Let the solution curve $$y = y(x)$$ of the differential equation, $$\left[\frac{x}{\sqrt{x^2-y^2}} + e^{y/x}\right]x\frac{dy}{dx} = x + \left[\frac{x}{\sqrt{x^2-y^2}} + e^{y/x}\right]y$$ pass through the points $$(1, 0)$$ and $$(2\alpha, \alpha), \alpha > 0$$. Then $$\alpha$$ is equal to

The differential equation is:

$$\left[\frac{x}{\sqrt{x^2-y^2}} + e^{y/x}\right]x\frac{dy}{dx} = x + \left[\frac{x}{\sqrt{x^2-y^2}} + e^{y/x}\right]y$$

Let $$g(x,y) = \frac{x}{\sqrt{x^2-y^2}} + e^{y/x}$$. The equation becomes:

$$g \cdot x \frac{dy}{dx} = x + g \cdot y$$

$$g\left(x\frac{dy}{dx} - y\right) = x$$

Substituting $$y = vx$$ so $$\frac{dy}{dx} = v + x\frac{dv}{dx}$$:

$$x\frac{dy}{dx} - y = x\left(v + x\frac{dv}{dx}\right) - vx = x^2 \frac{dv}{dx}$$

Also, $$g = \frac{x}{\sqrt{x^2 - v^2x^2}} + e^v = \frac{1}{\sqrt{1-v^2}} + e^v$$.

So: $$\left(\frac{1}{\sqrt{1-v^2}} + e^v\right) x^2 \frac{dv}{dx} = x$$

$$\left(\frac{1}{\sqrt{1-v^2}} + e^v\right) dv = \frac{dx}{x}$$

Integrating both sides:

$$\int \frac{dv}{\sqrt{1-v^2}} + \int e^v \, dv = \int \frac{dx}{x}$$

$$\sin^{-1}(v) + e^v = \ln|x| + C$$

$$\sin^{-1}\left(\frac{y}{x}\right) + e^{y/x} = \ln|x| + C$$

Using the initial condition $$(1, 0)$$:

$$\sin^{-1}(0) + e^0 = \ln 1 + C \implies 0 + 1 = 0 + C \implies C = 1$$

So the solution is: $$\sin^{-1}\left(\frac{y}{x}\right) + e^{y/x} = \ln x + 1$$

Now substituting the point $$(2\alpha, \alpha)$$ where $$\alpha > 0$$:

$$\sin^{-1}\left(\frac{1}{2}\right) + e^{1/2} = \ln(2\alpha) + 1$$

$$\frac{\pi}{6} + \sqrt{e} = \ln(2\alpha) + 1$$

$$\ln(2\alpha) = \frac{\pi}{6} + \sqrt{e} - 1$$

$$2\alpha = \exp\left(\frac{\pi}{6} + \sqrt{e} - 1\right)$$

$$\alpha = \frac{1}{2}\exp\left(\frac{\pi}{6} + \sqrt{e} - 1\right)$$

Hence the correct answer is Option A: $$\dfrac{1}{2}\exp\left(\dfrac{\pi}{6} + \sqrt{e} - 1\right)$$.