JEE Main 2022 · 28 Jun · Shift 1 · Mathematics Q10

← JEE Main 2022 · 28 Jun · Shift 1

MathematicsSingle correct+4 / -1Question 10 of 90

Let the solution curve $y=y(x)$ of the differential equation
$\left[\frac x{\sqrt{x^2-y^2}}+e^{\frac yx}\right]x\frac{dy}{dx}=x+\left[\frac x{\sqrt{x^2-y^2}}+e^{\frac yx}\right]y$
pass through the points $(1,0)$ and $(2\alpha,\alpha)$ , $\alpha>0$ . Then $\alpha$ is equal to

\dfrac12\exp\left(\dfrac\pi6+\sqrt e-1\right)

\dfrac12\exp\left(\dfrac\pi3+e-1\right)

\exp\left(\dfrac\pi6+\sqrt e+1\right)

2\exp\left(\dfrac\pi3+\sqrt e-1\right)

Answer and worked solution

Official answer: A

Rearrange. $F\,(x\,dy-y\,dx)=x\,dx$ with $F=\frac1{\sqrt{1-v^2}}+e^v$ , $v=\frac yx$ . Since $x\,dy-y\,dx=x^2dv$ :
$\left(\frac1{\sqrt{1-v^2}}+e^v\right)dv=\frac{dx}x\ \Rightarrow\ \sin^{-1}v+e^v=\ln x+C.$

Constants. $(1,0)$ gives $C=1$ . At $(2\alpha,\alpha)$ , $v=\frac12$ : $\frac\pi6+\sqrt e=\ln2\alpha+1$ .

So $\alpha=\frac12\exp\left(\frac\pi6+\sqrt e-1\right)$ .

(A)

Topics:Homogeneous equations

← Previous question Next question →

JEE Main 2022 · 28 Jun · Shift 1, Mathematics question 10