Let $$y = y(x)$$ be the solution of the differential equation $$(x^2 + 4)^2 dy + (2x^3 y + 8xy - 2)dx = 0$$. If $$y(0) = 0$$, then $$y(2)$$ is equal to

Rewriting: $$y' + \frac{2x}{x^2+4}y = \frac{2}{(x^2+4)^2}$$.

Integrating factor: $$e^{\int \frac{2x}{x^2+4}dx} = e^{\ln(x^2+4)} = x^2+4$$.

$$\frac{d}{dx}[y(x^2+4)] = \frac{2}{x^2+4}$$

Integrating: $$y(x^2+4) = \int \frac{2}{x^2+4}dx = \tan^{-1}(x/2) + C$$.

Using $$y(0) = 0$$: $$0 = 0 + C$$, so $$C = 0$$.

$$y(2) = \frac{\tan^{-1}(1)}{4+4} = \frac{\pi/4}{8} = \frac{\pi}{32}$$.

The correct answer is Option 1: $$\frac{\pi}{32}$$.