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The general solution of the differential equation
$$
y(1+\log x)\left(\frac{d x}{d y}\right)
$$
$-x \log x=0$ is
Options:
$$
y(1+\log x)\left(\frac{d x}{d y}\right)
$$
$-x \log x=0$ is
Solution:
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Verified Answer
The correct answer is:
$\mathrm{x} \log \mathrm{x}=\mathrm{yc}$
$$
\begin{aligned}
& y(1+\log x)\left(\frac{d y}{d x}\right)-x \log x=0 \\
& \therefore \int \frac{(1+\log x)}{x \log x} d x=\int \frac{d y}{y} \\
& \int \frac{d x}{x \log x}+\int \frac{d x}{x}=\int \frac{d y}{y} \\
& \therefore \log (\log x)+\log x=\log y+\log c \\
& \therefore \log [x \log x]=\log (y, c)
\Rightarrow x \log x=y c
\end{aligned}
$$
\begin{aligned}
& y(1+\log x)\left(\frac{d y}{d x}\right)-x \log x=0 \\
& \therefore \int \frac{(1+\log x)}{x \log x} d x=\int \frac{d y}{y} \\
& \int \frac{d x}{x \log x}+\int \frac{d x}{x}=\int \frac{d y}{y} \\
& \therefore \log (\log x)+\log x=\log y+\log c \\
& \therefore \log [x \log x]=\log (y, c)
\Rightarrow x \log x=y c
\end{aligned}
$$
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