JEE Main 2022 · 28 Jul · Shift 1 · Mathematics Q1

← JEE Main 2022 · 28 Jul · Shift 1

MathematicsSingle correct+4 / -1Question 1 of 90

Let the solution curve of the differential equation $x\,dy=\left(\sqrt{x^2+y^2}+y\right)dx$ , $x>0$ , intersect the line $x=1$ at $y=0$ and the line $x=2$ at $y=\alpha$ . Then the value of $\alpha$ is :

\dfrac12

\dfrac32

-\dfrac32

\dfrac52

Answer and worked solution

Official answer: B

Homogeneous. With $y=vx$ : $\frac{dv}{\sqrt{1+v^2}}=\frac{dx}x$ , so $y+\sqrt{x^2+y^2}=cx^2$ .

Through (1, 0): $c=1$ . At x = 2: $\sqrt{4+\alpha^2}=4-\alpha$ , so $\alpha=\frac32$ .

(B)

Topics:Homogeneous equations

Next question →

JEE Main 2022 · 28 Jul · Shift 1, Mathematics question 1