The solution of the differential equation, $y d x + (x + x^{2} y) d y = 0$ is

log y = c x

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log y - \frac{1}{x y} = c

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\frac{1}{x y} - log y = c

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\frac{1}{x y} + log y = c

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Solution

The correct option is D $log y - \frac{1}{x y} = c$
We have,
$y d x + (x + x^{2} y) d y = 0 y d x = - (x + x^{2} y) d y y d x + x d y = - x^{2} y d y \frac{y d x + x d y}{{(x y)}^{2}} = - \frac{d y}{y} \frac{d (x y)}{{(x y)}^{2}} = - \frac{d y}{y}$

On taking integrating both sides
$\begin{matrix} \frac{- 1}{x y} = - log y + c log y - \frac{1}{x y} = c \end{matrix}$

Hence, this is the answer.

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[EAMCET 2003]

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