JEE Main 2022 · 29 Jun · Shift 1 · Mathematics Q2

MathematicsSingle correct+4 / -1Question 2 of 90

Let the solution curve of the differential equation $x\dfrac{dy}{dx}-y=\sqrt{y^2+16x^2}$ , $y(1)=3$ be $y=y(x)$ . Then $y(2)$ is equal to :

Answer and worked solution

Official answer: A

Homogeneous. With $y=vx$ : $xv'=\sqrt{v^2+16}$ , so $\ln\left(v+\sqrt{v^2+16}\right)=\ln x+C$ ; $y(1)=3$ gives $C=\ln8$ .

At $x=2$ . $v+\sqrt{v^2+16}=16$ , so $v=7.5$ and $y=15$ .

(A)

Topics:Homogeneous equations

JEE Main 2022 · 29 Jun · Shift 1, Mathematics question 2