# Let y=y(x) be the solution of the differential equation

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

Solution:

$\mathrm{y}+1=\mathrm{Y} \Rightarrow \mathrm{dy}=\mathrm{d} \mathrm{Y}$

$x+2=X \Rightarrow d x=d X$

$\Rightarrow\left(X e^{\frac{Y}{X}}+Y\right) d X=X d Y$

$\Rightarrow \mathrm{Xd} \mathrm{Y}-\mathrm{YdX}=\mathrm{Xe}^{\mathrm{Y} / \mathrm{x}} \mathrm{dX}$

$\Rightarrow \mathrm{d}\left(\frac{\mathrm{Y}}{\mathrm{X}}\right) \mathrm{e}^{-\frac{\mathrm{Y}}{\mathrm{X}}}=\frac{\mathrm{dX}}{\mathrm{X}}$

$-\mathrm{e}^{-\mathrm{Y} / \mathrm{x}}=\ell|\mathrm{X}|+\mathrm{c}$

$(3,2) \rightarrow-\mathrm{e}^{-2 / 3}=\ell|3|+\mathrm{c}$

$-e^{-\frac{Y}{X}}=\ell n|X|-e^{-\frac{2}{3}}-\ell n 3$

$e^{-\frac{Y}{X}}=e^{2 / 3}+\ln 3-\ell n|X|>0$

$\ell \mathrm{n}|\mathrm{X}|<\left(\mathrm{e}^{2 / 3}+\ell \mathrm{n} 3\right)$

$\operatorname{Let} \lambda=\left(e^{2 / 3}+\ell n 3\right)$

$|x+2|$-\mathrm{e}^{\lambda}<\mathrm{x}+2<\mathrm{e}^{\lambda}-\mathrm{e}^{\lambda}-2<\mathrm{x}<\mathrm{e}^{\lambda}-2$Although$x=-2\$ should be excluded from domain but according to the given problem it will the most appropriate solution.