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The particular solution of the differential equation $y\left(\frac{\mathrm{d} x}{\mathrm{~d} y}\right)=x \log x$ at $x=\mathrm{e}$ and $\mathrm{y}=1$ is
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The correct answer is:
$x=\mathrm{e}^{\mathrm{y}}$
(D)
$y\left(\frac{d x}{d y}\right)=x \cdot \log x$
$\therefore \int \frac{1}{x \cdot \log x} d x=\int \frac{1}{y} d y$
$\therefore \log |\log x|=\log y+\log c$
We have $x=e$ and $y=1$
$\therefore \log |\log e|=\log 1+\log c \Rightarrow \log c=0$
$\therefore \log |\log x|=\log y \Rightarrow \log x=y \Rightarrow x=e^{y}$
$y\left(\frac{d x}{d y}\right)=x \cdot \log x$
$\therefore \int \frac{1}{x \cdot \log x} d x=\int \frac{1}{y} d y$
$\therefore \log |\log x|=\log y+\log c$
We have $x=e$ and $y=1$
$\therefore \log |\log e|=\log 1+\log c \Rightarrow \log c=0$
$\therefore \log |\log x|=\log y \Rightarrow \log x=y \Rightarrow x=e^{y}$
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