F.= $x^{-\frac{5}{2}}y^{-\frac{1}{2}}$
Multiplying equation (i) with the I.F. we get,
$$(x^{-\frac{5}{2}}y^\frac{3}{2}dx+2x^{-\frac{1}{2}}y^\frac{1}{2})dx+(2x^\frac{1}{2}y^{-\frac{1}{2}}-x^{-\frac{3}{2}}y^\frac{1}{2})dy=0$$
Therefore the solution is:
$$\int{Mdx}+\int{(\text{terms of N not containing x})dy}=c\\
\implies \int{x^{-\frac{5}{2}}y^\frac{3}{2}dx}+2\int{x^{-\frac{1}{2}}y^\frac{1}{2}dx}=c\\
\implies -\frac{2}{3}x^{-\frac{3}{2}}y^\frac{3}{2}+4(xy)^\frac{1}{2}=c\\
\implies 4(xy)^\frac{1}{2}-\frac{2}{3}\left(\frac{y}{x}\right)^\frac{3}{2}=c$$
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