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If $x \frac{d y}{d x}=y(\log y-\log x+1)$, then the solution of the equation is
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Verified Answer
The correct answer is:
$\log \left(\frac{y}{x}\right)=c x$
$\log \left(\frac{y}{x}\right)=c x$
$$
\begin{aligned}
& \frac{x d y}{d x}=y(\log y-\log x+1) \\
& \frac{d y}{d x}=\frac{y}{x}\left(\log \left(\frac{y}{x}\right)+1\right)
\end{aligned}
$$
Put $y=v x$
$\frac{d y}{d x}=v+\frac{x d v}{d x}$
$\Rightarrow v+\frac{x d v}{d x}=v(\log v+1)$
$\frac{x d v}{d x}=v \log v$
$\Rightarrow \frac{d v}{v \log v}=\frac{d x}{x}$
put $\log v=z$
$\frac{1}{v} d v=d z$
$\Rightarrow \frac{d z}{z}=\frac{d x}{x}$
$\ln z=\ln x+\ln c$
$z=c x$
$\log v=c x$
$\log \left(\frac{y}{x}\right)=c x$
\begin{aligned}
& \frac{x d y}{d x}=y(\log y-\log x+1) \\
& \frac{d y}{d x}=\frac{y}{x}\left(\log \left(\frac{y}{x}\right)+1\right)
\end{aligned}
$$
Put $y=v x$
$\frac{d y}{d x}=v+\frac{x d v}{d x}$
$\Rightarrow v+\frac{x d v}{d x}=v(\log v+1)$
$\frac{x d v}{d x}=v \log v$
$\Rightarrow \frac{d v}{v \log v}=\frac{d x}{x}$
put $\log v=z$
$\frac{1}{v} d v=d z$
$\Rightarrow \frac{d z}{z}=\frac{d x}{x}$
$\ln z=\ln x+\ln c$
$z=c x$
$\log v=c x$
$\log \left(\frac{y}{x}\right)=c x$
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