$(x^{2} y - 2 x y^{2}) d x = (x^{3} - 3 x^{2} y) d y .$

k y^{3} e^{x / y} = x^{2}

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k x^{3} e^{x / y} = x^{2}

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k x^{3} e^{x / y} = y^{2}

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k y^{3} e^{x / y} = y^{2}

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Solution

The correct option is A $k y^{3} e^{x / y} = x^{2}$
$(x^{2} y - 2 x y^{2}) d x = (x^{3} - 3 x^{2} y) d y$
$\Rightarrow \frac{d y}{d x} = \frac{x^{2} y - 2 x y^{2}}{x^{3} - 3 x^{2} y}$
Substitute $y = v x$
$\frac{d y}{d x} = v + x \frac{d v}{d x}$
$v + x \frac{d v}{d x} = \frac{v - 2 v^{2}}{1 - 3 v}$
$\Rightarrow x \frac{d v}{d x} = \frac{v^{2}}{1 - 3 v}$
$\Rightarrow (\frac{1 - 3 v}{v^{2}}) d v = \frac{d x}{x}$
Integrating both sides w.r.t. x, we get
$\frac{- 1}{v} - 3 log v = log x + log k$
$\Rightarrow \frac{- x}{y} - 3 log \frac{y}{x} = log x + log k$
$\Rightarrow k y^{3} e^{x / y} = x^{2}$

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