Given differential equation

$x\ln x\,\frac{dy}{dx}+y=x^2\ln x, \qquad x>1$

with initial condition

$y(2)=2.$

Rewrite in linear form

Divide both sides by $x\ln x$:

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

This is a linear differential equation of the form

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

where

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

Find the integrating factor

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

Since

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

we get

$\text{I.F.}=e^{\ln(\ln x)}=\ln x$

(valid because $x>1 \Rightarrow \ln x>0$).

Multiply the equation by the integrating factor

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

The left-hand side is

$\frac{d}{dx}(y\ln x).$

Hence,

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

Integrate both sides

$y\ln x=\int x\ln x\,dx + C.$

Use integration by parts for $\int x\ln x\,dx$:

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

So,

$\int x\ln x\,dx=\frac{x^2}{2}\ln x-\int \frac{x^2}{2}\cdot \frac{1}{x}dx$ $=\frac{x^2}{2}\ln x-\frac{1}{2}\int x\,dx$ $=\frac{x^2}{2}\ln x-\frac{x^2}{4}+C.$

Therefore,

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

Hence,

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

Use the initial condition $y(2)=2$

Substitute $x=2$:

$2=\frac{4}{2}-\frac{4}{4\ln 2}+\frac{C}{\ln 2}$ $2=2-\frac{1}{\ln 2}+\frac{C}{\ln 2}.$

Thus,

$-\frac{1}{\ln 2}+\frac{C}{\ln 2}=0$ $C=1.$

Find $y(e)$

Since $\ln e=1$,

$y(e)=\frac{e^2}{2}-\frac{e^2}{4}+1$ $=\frac{e^2}{4}+1$ $=\frac{4+e^2}{4}.$

Compare with options

$\frac{4+e^2}{4}$

This matches Option D.