JEE Main 2022 · 24 Jun · Shift 1 · Mathematics Q14

MathematicsSingle correct+4 / -1Question 14 of 90

If $x=x(y)$ is the solution of the differential equation $y\dfrac{dx}{dy}=2x+y^3(y+1)e^y$ , $x(1)=0$ ; then $x(e)$ is equal to :

e^3(e^e-1)

e^e(e^3-1)

e^2(e^e+1)

e^e(e^2-1)

Answer and worked solution

Official answer: A

Linear in $x$ . $\dfrac{dx}{dy}-\dfrac2yx=y^2(y+1)e^y$ ; integrating factor $y^{-2}$ .

Integrate. $\dfrac{d}{dy}\left(\dfrac x{y^2}\right)=(y+1)e^y$ , so $\dfrac x{y^2}=ye^y+C$ .

Constant. $x(1)=0$ gives $C=-e$ , so $x=y^2(ye^y-e)$ .

At $y=e$ . $x=e^2(e\cdot e^e-e)=e^3(e^e-1)$ .

(A)

Topics:Linear first-order equations Integration by parts

JEE Main 2022 · 24 Jun · Shift 1, Mathematics question 14