Question

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

Answer 1

$y+1=Y \Rightarrow d y=d Y$
$x+2=X \Rightarrow d x=d X$
$\Rightarrow\left(X e^{\frac{Y}{X}}+Y\right) d X=X d Y$
$\Rightarrow X d Y-Y d X=X e^{Y / x} d X$
$\Rightarrow d\left(\frac{Y}{X}\right) e^{-\frac{Y}{x}}=\frac{d X}{X}$
$-e^{-Y / x}=l|X|+c$
$(3,2) \rightarrow-e^{-\frac{2}{3}}=l|3|+c$
$-e^{-\frac{Y}{x}}=\ln|X|-e^{-\frac{2}{3}}- \ln 3$
$e^{-\frac{Y}{x}}=e^{2 / 3}+\ln 3-\ln |X| > 0$
$\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.