JEE Main 2016 · 10 Apr · Shift 1 · Mathematics Q77

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The solution of the differential equation $\dfrac{dy}{dx}+\dfrac y2\sec x=\dfrac{\tan x}{2y}$ , where $0\le x<\dfrac\pi2$ , and y(0) = 1, is given by :

y=1-\dfrac x{\sec x+\tan x}

y^2=1+\dfrac x{\sec x+\tan x}

y^2=1-\dfrac x{\sec x+\tan x}

y=1+\dfrac x{\sec x+\tan x}

Answer and worked solution

Official answer: C

Multiplying by 2y and putting $u=y^2$ : $u'+u\sec x=\tan x$ . The integrating factor is $\sec x+\tan x$ , and
$u(\sec x+\tan x)=\int(\sec x\tan x+\sec^2x-1)dx=\sec x+\tan x-x+C.$
y(0) = 1 gives C = 0, so $y^2=1-\frac x{\sec x+\tan x}$ .

Topics:Linear first-order equations

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JEE Main 2016 · 10 Apr · Shift 1, Mathematics question 77