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(B) Let $Y = y+1$ and $X = x+2$. Then $dY = dy$ and $dX = dx$.Substituting these into the equation: $(X e^{Y/X} + Y) dX = X dY$.Rearranging gives $X d

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$4$

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(B) Let $Y = y+1$ and $X = x+2$. Then $dY = dy$ and $dX = dx$.Substituting these into the equation: $(X e^{Y/X} + Y) dX = X dY$.Rearranging gives $X dY - Y dX = X e^{Y/X} dX$.Dividing by $X^2$: $\frac{X dY - Y dX}{X^2} = \frac{e^{Y/X}}{X} dX$.This simplifies to $d(\frac{Y}{X}) = e^{Y/X} \frac{dX}{X}$.Integrating both sides: $\int e^{-Y/X} d(Y/X) = \int \frac{dX}{X} \Rightarrow -e^{-Y/X} = \ln|X| + C$.Using $y(1)=1$,we have $Y=2$ and $X=3$: $-e^{-2/3} = \ln|3| + C$,so $C = -e^{-2/3} - \ln 3$.Thus,$-e^{-(y+1)/(x+2)} = \ln|x+2| - e^{-2/3} - \ln 3$.$e^{-(y+1)/(x+2)} = e^{-2/3} + \ln 3 - \ln|x+2|$.For the solution to exist,we require $e^{-2/3} + \ln 3 - \ln|x+2| > 0$,which means $\ln|x+2| Let $k = e^{-2/3} + \ln 3$. Then $|x+2| Thus,$\alpha = -e^k - 2$ and $\beta = e^k - 2$.Then $\alpha + \beta = -4$,so $|\alpha + \beta| = 4$.

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

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