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

Given the differential equation $$x \log_e x \dfrac{dy}{dx} + y = x^2 \log_e x$$, $$x > 1$$, with $$y(2) = 2$$.

Dividing both sides by $$x \log_e x$$:

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

This is a linear first-order ODE. The integrating factor is:

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

Multiplying both sides by $$\ln x$$:

$$\frac{d}{dx}[y \ln x] = x \ln x$$

Integrating both sides:

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

Applying $$y(2) = 2$$:

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

$$2\ln 2 = 2\ln 2 - 1 + C$$

$$C = 1$$

So $$y \ln x = \frac{x^2}{2}\ln x - \frac{x^2}{4} + 1$$.

At $$x = e$$ (where $$\ln e = 1$$):

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

The answer is Option B: $$\dfrac{1 + e^2}{4}$$.