Question

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 _______.

Answer 1

Answer: 4

$y+1=Y\Rightarrow dy=dY$
$x+2=X\Rightarrow dx=dX$
$\Rightarrow \left(X{e}^{\frac{Y}{X}}+Y\right)dX=XdY$
$\Rightarrow XdY-YdX=X{e}^{Y/x}dX$
$\Rightarrow d\left(\frac{Y}{X}\right)e^{-\frac{Y}{x}}=\frac{dX}{X}$
$-e^{-Y/x}=l|X|+c$
$(3,2)\to -e^{-\frac{2}{3}}=l|3|+c$
$-e^{-\frac{Y}{x}}=\ln |X|-e^{-\frac{2}{3}}-\ln 3$
$e^{-\frac{Y}{x}}=e^{2/3}+\ln 3-\ln |X|>0$
$\ln |X|<\left(e^{2/3}+\ln 3\right)$
Let $\lambda =\left(e^{2/3}+\ln 3\right)$
$|x+2|<e^{\lambda }$
$-e^{\lambda }<x+2<e^{\lambda }$
$-e^{\lambda }-2<x<e^{\lambda }-2$
$\alpha +\beta =-4\Rightarrow |\alpha +\beta |=4$
Although $x=-2$ should be excluded from domain but according to the given problem it will the most appropriate solution.