JEE Main 2022 · 28 Jul · Shift 2 · Mathematics Q71

← JEE Main 2022 · 28 Jul · Shift 2

MathematicsSingle correct+4 / -1Question 71 of 90

Let $y=y(x)$ be the solution curve of the differential equation $\dfrac{dy}{dx}+\dfrac1{x^2-1}y=\left(\dfrac{x-1}{x+1}\right)^{1/2}$ , $x>1$ passing through the point $\left(2,\sqrt{\dfrac13}\right)$ . Then $\sqrt7\,y(8)$ is equal to :

11+6\log_e3

12-2\log_e3

19-6\log_e3

Answer and worked solution

Official answer: D

Integrating factor $\sqrt{\frac{x-1}{x+1}}$ , so $\frac d{dx}\left(y\sqrt{\tfrac{x-1}{x+1}}\right)=\frac{x-1}{x+1}=1-\frac2{x+1}$ .

Integrate from 2: $y\sqrt{\tfrac79}=\frac13+6-2\ln3$ , hence $\sqrt7y(8)=19-6\ln3$ .

(D)

Topics:Linear first-order equations

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JEE Main 2022 · 28 Jul · Shift 2, Mathematics question 71