Let $$y = y(x)$$ be the solution of the differential equation $$\left((x+2)e^{\left(\frac{y+1}{x+2}\right)} + (y+1)\right)dx = (x+2)dy$$, $$y(1) = 1$$. If the domain of $$y = y(x)$$ is an open interval $$(\alpha, \beta)$$, then $$|\alpha + \beta|$$ is equal to ___.

Put

$$X=x+2,\qquad Y=y+1$$

Then,

$$dX=dx,\qquad dY=dy$$

and the initial condition becomes

$$X=3,\qquad Y=2$$

The differential equation becomes $$\left(Xe^{Y/X}+Y\right)dX=XdY$$

Hence,

$$\frac{dY}{dX}=e^{Y/X}+\frac{Y}{X}$$

Let

$$Y=vX$$

Then,

$$\frac{dY}{dX}=v+X\frac{dv}{dX}$$

Substituting,

$$v+X\frac{dv}{dX}=e^v+v$$

Thus,

$$X\frac{dv}{dX}=e^v$$

Separating variables,

$$e^{-v}dv=\frac{dX}{X}$$

Integrating,

$$-e^{-v}=\ln|X|+C$$

Using the initial condition

$$X=3,\qquad v=\frac23,$$

we get

$$-e^{-2/3}=\ln3+C$$

Hence,

$$C=-e^{-2/3}-\ln3$$

Therefore,

$$-e^{-v}=\ln|X|-\ln3-e^{-2/3}$$

or

$$e^{-v}=\ln\left(\frac3{|X|}\right)+e^{-2/3}$$

Since $$e^{-v}>0,$$ we must have

$$\ln\left(\frac3{|X|}\right)+e^{-2/3}>0$$

Thus,

$$\ln\left(\frac3{|X|}\right)>-e^{-2/3}$$

$$\frac3{|X|}>e^{-e^{-2/3}}$$

$$|X|<3e^{e^{-2/3}}$$

Hence,

$$-3e^{e^{-2/3}}<X<3e^{e^{-2/3}}$$

Substituting

$$X=x+2,$$

we get

$$-3e^{e^{-2/3}}-2<x<3e^{e^{-2/3}}-2$$

Therefore,

$$\alpha=-3e^{e^{-2/3}}-2,\qquad \beta=3e^{e^{-2/3}}-2$$

Hence,

$$\alpha+\beta=-4$$

Therefore,

$$|\alpha+\beta|=4$$