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The integrating factor of the differential equation y $\log _{y}\left(\frac{\mathrm{d} x}{\mathrm{dy}}\right)+x-\log \mathrm{y}=0$
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The correct answer is:
$\log y$
We have $y \log y\left(\frac{d x}{d y}\right)+x=\log y$
$\therefore \frac{\mathrm{dx}}{\mathrm{dy}}+\frac{\mathrm{x}}{\mathrm{y} \log \mathrm{y}}=\frac{1}{\mathrm{y}}$
$\therefore$ I.F. $=e^{\int \frac{1}{y \log y} d y}=e^{\log (\log y)}=\log y$
$\therefore \frac{\mathrm{dx}}{\mathrm{dy}}+\frac{\mathrm{x}}{\mathrm{y} \log \mathrm{y}}=\frac{1}{\mathrm{y}}$
$\therefore$ I.F. $=e^{\int \frac{1}{y \log y} d y}=e^{\log (\log y)}=\log y$
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