Let y = y(x) be the solution curve of the differential equation (dy)/(dx) + 1 x ^ 2 - 1 y = ((x - 1)/(x + 1))(1)/(2) , x > 1 passing through the point (2, (1)/(3) ) . Then 7 y(8) is equal to

Correct answer: D. 19 - 6 _ e 3 Given, (dy)/(dx) + y x ^ 2 - 1 = ((x - 1)/(x + 1))(1)/(2) is a linear differential equation. Here I . F . = e ^ ∫ dx x ^ 2 - 1 = e ^ (1)/(2) ((x - 1)/(x + 1)) = (x - 1)/(x + 1) General solution will be y (x - 1)/(x + 1) = ∫ (x - 1)/(x + 1)dx + C ⇒ y (x - 1)/(x + 1) = x - 2 (x + 1) + c Given y(2) = 1 3 ⇒ c = 2 3 - (5)/(3) So, the equation of the curve is y (x - 1)/(x + 1) = x - 2 (x + 1) + 2 3 - (5)/(3) Now putting x = 8 , we get 7 y(8) 3 = 8 - 4 3 + 2 3 - (5)/(3) ⇒ 7 y(8) = 19 - 6 3

JEE Main 202228 JulyEvening Shift

Exam: JEE Main
Year: 2022
Date: 28 July
Shift: Evening Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2022 · 28 July · Evening Shift · Mathematics · Differential Equations

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

Let

y = y\left(x\right)

be the solution curve of the differential equation

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

x > 1

passing through the point \left(2, \sqrt{\dfrac{1}{3}}\right)

. Then

\sqrt{7}y\left(8\right) is equal to

Options

A $11 + 6\log_{e}3$
B $19$
C $12 - 2\log_{e}3$
D $19 - 6\log_{e}3$

Correct answer

D. $19 - 6\log_{e}3$

Step-by-step solution

Given,
$\dfrac{dy}{dx} + \dfrac{y}{{x}^{2} - 1} =$ \left(\dfrac{x - 1}{x + 1}\right)\dfrac{1}{2} $is a linear differential equation.Here$ I . F . = {e}^{∫ \dfrac{dx}{{x}^{2} - 1}} = {e}^{\dfrac{1}{2}\ln \left(\dfrac{x - 1}{x + 1}\right)} = \sqrt{\dfrac{x - 1}{x + 1}} $General solution will be$ y\sqrt{\dfrac{x - 1}{x + 1}} = ∫ \dfrac{x - 1}{x + 1}dx + C $$ ⇒ y\sqrt{\dfrac{x - 1}{x + 1}} = x - 2\ln \left(x + 1\right) + c $Given$ y\left(2\right) = \dfrac{1}{\sqrt{3}} $$ ⇒ c = 2\ln 3 - \dfrac{5}{3} $So, the equation of the curve is$ y\sqrt{\dfrac{x - 1}{x + 1}} = x - 2\ln \left(x + 1\right) + 2\ln 3 - \dfrac{5}{3} $Now putting$ x = 8 $, we get$ \dfrac{\sqrt{7}y\left(8\right)}{3} = 8 - 4\ln 3 + 2\ln 3 - \dfrac{5}{3} $$ ⇒ \sqrt{7}y\left(8\right) = 19 - 6\ln 3

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