Verified Solution • 2025–26 Session moderate

QJEEM21Ma1fa1: JEE Main 2021 Mathematics - Class 12 JEEM21Ma1fa1 Solution: Let y = y(x) be the solution of the differential equation ,…

Let y = y(x) be the solution of the differential equation ((x+2)e(y+1x+2)+(y+1))dx=(x+2)dy\left( {(x + 2){e^{\left( {{{y + 1} \over {x + 2}}} \right)}} + (y + 1)} \right)dx = (x + 2)dy , y(1) = 1. If the domain of y = y(x) is an open interval ( α\alpha , β\beta ), then | α\alpha + β\beta | is equal to ______________.

Step-by-Step Verified Solution

Let y + 1 = Y and x + 2 = X

dy = dY

dx = dX

(XeXX+Y)dX=XdY\left( {X{e^{{X \over X}}} + Y} \right)dX = XdY

⇒XdY−YdXX2=eYXXdX\Rightarrow {{XdY - YdX} \over {{X^2}}} = {{{e^{{Y \over X}}}} \over X}dX

⇒e−YXd(YX)=dXX\Rightarrow {e^{ - {Y \over X}}}d\left( {{Y \over X}} \right) = {{dX} \over X}

⇒−e−YX=ln⁡∣X∣+c\Rightarrow - {e^{ - {Y \over X}}} = \ln |X| + c

⇒−e−(y+1x+2)=ln⁡∣x+2∣+c\Rightarrow - {e^{ - \left( {{{y + 1} \over {x + 2}}} \right)}} = \ln |x + 2| + c

∵\because (1, 1) satisfy this equation

So, c=−e−23−ln⁡3c = - {e^{ - {2 \over 3}}} - \ln 3

Now, y=−1−(x+2)ln⁡(ln⁡(∣3x+2∣)+e−23)y = - 1 - (x + 2)\ln \left( {\ln \left( {\left| {{3 \over {x + 2}}} \right|} \right) + {e^{ - {2 \over 3}}}} \right)

Domain :

\ln \left| {{3 \over {x + 2}}} \right| > {e^{ - {e^{ - {2 \over 3}}}}}

\Rightarrow {3 \over {\left| {x + 2} \right|}} > {e^{ - {e^{ - {2 \over 3}}}}}

\Rightarrow \left| {x + 2} \right| < 3{e^{{e^{ - {2 \over 3}}}}}

\Rightarrow - 3{e^{{e^{ - {2 \over 3}}}}} - 2 < x < 3{e^{{e^{ - {2 \over 3}}}}} - 2

So, α+β=−4\alpha + \beta = - 4

⇒∣α+β∣=4\Rightarrow \left| {\alpha + \beta } \right| = 4

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