Let the solution curve of the differential equation $$x dy = (\sqrt{x^2 + y^2} + y) dx$$, $$x > 0$$, intersect the line $$x = 1$$ at $$y = 0$$ and the line $$x = 2$$ at $$y = \alpha$$. Then the value of $$\alpha$$ is

We are given the differential equation $$x\, dy = \left(\sqrt{x^2 + y^2} + y\right) dx$$, $$x > 0$$. Dividing both sides by $$dx$$ and by $$x$$ yields the equivalent form $$\frac{dy}{dx} = \frac{\sqrt{x^2 + y^2} + y}{x}$$.

Next, we use the substitution $$y = vx$$ so that $$\frac{dy}{dx} = v + x\frac{dv}{dx}$$. Substituting into the equation gives $$v + x\frac{dv}{dx} = \frac{\sqrt{x^2 + v^2x^2} + vx}{x} = \sqrt{1 + v^2} + v$$, which simplifies to $$x\frac{dv}{dx} = \sqrt{1 + v^2}$$.

Separating variables, we obtain $$\frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x}$$. Integrating both sides leads to $$\ln\left|v + \sqrt{1 + v^2}\right| = \ln|x| + C$$. Exponentiation then gives $$v + \sqrt{1 + v^2} = kx$$ where $$k = e^C$$.

Applying the initial condition $$y(1) = 0$$, which corresponds to $$v = 0$$ when $$x = 1$$, we have $$0 + \sqrt{1 + 0} = k \cdot 1$$ and hence $$k = 1$$. Therefore the relationship simplifies to $$v + \sqrt{1 + v^2} = x$$.

To determine $$\alpha = y(2)$$, we set $$x = 2$$ in the equation, giving $$v + \sqrt{1 + v^2} = 2$$. It follows that $$\sqrt{1 + v^2} = 2 - v$$. Squaring both sides yields $$1 + v^2 = 4 - 4v + v^2$$, which simplifies to $$1 = 4 - 4v$$ and thus $$v = \frac{3}{4}$$. Since $$y = vx$$, it follows that $$\alpha = y(2) = \frac{3}{4} \times 2 = \frac{3}{2}$$.

The answer is Option B.