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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?

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3 Answers 3

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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 $1/y^2$.

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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.

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

$$ \implies 3x^2y^4dx + 2x^3y^3dy + 2xydx - x^2dy = 0 \tag{$1$} $$

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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