Solve $(3x^2y^4 +2xy)dx + (2x^3y^3 - x^2)dy=0$ 
Solve $(3x^2y^4 +2xy)dx + (2x^3y^3 - x^2)dy$  

This is not one of the standard forms and neither is it an exact form. How do I go about doing this question?
 A: I tried an integrating factor of the form $\mu(x)$ and failed, but when I tried a function of $y$ it worked! Your equation becomes exact upon multiplication by $1/y^2$.
A: The rearrangement $y'=\frac{y(3xy^3+2)}{x(1-2xy^3)}$ suggests we should consider $z=xy^3$. Rearranging $z'$ into a result of the form $\int\frac{7dx}{x}=\frac{1-2z}{z(z+1)}dz$, and integrating to $x^7=\frac{kz}{(z+1)^3}$ and rewriting in terms of $x,\,y$, we find $x^3y^3-ky+x^2=0$ with $k$ an integration constant. This is the same result as user1337's method gets, but it doesn't require you to spot which factor to use.
A: $$
(3x^2y^4 + 2xy)dx + (2x^3y^3 - x^2)dy = 0
$$
$$
\implies 3x^2y^4dx + 2x^3y^3dy + 2xydx - x^2dy = 0 \tag{$1$}
$$
We know that:
$$
d\left(\frac{x^2}{y}\right) = \frac{2xydx - x^2dy}{y^2}
$$
Substituting this back into $(1)$:
$$
\implies 3x^2y^4dx + 2x^3y^3dy + d\left(\frac{x^2}{y}\right)y^2 = 0
$$
Assuming $y \not= 0$:
$$
\implies 3x^2y^2dx + 2x^3ydy + d\left(\frac{x^2}{y}\right) = 0 \tag{$2$}
$$
Also:
$$
d(x^3y^2) = 3x^2y^2dx + 2x^3ydy
$$
Substituting this into $(2)$:
$$
\implies d(x^3y^2) + d\left(\frac{x^2}{y}\right) = 0\\
\implies x^3y^2 + \frac{x^2}{y} = c
$$
I don't know if this is formally legal in mathematics (I only have high school mathematics under my belt), but this is how I would have done it.
