Let the solution curve y = y(x) of the differential equation, [ x x ^ 2 - y ^ 2 + e(y)/(x)]x(dy)/(dx) = x + [ x x ^ 2 - y ^ 2 + e(y)/(x)]y pass through the points (1, 0) and (2α, α), α > 0 . Then α is equal to

Correct answer: A. (1)/(2)exp((π)/(6) + e - 1) Given x( x x ^ 2 - y ^ 2 + e(y)/(x))(dy)/(dx) = y( x x ^ 2 - y ^ 2 + e(y)/(x)) + x Taking x common & cancelling them we get, (dy)/(dx) × ( 1 1 - ((y)/(x)) ^ 2 + e(y)/(x)) = (y)/(x)( 1 1 - ((y)/(x)) ^ 2 + e(y)/(x)) + 1 Let y = vx⇒ (dy)/(dx) = v + x(dv)/(dx) (v + x(dv)/(dx))( 1 1 - v ^ 2 + e ^ v ) = v( 1 1 - v ^ 2 + e ^ v ) + 1 v + x(dV)/(dx) = v + 1 ( 1 1 - v ^ 2 + e ^ v ) x(dv)/(dx) = 1 ( 1 1 - v ^ 2 + e ^ v ) ⇒ ( 1 1 - v ^ 2 + e ^ v )dv = (dx)/(x) Integrating both side we get, ⇒ ∫ ( 1 1 - v ^ 2 + e ^ v )dv = ∫ (dx)/(x) ⇒ ^ - 1 v + e ^ v = x + c ⇒ ^ - 1 ((y)/(x)) + e(y)/(x) = x + c Now y(1) = 0 ⇒ ^ - 1 ((0)/(1)) + e ^ 0 = 1 + c ⇒ c = 1 ⇒ ^ - 1 ((y)/(x)) + e(y)/(x) = x + 1 ........(i) Now y(2α) = α putting in equation (i) we get, ⇒ ^ - 1 ((α)/(2α)) + e(α)/(2α) = 2α + 1 ⇒ (π)/(6) + e(1)/(2) = 2α + 1 ⇒ α = (1)/(2) e ^ (π)/(6) + e - 1

JEE Main 202228 JuneMorning Shift

Exam: JEE Main
Year: 2022
Date: 28 June
Shift: Morning Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2022 · 28 June · Morning Shift · Mathematics · Differential Equations

Let the solution curve y = y(x) of the differential equation, [ x x...

Let the solution curve

y = y\left(x\right)

of the differential equation, \left[\dfrac{x}{\sqrt{{x}^{2} - {y}^{2}}} + e\dfrac{y}{x}\right]x\dfrac{dy}{dx} = x +

\left[\dfrac{x}{\sqrt{{x}^{2} - {y}^{2}}} + e\dfrac{y}{x}\right]y

pass through the points \left(1, 0\right)

and \left(2\alpha, \alpha\right), \alpha > 0

. Then

\alpha

is equal to

Options

A $\dfrac{1}{2}exp\left(\dfrac{\pi}{6} + \sqrt{e} - 1\right)$
B $\dfrac{1}{2}exp\left(\dfrac{\pi}{3} + \sqrt{e} - 1\right)$
C $exp\left(\dfrac{\pi}{6} + \sqrt{e} + 1\right)$
D $2exp\left(\dfrac{\pi}{3} + \sqrt{e} - 1\right)$

Correct answer

A. $\dfrac{1}{2}exp\left(\dfrac{\pi}{6} + \sqrt{e} - 1\right)$

Step-by-step solution

Given $x\left(\dfrac{x}{\sqrt{{x}^{2} - {y}^{2}}} + e\dfrac{y}{x}\right)\dfrac{dy}{dx} = y\left(\dfrac{x}{\sqrt{{x}^{2} - {y}^{2}}} + e\dfrac{y}{x}\right) + x$

Taking $x$ common & cancelling them we get,

$\dfrac{dy}{dx} \times \left(\dfrac{1}{\sqrt{1 - {\left(\dfrac{y}{x}\right)}^{2}}} + e\dfrac{y}{x}\right) = \dfrac{y}{x}\left(\dfrac{1}{\sqrt{1 - {\left(\dfrac{y}{x}\right)}^{2}}} + e\dfrac{y}{x}\right) + 1$

Let $y = vx⇒ \dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$

\left(v + x\dfrac{dv}{dx}\right)\left(\dfrac{1}{\sqrt{1 - {v}^{2}}} + {e}^{v}\right) = v\left(\dfrac{1}{\sqrt{1 - {v}^{2}}} + {e}^{v}\right) + 1 $$ v + x\dfrac{dV}{dx} = v + \dfrac{1}{\left(\dfrac{1}{\sqrt{1 - {v}^{2}}} + {e}^{v}\right)} $$ x\dfrac{dv}{dx} = \dfrac{1}{\left(\dfrac{1}{\sqrt{1 - {v}^{2}}} + {e}^{v}\right)} ⇒ \left(\dfrac{1}{\sqrt{1 - {v}^{2}}} + {e}^{v}\right)dv = \dfrac{dx}{x} $Integrating both side we get,$ ⇒ ∫ \left(\dfrac{1}{\sqrt{1 - {v}^{2}}} + {e}^{v}\right)dv = ∫ \dfrac{dx}{x} $$ ⇒ \sin^{- 1}v + {e}^{v} = \ln x + c ⇒ \sin^{- 1}\left(\dfrac{y}{x}\right) + e\dfrac{y}{x} = \ln x + c $Now$ y\left(1\right) = 0 $$ ⇒ \sin^{- 1}\left(\dfrac{0}{1}\right) + {e}^{0} = \ln 1 + c $$ ⇒ c = 1 $$ ⇒ \sin^{- 1}\left(\dfrac{y}{x}\right) + e\dfrac{y}{x} = \ln x + 1 $........(i)Now$ y\left(2\alpha\right) = \alpha $putting in equation (i) we get,$ ⇒ \sin^{- 1}\left(\dfrac{\alpha}{2\alpha}\right) + e\dfrac{\alpha}{2\alpha} = \ln 2\alpha + 1 $$ ⇒ \dfrac{\pi}{6} + e\dfrac{1}{2} = \ln 2\alpha + 1 ⇒ \alpha = \dfrac{1}{2}{e}^{\dfrac{\pi}{6} + \sqrt{e} - 1}$

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