JEE Main 2021 — Mathematics Calculus

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

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

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

$$\Rightarrow XdY-YdX=X{e}^{\frac{Y}{X}}dX$$⇒XdY−YdX=XeXY​dX

$$\Rightarrow \frac{(XdY-YdX)}{{X}^{2}}=(\frac{X{e}^{\frac{Y}{X}}}{{X}^{2}})dX$$⇒X2(XdY−YdX)​=(X2XeXY​​)dX

$$\Rightarrow \frac{d}{dX}(\frac{Y}{X})=(\frac{X{e}^{\frac{Y}{X}}}{{X}^{2}})dX$$⇒dXd​(XY​)=(X2XeXY​​)dX

$$\Rightarrow d(\frac{Y}{X}){e}^{-\frac{Y}{X}}=\frac{dX}{X}$$⇒d(XY​)e−XY​=XdX​

Integrating both sides w.r.to $X,$X, we get $\displaystyle \int d(\frac{Y}{X}){e}^{-\frac{Y}{X}}=\int \frac{dX}{X}$∫d(XY​)e−XY​=∫XdX​

$$\Rightarrow -{e}^{-\frac{Y}{X}}={\mathrm{log}}_{e}|X|+c$$⇒−e−XY​=loge​∣X∣+c

Given, at $X=3,Y=2$X=3,Y=2

$$\Rightarrow -{e}^{-\frac{2}{3}}={\mathrm{log}}_{e}|3|+c$$⇒−e−32​=loge​∣3∣+c

$$\Rightarrow c=-{e}^{-\frac{2}{3}}-{\mathrm{log}}_{e}|3|$$⇒c=−e−32​−loge​∣3∣

$$\Rightarrow -{e}^{-\frac{Y}{X}}={\mathrm{log}}_{e}|X|-{e}^{-\frac{2}{3}}-{\mathrm{log}}_{e}3$$⇒−e−XY​=loge​∣X∣−e−32​−loge​3

$${e}^{-\frac{Y}{X}}={e}^{\frac{2}{3}}+{\mathrm{log}}_{e}3-{\mathrm{log}}_{e}|X|>0$$e−XY​=e32​+loge​3−loge​∣X∣>0

$${\mathrm{log}}_{e}|X|<{e}^{\frac{2}{3}}+{\mathrm{log}}_{e}3$$loge​∣X∣<e32​+loge​3

Let, $\lambda =({e}^{\frac{2}{3}}+{\mathrm{log}}_{e}3)$λ=(e32​+loge​3) then, we have $|x+2|<{e}^{\lambda }$∣x+2∣<eλ

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

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

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

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

$$\Rightarrow \alpha +\beta =-{e}^{\lambda }-2+{e}^{\lambda }-2=-4$$⇒α+β=−eλ−2+eλ−2=−4

$$\Rightarrow |\alpha +\beta |=4.$$⇒∣α+β∣=4.