JEE Main 201910 Apr, Morning shiftMathematicsMediumDifferential Equations163× since 2002If y = y(x) is the solution of the differential equation dydx=(tanx−y)sec2x{{dy} \over {dx}} = \left( {\tan x - y} \right){\sec ^2}xdxdy=(tanx−y)sec2x, x∈(−π2,π2)x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)x∈(−2π,2π), such that y (0) = 0, then y(−π4)y\left( { - {\pi \over 4}} \right)y(−4π) is equal to :AOption A. 12−e{1 \over 2} - e21−eBOption B. e−2e - 2e−2COption C. 2+1e2 + {1 \over e}2+e1DOption D. 1e−2{1 \over e} - 2e1−2CheckSkip · show solution
