The solution of the differential equation $x c o s y d y = (x e^{x} l o g x + e^{x}) d x$ on is
[DSSE 1988]

$s i n y = \frac{1}{x} e^{x} + c$

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s i n y + e^{x} l o g x + c = 0

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$s i n y = e^{x} l o g x + c$

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None of these

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Solution

The correct option is C
$s i n y = e^{x} l o g x + c$

$x c o s y d y = (x e^{x} l o g x + e^{x}) d x \Rightarrow c o s y d y = (e^{x} l o g x + \frac{e^{x}}{x}) d x$
On integrating, sin $y = e^{x} l o g x + c .$

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[DSSE 1988]

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