Find the solution of given differential equation
$y d x - x d y + 3 x^{2} y^{2} e^{x^{3}} d x = 0$

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Solution

$\frac{y d x - x d y}{y^{2}} = - 3 x^{2} e^{x^{3}} d x$
$d \frac{x}{y} = - 3 x^{2} e^{x^{3}} d x$
On RHS, Let $x^{3} = t$
$3 x^{2} d x = d t$
Now integrate,
$\frac{x}{y} = \int - e^{t} d t = - e^{t} + C$
$\frac{x}{y} = - e^{x^{3}} + C$

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