Let the solution curve $$y = y(x)$$ of the differential equation $$(4 + x^2)dy - 2x(x^2 + 3y + 4)dx = 0$$ pass through the origin. Then $$y(2)$$ is equal to ______

The differential equation: $$(4 + x^2)dy - 2x(x^2 + 3y + 4)dx = 0$$ can be rewritten in the form $$(4 + x^2)\frac{dy}{dx} = 2x^3 + 6xy + 8x$$. Dividing by $$4+x^2$$ yields $$\frac{dy}{dx} - \frac{6x}{4+x^2}y = \frac{2x^3 + 8x}{4+x^2} = 2x$$.

Now the integrating factor corresponds to $$P(x) = -\frac{6x}{4+x^2}$$, so that $$\int P\,dx = -3\ln(4+x^2)$$ and hence $$\text{IF} = e^{-3\ln(4+x^2)} = (4+x^2)^{-3}$$.

Substituting this into the equation gives $$y\cdot(4+x^2)^{-3} = \int 2x\cdot(4+x^2)^{-3}\,dx$$. Letting $$u = 4+x^2$$ and noting that $$du = 2x\,dx$$ transforms the integral into $$\int u^{-3}\,du = \frac{u^{-2}}{-2} = \frac{-1}{2(4+x^2)^2} + C$$. This yields $$\frac{y}{(4+x^2)^3} = \frac{-1}{2(4+x^2)^2} + C$$ and therefore $$y = \frac{-(4+x^2)}{2} + C(4+x^2)^3$$.

Applying the initial condition $$y(0) = 0$$ gives $$0 = \frac{-4}{2} + C\cdot64 = -2 + 64C$$, from which $$C = \frac{1}{32}$$.

Finally, evaluating at $$x=2$$ yields $$y(2) = \frac{-(4+4)}{2} + \frac{1}{32}(4+4)^3 = -4 + \frac{512}{32} = -4 + 16 = 12$$. The answer is $$\boxed{12}$$.