JEE Main 2026 · 2 Apr · Shift 2 · Mathematics Q20

MathematicsSingle correct+4 / -1Question 20 of 75

Let $x=x(y)$ be the solution of the differential equation $2y^2\dfrac{dx}{dy}-2xy+x^2=0$ , $y>1$ , $x(e)=e$ . Then $x\left(e^2\right)$ is equal to :

\dfrac32e^2

\dfrac23e^2

e^2

2e^2

Answer and worked solution

Official answer: B

Homogeneous. $\dfrac{dx}{dy}=\dfrac{2xy-x^2}{2y^2}$ . With $x=vy$ : $y\frac{dv}{dy}=-\frac{v^2}2$ , so $\frac1v=\frac12\ln y+C$ .

Constant. $x(e)=e$ means $v=1$ at $y=e$ : $C=\frac12$ .

At $y=e^2$ : $\frac1v=\frac32$ , so $x=\frac23e^2$ .

(B)

Topics:Homogeneous equations

JEE Main 2026 · 2 Apr · Shift 2, Mathematics question 20