The Solution curve of the differential equation y d x d y = x ( log e x - log e y + 1 ) , x > 0, y > 0 passing through the point (e, 1) is [2024]

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Solution

Q.
The Solution curve of the differential equation $y \frac{d x}{d y} = x (\log_{e} x - \log_{e} y + 1)$ , x > 0, y > 0 passing through the point (e, 1) is [2024]

1 $| \log_{e} \frac{x}{y} |$ $= y$

2 $| \log_{e} \frac{y}{x} |$ $= y^{2}$

3 $2$ $| \log_{e} \frac{x}{y} |$ $= y + 1$

4 $| \log_{e} \frac{y}{x} |$ $= x$

Ans.
(1)

$y \frac{d y}{d x} = x (\log_{e} x - \log_{e} y + 1)$

$\frac{d x}{d y} = \frac{x}{y} (\log x - \log y + 1)$

$\frac{d x}{d y} = \frac{x}{y} (\log \frac{x}{y} + 1) (∵ \log x - \log y = \log \frac{x}{y})$

Let $\frac{x}{y} = t \Rightarrow x = t y$

Differentiating w.r.t y

$\Rightarrow \frac{d x}{d y} = t + y \frac{d t}{d y} \Rightarrow t + y \frac{d t}{d y} = t \log t + t$

$\Rightarrow \frac{d t}{t \log t} = \frac{d y}{y} \Rightarrow \log (\log t) = \log y + c$

$\Rightarrow \log (\log \frac{x}{y}) = \log y + c$

at x = e and y = 1 (passing point)

$\Rightarrow \log (\log \frac{e}{1}) = \log 1 + c \Rightarrow c = 0$

$\Rightarrow \log (\log \frac{x}{y}) = \log y + 0$

$\Rightarrow \log (\log \frac{x}{y}) = \log y \Rightarrow \log (\frac{x}{y}) = y$

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