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If $y d x - x d y + \ln x d x = 0, y (1) = - 1$ , then

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$2 (y + 1) + \ln x = 0$

$y + 1 + 2 \ln x = 0$

$y + 1 + \ln x = 0$

$y + 1 - y \ln x = 0$

answer is A.

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Detailed Solution

$y d x - x d y + \ln x d x = 0$

$\Rightarrow \frac{x d y - y d x}{x^{2}} = \frac{\ln x d x}{x^{2}}$

$\Rightarrow d (\frac{y}{x}) = \frac{\ln x d x}{x^{2}} \Rightarrow \frac{y}{x} = \int \frac{\ln x}{x^{2}} d x \Rightarrow \frac{y}{x} = - \frac{1}{x} - \frac{\ln x}{x} + c$

$x = 1, y = - 1 \Rightarrow c = 0 \Rightarrow y + 1 + \ln x = 0$

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