1. Given differential equation

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

   We need to find $y(e)$.

2. Rewrite in standard linear form

   Divide both sides by $x\log x$ (for $x>1$):

   $\frac{dy}{dx} + \frac{1}{x\log x}y = 2$

   This is a linear differential equation of the form

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

   where

   $P(x)=\frac{1}{x\log x}, \qquad Q(x)=2.$

3. Find the integrating factor

   $IF = e^{\int P(x)\,dx} = e^{\int \frac{1}{x\log x}\,dx}$

   Now,

   $\int \frac{1}{x\log x}\,dx = \log(\log x)$

   so

   $IF = e^{\log(\log x)} = \log x$

   (since $x>1$, $\log x>0$).

4. Multiply the equation by the integrating factor

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

   Observe that the left side is:

   $\frac{d}{dx}(y\log x)$

   because

   $\frac{d}{dx}(y\log x)=\log x\frac{dy}{dx}+y\cdot\frac{1}{x}.$

   Hence,

   $\frac{d}{dx}(y\log x)=2\log x.$

5. Integrate both sides

   $y\log x = \int 2\log x\,dx + C$

   Using

   $\int \log x\,dx = x\log x - x,$

   we get

   $y\log x = 2(x\log x - x) + C.$

   Therefore,

   $y = \frac{2x\log x - 2x + C}{\log x}.$

6. Use the condition $x\ge 1$ to determine the solution

   Since the solution is given for $x\ge 1$, it must be defined at $x=1$ as well. But $\log 1 = 0$, so for $y$ to remain finite at $x=1$, the numerator must also vanish at $x=1$.

   At $x=1$:

   $2(1)\log 1 - 2(1) + C = 0 - 2 + C = 0$

   so

   $C=2.$

   Thus,

   $y = \frac{2x\log x - 2x + 2}{\log x}.$

7. Find $y(e)$

   Since $\log e = 1$,

   $y(e)=\frac{2e(1)-2e+2}{1}=2.$

8. Check options

   • A: $2$ ✓
   • B: $2e$ ✗
   • C: $e$ ✗
   • D: $0$ ✗

Therefore, the correct answer is A.