Let y(x) be the solution of the differential equation  $\left(x\log x\right)\frac{dy}{dx}+y=2x\log x,(x\ge 1)$. Then, y(e) is equal to

Given differential equation is
                     $$\begin{gathered} \left(x\log x\right)\frac{dy}{dx}+y=2x\log x, \\ \Rightarrow \frac{dy}{dx}+\frac{y}{x\log x}=2 \end{gathered}$$

This is a linear differential rqaution.
$\therefore IF=e^{\int \frac{1}{x\log x}dx}=e^{\log \left(\log x\right)}=\log x$
Now, the solution of given differential equation is given by
                          $y.\log x=\int \log x.2dx\Rightarrow y.\log x=2\int \log xdx$
$\Rightarrow y.\log x=2[x\log x-x]+c$

At                      x = 1, c = 2
$\Rightarrow y.\log x=2[x\log x-x]+2$
At         x = e,
            Y = 2(e – e) +2
$\Rightarrow y=2$