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

$\left(x\log x\right)\frac{dy}{dx}+y=2x\log x,(x\ge 1)$

$\frac{dy}{dx}+\frac{y}{x\log x}=2\Rightarrow P\left(x\right)=\frac{1}{x\log x}\&Q\left(x\right)=2$

$\text{IL}{e}^{\int Pdx}=e^{\int \frac{1}{x\log x}dx}=e^{\log \left(\log x\right)}=\log x$

G.S is y(IF)=$\int Q\left(x\right)\left(I.F\right)dx$

$y\log x=\int 2\log xdx=2x(\log x-1)+c$

Put x=1 we get c=2

$y\log x=\int 2\log xdx=2x(\log x-1)+2$

Put x=e we get y(e)=2