JEE Main 202122 Jul, Evening shiftMathematicsMediumDifferential Equations163× since 2002Let y = y(x) be the solution of the differential equation cosec2xdy+2dx=(1+ycos2x)cosec2xdx\cos e{c^2}xdy + 2dx = (1 + y\cos 2x)\cos e{c^2}xdxcosec2xdy+2dx=(1+ycos2x)cosec2xdx, with y(π4)=0y\left( {{\pi \over 4}} \right) = 0y(4π)=0. Then, the value of (y(0)+1)2{(y(0) + 1)^2}(y(0)+1)2 is equal to :AOption A. e1/2BOption B. e^−-−1/2COption C. e^−-−1DOption D. eCheckSkip · show solution
