Let $$y(x)$$ be the solution of the differential equation $$(x \log x)\frac{dy}{dx} + y = 2x \log x$$, $$(x \geq 1)$$. Then $$y(e)$$ is equal to

The given differential equation is $$(x \log x) \frac{dy}{dx} + y = 2x \log x$$ for $$x \geq 1$$. We need to find $$y(e)$$.

First, rewrite the equation in standard linear form $$\frac{dy}{dx} + P(x)y = Q(x)$$. Divide both sides by $$x \log x$$ (for $$x > 1$$ since $$\log x \neq 0$$):

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

Here, $$P(x) = \frac{1}{x \log x}$$ and $$Q(x) = 2$$. To solve, use the integrating factor method. The integrating factor (IF) is $$e^{\int P(x) dx}$$. Compute $$\int P(x) dx = \int \frac{1}{x \log x} dx$$.

Set $$u = \log x$$, so $$du = \frac{1}{x} dx$$. Then:

$$\int \frac{1}{x \log x} dx = \int \frac{1}{u} du = \log |u| + C = \log |\log x| + C$$

Since $$x \geq 1$$, $$\log x \geq 0$$, so $$|\log x| = \log x$$. Thus, IF = $$e^{\log \log x} = \log x$$.

Multiply both sides of the differential equation by the integrating factor:

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

Simplify:

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

The left side is the derivative of $$y \log x$$:

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

Integrate both sides with respect to $$x$$:

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

Compute $$\int \log x dx$$ using integration by parts. Set $$u = \log x$$, $$dv = dx$$, so $$du = \frac{1}{x} dx$$, $$v = x$$:

$$\int \log x dx = x \log x - \int x \cdot \frac{1}{x} dx = x \log x - \int dx = x \log x - x + C$$

Thus,

$$\int 2 \log x dx = 2(x \log x - x) + C = 2x \log x - 2x + C$$

So,

$$y \log x = 2x \log x - 2x + C$$

Solve for $$y$$:

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

At $$x = 1$$, the original equation gives $$y(1) = 0$$ because substituting $$x = 1$$ yields $$(1 \cdot \log 1) \frac{dy}{dx} + y = 2(1) \log 1$$, and $$\log 1 = 0$$, so $$0 \cdot \frac{dy}{dx} + y = 0$$, implying $$y(1) = 0$$.

Use the initial condition $$y(1) = 0$$. Substitute $$x = 1$$ into the equation $$y \log x = 2x \log x - 2x + C$$:

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

Since $$\log 1 = 0$$,

$$0 \cdot 0 = 0 - 2 + C \implies 0 = C - 2 \implies C = 2$$

Substitute $$C = 2$$ back:

$$y \log x = 2x \log x - 2x + 2$$

So,

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

Now evaluate at $$x = e$$:

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

Since $$\log e = 1$$,

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

Hence, the correct answer is Option D.