Let \(y=y(x)\) be the solution of the differential equation \((x \log x) \frac{\mathrm{d} y}{\mathrm{~d} x}+y=2 x \log x(x \geq 1)\) then \(y(\mathrm{e})\) is equal to

\(\begin{aligned}
& \text { Given, }(x \log x) \frac{\mathrm{d} y}{\mathrm{~d} x}+y=2 x \log x \\
& \text { When } x=1, y=0 \\
& (x \log x) \frac{\mathrm{d} y}{\mathrm{~d} x}+y=2 x \log x \\
\therefore \quad & \frac{\mathrm{~d} y}{\mathrm{~d} x}+\frac{y}{x \log x}=2 \\
\therefore \quad & \text { I.F. }=\mathrm{e}^{\int \frac{1}{x \log x} \mathrm{~d} x}=\mathrm{e}^{\log (\log x)}=\log x
\end{aligned}\)
\(\therefore \quad\) Solution of the given equation is
\(\begin{array}{ll}
& y \cdot \log x=\int 2 \log x \mathrm{~d} x+\mathrm{c} \\
\therefore \quad & y \log x=2(x \log x-x)+\mathrm{c} \\
& \text { When } x=1, y=0 \\
\therefore \quad & 0=-2+\mathrm{c} \Rightarrow \mathrm{c}=2 \\
\therefore \quad & y \log x=2(x \log x-x)+2 \\
\therefore \quad & y(\mathrm{e})=2(\mathrm{e}-\mathrm{e})+2=2
\end{array}\)