JEE Main 201610 Apr, Morning shiftMathematicsMediumDifferential Equations163× since 2002The solution of the differential equation dydx + y2 secx=tanx2y, {{dy} \over {dx}}\, + \,{y \over 2}\,\sec x = {{\tan x} \over {2y}},\,\,dxdy+2ysecx=2ytanx, where 0 ≤\le≤ x < π2{\pi \over 2}2π, and y (0) = 1, is given by :AOption A. y = 1 −-− xsecx+tanx{x \over {\sec x + \tan x}}secx+tanxxBOption B. y² = 1 + xsecx+tanx{x \over {\sec x + \tan x}}secx+tanxxCOption C. y² = 1 −-− xsecx+tanx{x \over {\sec x + \tan x}}secx+tanxxDOption D. y = 1 + xsecx+tanx{x \over {\sec x + \tan x}}secx+tanxxCheckSkip · show solution
