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Differential Equations question

2016 · 10 Apr · Shift 1 · Q35
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  5. /2016 · 10 Apr · Shift 1 · Q35

Differential Equations question

2016 · 10 Apr · Shift 1 · Q35

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
The solution of the differential equation dydx + y2 sec⁡x=tan⁡x2y,  {{dy} \over {dx}}\, + \,{y \over 2}\,\sec x = {{\tan x} \over {2y}},\,\,dxdy​+2y​secx=2ytanx​, where 0 ≤\le≤ x <π2{\pi \over 2}2π​, and y (0) = 1, is given by :
  1. A
    y = 1 −xsec⁡x+tan⁡x-{x \over {\sec x + \tan x}}−secx+tanxx​
  2. B
    y2 = 1 + xsec⁡x+tan⁡x{x \over {\sec x + \tan x}}secx+tanxx​
  3. C
    y2 = 1 −xsec⁡x+tan⁡x-{x \over {\sec x + \tan x}}−secx+tanxx​
  4. D
    y = 1 + xsec⁡x+tan⁡x{x \over {\sec x + \tan x}}secx+tanxx​
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