JEE Main 202117 Mar, Evening shiftMathematicsMediumDifferential Equations163× since 2002Let y = y(x) be the solution of the differential equation cosx(3sinx+cosx+3)dy=(1+ysinx(3sinx+cosx+3))dx,0≤x≤π2,y(0)=0\cos x(3\sin x + \cos x + 3)dy = (1 + y\sin x(3\sin x + \cos x + 3))dx,0 \le x \le {\pi \over 2},y(0) = 0cosx(3sinx+cosx+3)dy=(1+ysinx(3sinx+cosx+3))dx,0≤x≤2π,y(0)=0. Then, y(π3)y\left( {{\pi \over 3}} \right)y(3π) is equal to :AOption A. 2loge(3+72)2{\log _e}\left( {{{\sqrt 3 + 7} \over 2}} \right)2loge(23+7)BOption B. 2loge(33−84)2{\log _e}\left( {{{3\sqrt 3 - 8} \over 4}} \right)2loge(433−8)COption C. 2loge(23+1011)2{\log _e}\left( {{{2\sqrt 3 + 10} \over {11}}} \right)2loge(1123+10)DOption D. 2loge(23+96)2{\log _e}\left( {{{2\sqrt 3 + 9} \over 6}} \right)2loge(623+9)CheckSkip · show solution
