JEE Main 2014 · 12 Apr · Shift 1 · Mathematics Q77

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MathematicsSingle correct+4 / -1Question 77 of 90

The general solution of the differential equation, $\sin2x\left(\dfrac{dy}{dx}-\sqrt{\tan x}\right)-y=0$ , is :

y\sqrt{\tan x}=x+c

y\sqrt{\cot x}=\tan x+c

y\sqrt{\tan x}=\cot x+c

y\sqrt{\cot x}=x+c

Answer and worked solution

Official answer: D

$\dfrac{dy}{dx}-\dfrac y{\sin2x}=\sqrt{\tan x}$ . The integrating factor is
$e^{-\int\csc2x\,dx}=e^{-\frac12\ln\tan x}=\sqrt{\cot x}.$
Then $y\sqrt{\cot x}=\int\sqrt{\tan x}\sqrt{\cot x}\,dx=x+c$ .

Topics:Linear first-order equations

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JEE Main 2014 · 12 Apr · Shift 1, Mathematics question 77