<s>[INST] 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?
 [/INST] $$
(3x^2y^4 + 2xy)dx + (2x^3y^3 - x^2)dy = 0
$$
$$
\implies 3x^2y^4dx + 2x^3y^3dy + 2xydx - x^2dy = 0 \tag{$1$}
$$
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.
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