MathematicsMedium38×since 2002Q3509The 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 :Ay = 1 −-− xsecx+tanx{x \over {\sec x + \tan x}}secx+tanxxBy² = 1 + xsecx+tanx{x \over {\sec x + \tan x}}secx+tanxxCy² = 1 −-− xsecx+tanx{x \over {\sec x + \tan x}}secx+tanxxDy = 1 + xsecx+tanx{x \over {\sec x + \tan x}}secx+tanxxCheck answerSkip