1. Given differential equation

   $x\frac{dy}{dx}+y=x\ln x, \qquad x>1$

2. Rewrite in standard linear form

   Divide throughout by $x$:

   $\frac{dy}{dx}+\frac{1}{x}y=\ln x$

   This is a linear differential equation:

   $\frac{dy}{dx}+P(x)y=Q(x)$

   where

   $P(x)=\frac{1}{x}, \qquad Q(x)=\ln x$

3. Find the integrating factor

   $\text{I.F.}=e^{\int \frac{1}{x}\,dx}=e^{\ln x}=x$

   Since $x>1$, we can take $\text{I.F.}=x$.

4. Multiply the equation by the integrating factor

   $x\frac{dy}{dx}+y=x\ln x$

   Observe that the left side is:

   $\frac{d}{dx}(xy)=x\ln x$

5. Integrate both sides

   $xy=\int x\ln x\,dx + C$

   Now evaluate:

   $\int x\ln x\,dx$

   Use integration by parts:

   Let $u=\ln x, \quad dv=x\,dx$ Then $du=\frac{1}{x}dx, \quad v=\frac{x^2}{2}$

   So,

   $\int x\ln x\,dx=\frac{x^2}{2}\ln x-\int \frac{x^2}{2}\cdot \frac{1}{x}\,dx$

   $=\frac{x^2}{2}\ln x-\frac{1}{2}\int x\,dx$

   $=\frac{x^2}{2}\ln x-\frac{x^2}{4}$

   Hence,

   $xy=\frac{x^2}{2}\ln x-\frac{x^2}{4}+C$

   Therefore,

   $y=\frac{x}{2}\ln x-\frac{x}{4}+\frac{C}{x}$

6. Use the initial condition

   Given:

   $2y(2)=\ln 4-1$

   So,

   $y(2)=\frac{\ln 4-1}{2}$

   Substitute $x=2$ into the general solution:

   $y(2)=\frac{2}{2}\ln 2-\frac{2}{4}+\frac{C}{2}$

   $y(2)=\ln 2-\frac{1}{2}+\frac{C}{2}$

   Also,

   $\frac{\ln 4-1}{2}=\frac{2\ln 2-1}{2}=\ln 2-\frac{1}{2}$

   Therefore,

   $\ln 2-\frac{1}{2}+\frac{C}{2}=\ln 2-\frac{1}{2}$

   which gives

   $\frac{C}{2}=0 \implies C=0$

7. Find $y(e)$

   Now,

   $y=\frac{x}{2}\ln x-\frac{x}{4}$

   At $x=e$:

   $y(e)=\frac{e}{2}\ln e-\frac{e}{4}$

   Since $\ln e=1$,

   $y(e)=\frac{e}{2}-\frac{e}{4}=\frac{e}{4}$

8. Option check

   • A: $-\frac{e}{2}$ ❌
   • B: $-\frac{e^2}{2}$ ❌
   • C: $\frac{e^2}{4}$ ❌
   • D: $\frac{e}{4}$ ✅

Therefore, the correct answer is D.