Let \(y=y(x)\) be the solution of the differential equation \(x \frac{\mathrm{~d} y}{\mathrm{~d} x}+y=x \log x,(x\gt1)\)
If \(2(y(2))=\log 4-1\) then the value of \(y(\mathrm{e})\) is

\(\begin{aligned}
& x \frac{\mathrm{~d} y}{\mathrm{~d} x}+y=x \log x \\
\therefore \quad & \frac{\mathrm{~d} y}{\mathrm{~d} x}+\frac{1}{x} y=\log x
\end{aligned}\)
Here, \(\mathrm{P}(x)=\frac{1}{x}, \mathrm{Q}(x)=\log x\)
\(\begin{array}{ll}
\therefore & \text { I.F. }=\mathrm{e}^{\int \mathrm{P}(x) \mathrm{dx}}=\mathrm{e}^{\int \frac{1}{x} \mathrm{dx}}=\mathrm{e}^{\log x}=x \\
\therefore & y(\text { I.F. })=\int \mathrm{Q}(\text { I.F. }) \mathrm{d} x+\mathrm{c} \\
\therefore & x y=\int x \log x \mathrm{~d} x+\mathrm{c} \\
\therefore & x y=\frac{x^2}{2} \log x-\left(\int \frac{x^2}{2} \times \frac{1}{x} \mathrm{~d} x\right)+\mathrm{c} \\
\therefore & x y=\frac{x^2}{2} \log x-\frac{1}{2} \int x \mathrm{~d} x+\mathrm{c} \\
\therefore & x y=\frac{x^2}{2} \log x-\frac{1}{4} x^2+\mathrm{c}...(i)
\end{array}\)
Given that \(2(y(2))=\log 4-1\)
\(\therefore \quad y(2)=\log 2-\frac{1}{2}\)
\(\therefore \quad\) From equation (i), we get
\(\begin{array}{ll}
& 2\left(\log 2-\frac{1}{2}\right)=\frac{(2)^2}{2} \log 2-\frac{1}{4}(2)^2+\mathrm{c} \\
\therefore & 2 \log 2-1=2 \log 2-1+c \\
\therefore \quad & \mathrm{c}=0...(ii)
\end{array}\)
\(\therefore \quad\) From (i) and (ii), we get
\(y(\mathrm{e})=\frac{\mathrm{e}}{2}-\frac{\mathrm{e}}{4}=\frac{\mathrm{e}}{4}\)