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JEE Main 20205 SeptemberEvening Shift
Exam
JEE Main
Year
2020
Date
5 September
Shift
Evening Shift
Subject
Mathematics
Chapter
Differential Equations

JEE Main 2020 · 5 September · Evening Shift · Mathematics · Differential Equations

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

Let y=y(x)y = y ( x ) be the solution of the differential equation cos⁡xdydx+2ysinx=sin⁡2x,x∈(0,π2)\cos x\dfrac{dy}{dx} + 2ysinx = \sin 2x, x \in \left(0, \dfrac{\pi}{2}\right)
If y(π/3)=0,y ( \pi / 3 ) = 0, then y(π/4)y ( \pi / 4 ) is equal to :

Options

  1. A2−22 - \sqrt{2}
  2. B2+22 + \sqrt{2}
  3. C2−2\sqrt{2} - 2
  4. D12−1\dfrac{1}{\sqrt{2}} - 1

Correct answer

C. 2−2\sqrt{2} - 2

Step-by-step solution

dydx+2tan⁡x⋅y=2sin⁡x\dfrac{dy}{dx} + 2\tan x \cdot y = 2\sin x

I.F. =e∫2tanxdx=sec⁡2x= {e}^{∫ 2tanxdx} = \sec^{2}x

solution is
y⋅sec⁡2x=∫2sin⁡x⋅sec⁡2xdx+Cy \cdot \sec^{2}x = ∫ 2\sin x \cdot \sec^{2}xdx + C

ysec2x=2sec⁡x+Cy{sec}^{2}x = 2\sec x + C
⇒0=2.2+C⇒C=−4⇒ 0 = 2 . 2 + C ⇒ C = - 4
ysec2x=2sec⁡x−4y{sec}^{2}x = 2\sec x - 4
y(π4)=2−2y\left(\dfrac{\pi}{4}\right) = \sqrt{2} - 2

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