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

The given differential equation is $\left(\left(x + 2\right){e}^{\left(\dfrac{y + 1}{x + 2}\right)} + \left(y + 1\right)\right)dx = \left(x + 2\right)dy, y\left(1\right) = 1$

Let, $y + 1 = Y ⇒ dy = dY$ and $x + 2 = X ⇒ dx = dX$ and at $y = 1, Y = 2$ and at $x = 1, X = 3 .$

Thus, the differential equation becomes \left(Xe\dfrac{Y}{X} + Y\right)dX = XdY$</p><p>$⇒ XdY - YdX = Xe\dfrac{Y}{X}dX$</p><p>$⇒ \dfrac{\left(XdY - YdX\right)}{{X}^{2}} = \left(\dfrac{X{e}^{\dfrac{Y}{X}}}{{X}^{2}}\right)dX$</p><p>$⇒ \dfrac{d}{dX}\left(\dfrac{Y}{X}\right) = \left(\dfrac{X{e}^{\dfrac{Y}{X}}}{{X}^{2}}\right)dX$</p><p>$⇒ d\left(\dfrac{Y}{X}\right){e}^{- \dfrac{Y}{X}} = \dfrac{dX}{X}$</p><p>Integrating both sides w.r.to$X,$we get$∫ d\left(\dfrac{Y}{X}\right){e}^{- \dfrac{Y}{X}} = ∫ \dfrac{dX}{X}$</p><p>$⇒ - {e}^{- \dfrac{Y}{X}} = \log_{e}\left|X\right| + c$</p><p>Given, at$X = 3, Y = 2$</p><p>$⇒ - {e}^{- \dfrac{2}{3}} = \log_{e}\left|3\right| + c$</p><p>$⇒ c = - {e}^{- \dfrac{2}{3}} - \log_{e}\left|3\right|$</p><p>$⇒ - {e}^{- \dfrac{Y}{X}} = \log_{e}\left|X\right| - {e}^{- \dfrac{2}{3}} - \log_{e}3$</p><p>${e}^{- \dfrac{Y}{X}} = e\dfrac{2}{3} + \log_{e}3 - \log_{e}\left|X\right| > 0$</p><p>$\log_{e}\left|X\right| < e\dfrac{2}{3} + \log_{e}3$</p><p>Let,$\lambda = \left(e\dfrac{2}{3} + \log_{e}3\right)$then, we have \left|x + 2\right| < {e}^{\lambda}$

$⇒ - {e}^{\lambda} < x + 2 < {e}^{\lambda}$

$⇒ - {e}^{\lambda} - 2 < x < {e}^{\lambda} - 2$

Thus, the domain of the function is \left(- {e}^{\lambda} - 2, {e}^{\lambda} - 2\right) .$</p><p>Given, the domain of the function is \left(\alpha, \beta\right),$ hence $\alpha = - {e}^{\lambda} - 2 \text{ and } \beta = {e}^{\lambda} - 2$

$⇒ \alpha + \beta = - {e}^{\lambda} - 2 + {e}^{\lambda} - 2 = - 4$

$⇒ | \alpha + \beta | = 4 .$