Let the solution curve of the differential equation $$x\frac{dy}{dx} - y = \sqrt{y^2 + 16x^2}$$, $$y(1) = 3$$ be $$y = y(x)$$. Then $$y(2)$$ is equal to

We need to solve $$x\frac{dy}{dx} - y = \sqrt{y^2 + 16x^2}$$ with the initial condition $$y(1) = 3$$.

To simplify the equation, we use the substitution $$y = vx$$, which gives $$\frac{dy}{dx} = v + x\frac{dv}{dx}$$. Substituting into the original differential equation yields $$x\left(v + x\frac{dv}{dx}\right) - vx = \sqrt{v^2x^2 + 16x^2}$$, and this simplifies to $$x^2\frac{dv}{dx} = x\sqrt{v^2 + 16}$$ or equivalently $$x\frac{dv}{dx} = \sqrt{v^2 + 16}\,.$$

Separating variables then gives $$\frac{dv}{\sqrt{v^2 + 16}} = \frac{dx}{x}\,.$$

Integrating both sides yields $$\ln\left|v + \sqrt{v^2 + 16}\right| = \ln|x| + C\,, $$ which implies $$v + \sqrt{v^2 + 16} = kx$$ for some constant $$k$$.

Applying the initial condition $$y(1) = 3$$ means $$v(1)=3$$, so $$3 + \sqrt{9 + 16} = k\cdot 1$$ and hence $$k = 3 + 5 = 8\,. $$ Therefore $$v + \sqrt{v^2 + 16} = 8x\,. $$

Substituting back $$v = \frac{y}{x}$$ leads to $$\frac{y}{x} + \sqrt{\frac{y^2}{x^2} + 16} = 8x\,, $$ which simplifies to $$y + \sqrt{y^2 + 16x^2} = 8x^2\,.$$

To find $$y(2)$$ we set $$x=2$$ so that $$y + \sqrt{y^2 + 64} = 32\,. $$ Isolating the square root gives $$\sqrt{y^2 + 64} = 32 - y\,, $$ and squaring both sides yields $$y^2 + 64 = 1024 - 64y + y^2\,. $$ This leads to $$64 = 1024 - 64y$$ and hence $$64y = 960\,, $$ giving $$y = 15\,. $$

Checking, we have $$15 + \sqrt{225 + 64} = 15 + \sqrt{289} = 15 + 17 = 32 = 8(4)\,, $$ which verifies the result.

The answer is Option A: $$15$$.