If the solution curve $$f(x, y) = 0$$ of the differential equation $$(1 + \log_e x)\dfrac{dx}{dy} - x\log_e x = e^y$$, $$x > 0$$, passes through the points (1, 0) and $$(a, 2)$$, then $$a^a$$ is equal to

Given the differential equation $$(1 + \log_e x)\frac{dx}{dy} - x\log_e x = e^y$$, $$x > 0$$, passing through $$(1, 0)$$ and $$(a, 2)$$.

Substitute $$t = x \log_e x$$.

Differentiating with respect to $$x$$: $$\frac{dt}{dx} = \log_e x + 1 = 1 + \log_e x$$.

So $$\frac{dt}{dy} = (1 + \log_e x)\frac{dx}{dy}$$.

The given equation becomes:

$$\frac{dt}{dy} - t = e^y$$

Solve the linear ODE.

This is a first-order linear ODE in $$t$$ and $$y$$. The integrating factor is:

$$\text{I.F.} = e^{\int -1 \, dy} = e^{-y}$$

Multiplying both sides by $$e^{-y}$$:

$$\frac{d}{dy}(t \cdot e^{-y}) = e^y \cdot e^{-y} = 1$$

Integrating: $$t \cdot e^{-y} = y + C$$

$$x \log_e x = (y + C)e^y$$

Apply the initial condition $$(1, 0)$$.

$$1 \cdot \log_e 1 = (0 + C)e^0$$

$$0 = C$$

So the solution curve is: $$x \log_e x = y \cdot e^y$$.

Find $$a^a$$ using the point $$(a, 2)$$.

$$a \log_e a = 2 \cdot e^2$$

$$\log_e(a^a) = a \log_e a = 2e^2$$

$$a^a = e^{2e^2}$$

Therefore, $$a^a = e^{2e^2}$$, which is Option A.