If $x \frac{d y}{d x}=y(\log y-\log x+1)$, then the solution of the equation is
Answer & explanation
Correct answer: option 2
We have,
$x \frac{d y}{d x}=y \log \left(\frac{y}{x}\right)+y$
Putting $y=v x$ and $\frac{d y}{d x}=v+x \frac{d v}{d x}$, we get
$v+x \frac{d v}{d x}=v \log v+v$
$\Rightarrow \frac{1}{v \log v} d v=\frac{1}{x} d x$
$\Rightarrow \log (\log v)=\log x+\log C$ [On integrating]
$\Rightarrow \log v=C x$
$\Rightarrow \log \left(\frac{y}{x}\right)=C x$, which is the required solution.