Let $$y = y(x)$$ be the solution of the differential equation, $$x\frac{dy}{dx} + y = x\log_e x$$, $$(x > 1)$$. If $$2y(2) = \log_e 4 - 1$$, then $$y(e)$$ is equal to

We have the differential equation $$x\dfrac{dy}{dx}+y=x\log_e x,$$ with the condition $$x>1.$$

First we convert the equation to the standard linear form. Dividing every term by $$x$$ gives

$$\dfrac{dy}{dx}+\dfrac{1}{x}\,y=\log_e x.$$

For a linear equation $$\dfrac{dy}{dx}+P(x)\,y=Q(x),$$ the integrating factor is $$\mu(x)=e^{\int P(x)\,dx}.$$

Here $$P(x)=\dfrac{1}{x},$$ so

$$\mu(x)=e^{\int \frac{1}{x}\,dx}=e^{\log_e x}=x.$$

Multiplying every term of the standard-form equation by this integrating factor $$x$$ we obtain

$$x\left(\dfrac{dy}{dx}\right)+y=x\log_e x.$$

We notice that the left side is exactly the derivative of the product $$x\,y,$$ because

$$\dfrac{d}{dx}(x\,y)=x\dfrac{dy}{dx}+y.$$

So the differential equation becomes

$$\dfrac{d}{dx}(x\,y)=x\log_e x.$$

Now we integrate both sides with respect to $$x$$:

$$\int d(x\,y)=\int x\log_e x\,dx.$$

This gives

$$x\,y=\int x\log_e x\,dx + C,$$

where $$C$$ is the constant of integration. We must calculate the integral on the right. We use integration by parts. Let

$$u=\log_e x\quad\text{and}\quad dv=x\,dx.$$

Then

$$du=\dfrac{1}{x}\,dx,\qquad v=\dfrac{x^{2}}{2}.$$

By the integration-by-parts formula $$\int u\,dv = uv-\int v\,du,$$ we have

$$\int x\log_e x\,dx = \dfrac{x^{2}}{2}\log_e x-\int \dfrac{x^{2}}{2}\cdot\dfrac{1}{x}\,dx.$$

Simplifying inside the remaining integral,

$$\int x\log_e x\,dx = \dfrac{x^{2}}{2}\log_e x-\int \dfrac{x}{2}\,dx = \dfrac{x^{2}}{2}\log_e x-\dfrac{x^{2}}{4}+K,$$

where $$K$$ is an integration constant that can be absorbed into $$C.$$ Hence we can write

$$x\,y=\dfrac{x^{2}}{2}\log_e x-\dfrac{x^{2}}{4}+C.$$

Dividing every term by $$x$$ gives the explicit solution

$$y(x)=\dfrac{x}{2}\log_e x-\dfrac{x}{4}+\dfrac{C}{x}.$$

We now use the given condition. It is stated that

$$2y(2)=\log_e 4-1.$$

Because $$\log_e 4=2\log_e 2,$$ the right side becomes $$2\log_e 2-1,$$ so

$$y(2)=\log_e 2-\dfrac{1}{2}.$$

Substituting $$x=2$$ into the general solution, we get

$$y(2)=\dfrac{2}{2}\log_e 2-\dfrac{2}{4}+\dfrac{C}{2}=\log_e 2-\dfrac{1}{2}+\dfrac{C}{2}.$$

Comparing this with the required value $$\log_e 2-\dfrac{1}{2},$$ we must have

$$\log_e 2-\dfrac{1}{2}+\dfrac{C}{2}=\log_e 2-\dfrac{1}{2}\quad\Longrightarrow\quad\dfrac{C}{2}=0,$$

so $$C=0.$$

Therefore the particular solution simplifies to

$$y(x)=\dfrac{x}{2}\log_e x-\dfrac{x}{4}.$$

We now evaluate this at $$x=e.$$ Because $$\log_e e=1,$$ we have

$$y(e)=\dfrac{e}{2}\cdot 1-\dfrac{e}{4}= \dfrac{e}{2}-\dfrac{e}{4}= \dfrac{e}{4}.$$

Hence, the correct answer is Option B.