Let $\alpha|x|=|y| \mathrm{e}^{x y-\beta}, \alpha, \beta \in \mathbf{N}$ be the solution of the differential equation $x \mathrm{~d} y-y \mathrm{~d} x+x y(x \mathrm{~d} y+y \mathrm{~d} x)=0,y(1)=2$. Then $\alpha+\beta$ is equal to ________

<p>$$\begin{aligned} & \alpha|x|=|y| e^{x y-\beta} \\ & \frac{x d y-y d x}{y^2}+\frac{x y(x d y+y d x)}{y^2}=0 \\ & -d\left(\frac{x}{y}\right)+\frac{x}{y} d(x y)=0 \end{aligned}$$</p> <p>$$\begin{aligned} & \int d(x y)=\int \frac{d\left(\frac{x}{y}\right)}{\frac{x}{y}} \\ & x y=\ln \left|\frac{x}{y}\right|+\ln c \\ & x y=\ln \left(\left|\frac{x}{y}\right| \cdot c\right) \\ & \because y(1)=2 \\ & 2=\ln \left|\frac{1}{2}\right| c \Rightarrow c=2 e^2 \\ & \therefore \quad \operatorname{solution} x y=\ln \left(\left|\frac{x}{y}\right| \cdot 2 e^2\right) \\ & e^{x y}=\frac{|x|}{|y|} \cdot 2 e^2 \\ & 2|x|=|y| e^{x y-2} \\ & \Rightarrow \alpha=2, \beta=2, \alpha+\beta=4 \end{aligned}$$</p>