JEE Main 2021 — Differential Equations Question with Solution

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

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

Thus, the differential equation becomes $\left(X{e}^{\frac{Y}{X}}+Y\right)dX=XdY$

$\Rightarrow XdY-YdX=X{e}^{\frac{Y}{X}}dX$

$\Rightarrow \frac{\left(XdY-YdX\right)}{X^2}=\left(\frac{X{e}^{{\displaystyle \frac{Y}{X}}}}{X^2}\right)dX$

$\Rightarrow \frac{d}{dX}\left(\frac{Y}{X}\right)=\left(\frac{X{e}^{{\displaystyle \frac{Y}{X}}}}{X^2}\right)dX$

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

Integrating both sides w.r.to $X,$ we get $\int d\left(\frac{Y}{X}\right)e^{-\frac{Y}{X}}=\int \frac{dX}{X}$

$\Rightarrow -e^{-\frac{Y}{X}}={\log }_e\left|X\right|+c$

Given, at $X=3, Y=2$

$\Rightarrow -e^{-\frac{2}{3}}={\log }_e\left|3\right|+c$

$\Rightarrow c=-e^{-\frac{2}{3}}-{\log }_e\left|3\right|$

$\Rightarrow -e^{-\frac{Y}{X}}={\log }_e\left|X\right|-e^{-\frac{2}{3}}-{\log }_{e}3$

$e^{-\frac{Y}{X}}=e^{\frac{2}{3}}+{\log }_{e}3-{\log }_e\left|X\right|>0$

${\log }_e\left|X\right|<e^{\frac{2}{3}}+{\log }_{e}3$

Let, $\lambda =\left(e^{\frac{2}{3}}+{\log }_{e}3\right)$ then, we have $\left|x+2\right|<e^{\lambda }$

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

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

Thus, the domain of the function is $\left(-e^{\lambda }-2, e^{\lambda }-2\right).$

Given, the domain of the function is $\left(\alpha , \beta \right),$ hence $\alpha =-e^{\lambda }-2 \& \beta =e^{\lambda }-2$

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

$\Rightarrow |\alpha +\beta |=4.$