The differential equation satisfied by the system of parabolas $$y^2 = 4a(x+a)$$ is:

The family of parabolas is $$y^2 = 4a(x + a)$$, which expands to $$y^2 = 4ax + 4a^2$$. Differentiating both sides with respect to $$x$$: $$2y\frac{dy}{dx} = 4a$$, so $$a = \frac{y}{2}\frac{dy}{dx}$$. Write $$y' = \frac{dy}{dx}$$ for brevity.

Substituting $$a = \frac{yy'}{2}$$ back into the original equation: $$y^2 = 4 \cdot \frac{yy'}{2} \cdot x + 4\left(\frac{yy'}{2}\right)^2 = 2xyy' + y^2(y')^2$$.

Dividing through by $$y$$ (for $$y \neq 0$$): $$y = 2xy' + y(y')^2$$, which rearranges to $$y(y')^2 + 2xy' - y = 0$$.

The correct answer is (C): $$y\left(\frac{dy}{dx}\right)^2 + 2x\left(\frac{dy}{dx}\right) - y = 0$$.