If $x \frac{d y}{d x} = y (l o g y - l o g x + 1)$ , then the solution of the equation is

y l o g (\frac{x}{y}) = c x

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x l o g (\frac{y}{x}) = c y

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l o g (\frac{y}{x}) = c x

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l o g (\frac{x}{y}) = c y

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Solution

The correct option is C $l o g (\frac{y}{x}) = c x$
$\frac{x d y}{d x} = y (l o g y - l o g x + 1)$
$\frac{d y}{d x} = \frac{y}{x} (l o g \frac{y}{x} + 1)$
Put y=vx
$\frac{d y}{d x} = v + \frac{x d v}{d x} \Rightarrow v + \frac{x d v}{d x} = v (l o g v + 1)$
$\frac{x d v}{d x} = v l o g v \Rightarrow \frac{d v}{v l o g v} = \frac{d x}{x}$
Put log v=z
$\frac{1}{v} = d z \Rightarrow \frac{d z}{z} = \frac{d x}{x} l n z = l n x + l n c$
$z = c x o r l o g v = c x o r l o g (\frac{y}{x}) = c x$ .

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