Answer:
Option A
Explanation:
Given, differential equation is
$y(1+\log x)\frac{dx}{dy}-x\log x=0$
$\Rightarrow \frac{(1+\log x)dx}{x\log x}=\frac{dy}{y}$
$\Rightarrow$ $ \left(\frac{1}{x \log x}+\frac{1}{x}\right)dx=\frac{1}{y}dy$
On integrating both sides , we get
$\Rightarrow \int \left(\frac{1}{x \log x}+\frac{1}{x}\right)dx=\int \frac{1}{y}dy$
Put log x=t
$\Rightarrow$ $\frac{1}{x}dx=dt$
$\therefore$ $ \int \frac{1}{t}dt+\int\frac{1}{x} dx=\int\frac{1}{y} dy$
$\Rightarrow$ log t+log x=log y+log c
$\Rightarrow$ log tx=log yc
$\Rightarrow$ tx=yc
$\Rightarrow$ x log x= yc
when x= e and y = $e^{2}$
$\therefore$ $e \log e=e^{2} c$
$\Rightarrow$ e x 1=$ e^{2}c$
$\Rightarrow$ c $\Rightarrow$ $ \frac{1}{e}$
$\therefore$ x log x= $\frac{y}{e}$
$\Rightarrow$ y= ex log x