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

$\frac{dy}{dx}+\frac{y}{x\log x}=2$

$P=\frac{1}{x\log x},Q=2$

I.F = $e^{\int Pdx}=e^{\int \frac{1}{x\log x}dx}=e^{\log \left(\log x\right)}=\log x$

Solution is y (I.F) = $\int Q.(I.F)dx+c$

$y\left(\log x\right)=\int 2.\left(\log x\right)dx+c$

$y\left(\log x\right)=2[x\log x-x]+c\to \left(1\right)$

If x = 1, y = 0 then c = 2

$\left(1\right)\Rightarrow y\left(\log x\right)=2[x\log x-x]+2$

At x = e

$y\left(\log e\right)=2[e\left(1\right)-e]+2$

$y\left(1\right)=2\left(0\right)+2 \Rightarrow y=2$