JEE Main 2022 · 25 Jun · Shift 2 · Mathematics Q13

MathematicsSingle correct+4 / -1Question 13 of 90

If $y=y(x)$ is the solution of the differential equation $2x^2\dfrac{dy}{dx}-2xy+3y^2=0$ such that $y(e)=\dfrac e3$ , then $y(1)$ is equal to

\dfrac13

\dfrac23

\dfrac32

Answer and worked solution

Official answer: B

Homogeneous. $y=vx$ : $2x^2(v+xv')-2x^2v+3v^2x^2=0$ , i.e. $2x\,v'=-3v^2$ .

Integrate. $-\dfrac1v=-\dfrac32\log_ex+C$ ; at $x=e$ , $v=\frac13$ : $C=-\frac32$ .

At $x=1$ . $-\frac1v=-\frac32$ , $v=\frac23$ , so $y(1)=\frac23$ .

(B)

Topics:Homogeneous equations

JEE Main 2022 · 25 Jun · Shift 2, Mathematics question 13