If $k x y = e^{x / y}$ then show that its differential equation is $(x y + y^{2}) d x + (x y - x^{2}) d y = 0$

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Solution

$k x y = e^{x / y} . . . . . . (i)$
Differentiating both sides,
$k (y d x + x d y) = e^{x / y} (\frac{y d x - x d y}{y^{2}})$
$\Rightarrow \frac{e^{x / y}}{x y} (y d x + x d y) = e^{x / y} (\frac{y d x - x d y}{y^{2}})$ (using (i))
$\Rightarrow y (y d x + x d y) = x (y d x - x d y)$
$\Rightarrow (y^{2} - x y) d x + (x y + x^{2}) d y = 0$
which is the required differential equation of the given curve.

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