JEE Main 201910 Apr, Evening shiftMathematicsMediumDifferential Equations163× since 2002Let y = y(x) be the solution of the differential equation, dydx+ytanx=2x+x2tanx{{dy} \over {dx}} + y\tan x = 2x + {x^2}\tan xdxdy+ytanx=2x+x2tanx, x∈(−π2,π2)x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)x∈(−2π,2π), such that y(0) = 1. Then :AOption A. y(π4)−y(−π4)=2y\left( {{\pi \over 4}} \right) - y\left( { - {\pi \over 4}} \right) = \sqrt 2y(4π)−y(−4π)=2BOption B. y′(π4)−y′(−π4)=π−2y'\left( {{\pi \over 4}} \right) - y'\left( { - {\pi \over 4}} \right) = \pi - \sqrt 2y′(4π)−y′(−4π)=π−2COption C. y(π4)+y(−π4)=π22+2y\left( {{\pi \over 4}} \right) + y\left( { - {\pi \over 4}} \right) = {{{\pi ^2}} \over 2} + 2y(4π)+y(−4π)=2π2+2DOption D. y′(π4)+y′(−π4)=−2y'\left( {{\pi \over 4}} \right) + y'\left( { - {\pi \over 4}} \right) = - \sqrt 2y′(4π)+y′(−4π)=−2CheckSkip · show solution
