Let $$\alpha|x| = |y|e^{xy - \beta}$$, $$\alpha, \beta \in \mathbb{N}$$ be the solution of the differential equation $$x\,dy - y\,dx + xy(x\,dy + y\,dx) = 0$$, $$y(1) = 2$$. Then $$\alpha + \beta$$ is equal to _____

The differential equation is: $$x\,dy - y\,dx + xy(x\,dy + y\,dx) = 0$$, with $$y(1) = 2$$.

Rearranging: $$x\,dy - y\,dx + xy \cdot d(xy) = 0$$ (since $$x\,dy + y\,dx = d(xy)$$).

Dividing by $$xy$$: $$\frac{x\,dy - y\,dx}{xy} + d(xy) = 0$$

Note that $$\frac{x\,dy - y\,dx}{xy} = \frac{dy}{y} - \frac{dx}{x} = d\left(\ln\left|\frac{y}{x}\right|\right)$$.

So: $$d\left(\ln\left|\frac{y}{x}\right|\right) + d(xy) = 0$$

Integrating: $$\ln\left|\frac{y}{x}\right| + xy = C$$

Using $$y(1) = 2$$: $$\ln|2| + 1 \cdot 2 = C \implies C = \ln 2 + 2$$.

$$\ln\left|\frac{y}{x}\right| + xy = \ln 2 + 2$$

$$\ln|y| - \ln|x| + xy = \ln 2 + 2$$

$$\ln|y| - \ln 2 = \ln|x| - xy + 2$$

Rearranging: $$\ln\left|\frac{y}{2}\right| = \ln|x| + 2 - xy$$

$$\left|\frac{y}{2}\right| = |x| \cdot e^{2-xy}$$

$$|y| = 2|x| \cdot e^{2-xy}$$

Comparing with $$\alpha|x| = |y|e^{xy-\beta}$$:

$$|y| = \alpha|x| \cdot e^{\beta - xy}$$

So $$\alpha = 2$$ and $$\beta = 2$$.

$$\alpha + \beta = 2 + 2 = 4$$.

The answer is $$\boxed{4}$$.