Let $$y = y(x)$$ be the solution of the differential equation $$(x + y + 2)^2 dx = dy$$, $$y(0) = -2$$. Let the maximum and minimum values of the function $$y = y(x)$$ in $$\left[0, \frac{\pi}{3}\right]$$ be $$\alpha$$ and $$\beta$$, respectively. If $$(3\alpha + \pi)^2 + \beta^2 = \gamma + \delta\sqrt{3}$$, $$\gamma, \delta \in \mathbb{Z}$$, then $$\gamma + \delta$$ equals _____

The differential equation is $$(x + y + 2)^2 dx = dy$$ with the initial condition $$y(0) = -2$$, which can be rewritten as $$\frac{dy}{dx} = (x + y + 2)^2$$. This suggests the substitution $$v = x + y + 2$$, and differentiating with respect to $$x$$ gives $$\frac{dv}{dx} = 1 + \frac{dy}{dx}$$. Substituting the expression for $$\frac{dy}{dx}$$ leads to $$\frac{dv}{dx} = 1 + v^2$$.

The resulting equation is separable since $$\frac{dv}{1 + v^2} = dx$$, so integrating both sides as $$\int \frac{dv}{1 + v^2} = \int dx$$ yields $$\tan^{-1} v = x + C$$, where $$C$$ is the constant of integration. Therefore $$v = \tan(x + C)$$, and substituting back for $$v$$ gives $$x + y + 2 = \tan(x + C)$$, which implies $$y = \tan(x + C) - x - 2$$.

Imposing the initial condition $$y(0) = -2$$ leads to $$-2 = \tan(0 + C) - 0 - 2\implies \tan C = 0\implies C = n\pi,\;n\in\mathbb{Z}$$. Choosing $$C = 0$$ (since other integer multiples of $$\pi$$ yield the same solution by periodicity) gives $$y(x) = \tan x - x - 2$$, which indeed satisfies $$y(0) = -2$$.

Next, to find the maximum and minimum of $$y(x) = \tan x - x - 2$$ on the interval $$[0, \tfrac{\pi}{3}]$$, one computes the derivative which gives $$y'(x) = \sec^2 x - 1 = \tan^2 x$$. Setting $$y'(x) = 0$$ implies $$\tan^2 x = 0\implies \tan x = 0\implies x = n\pi,\;n\in\mathbb{Z}$$, so within $$[0, \tfrac{\pi}{3}]$$ the only critical point is $$x = 0$$. Since $$y'(x) = \tan^2 x \ge 0$$ and vanishes only at isolated points, $$y(x)$$ is strictly increasing on this interval and therefore its minimum and maximum occur at the endpoints.

Evaluating at the endpoints yields at $$x = 0$$ that $$y(0) = \tan 0 - 0 - 2 = -2$$ and at $$x = \tfrac{\pi}{3}$$ that $$y\bigl(\tfrac{\pi}{3}\bigr) = \tan\bigl(\tfrac{\pi}{3}\bigr) - \tfrac{\pi}{3} - 2 = \sqrt{3} - \tfrac{\pi}{3} - 2$$. Hence the minimum value $$\beta$$ is $$-2$$ and the maximum value $$\alpha$$ is $$\sqrt{3} - \tfrac{\pi}{3} - 2$$.

Finally, one computes $$(3\alpha + \pi)^2 + \beta^2$$ by first finding $$3\alpha + \pi = 3\bigl(\sqrt{3} - \tfrac{\pi}{3} - 2\bigr) + \pi = 3\sqrt{3} - \pi - 6 + \pi = 3\sqrt{3} - 6$$. Then $$(3\alpha + \pi)^2 = (3\sqrt{3} - 6)^2 = (3\sqrt{3})^2 - 2\cdot 3\sqrt{3}\cdot 6 + 6^2 = 27 - 36\sqrt{3} + 36 = 63 - 36\sqrt{3}$$ and $$\beta^2 = (-2)^2 = 4$$, so that $$(3\alpha + \pi)^2 + \beta^2 = (63 - 36\sqrt{3}) + 4 = 67 - 36\sqrt{3}$$. Writing this in the form $$\gamma + \delta\sqrt{3}$$ gives $$\gamma = 67$$ and $$\delta = -36$$, hence $$\gamma + \delta = 67 + (-36) = 31$$.

Therefore, the final answer is 31.