JEE Main 2018Session, Shift 1MathematicsHardDifferential Equations163× since 2002Let y = y(x) be the solution of the differential equation sinxdydx+ycosx=4x\sin x{{dy} \over {dx}} + y\cos x = 4xsinxdxdy+ycosx=4x, x∈(0,π)x \in \left( {0,\pi } \right)x∈(0,π). If y(π2)=0y\left( {{\pi \over 2}} \right) = 0y(2π)=0, then y(π6)y\left( {{\pi \over 6}} \right)y(6π) is equal to :AOption A. −49π2- {4 \over 9}{\pi ^2}−94π2BOption B. 493π2{4 \over {9\sqrt 3 }}{\pi ^2}934π2COption C. −893π2- {8 \over {9\sqrt 3 }}{\pi ^2}−938π2DOption D. −89π2- {8 \over 9}{\pi ^2}−98π2CheckSkip · show solution
