If `2y sin x\ dy/dx = 2 sin x cos x-y^2 cos x; y(pi/2) = 1,` then solution of the equation is

2y sin x backslash(dy)/(dx)=2sin x cos x-y^(2)cos x;y((pi)/(2))=1 then solution of the equation is

Solution of the differential equation 2 y sin x (dy)/(dx) = 2 sin x cos - y^(2) cos x satisfying y(pi//2) = 1 is given by

Solution of the differential equation 2y sin x (dy)/(dx)=2 sin x cos x -y^(2) cos x satisfying y((pi)/(2))=1 is given by :

Solution of the differential equation 2y sin x(dy)/(dx)=2sin x cos x-y^(2)cos x satisfying y((pi)/(2))=1 is given by

Solution of the differential equation 2y sin x(dy)/(dx)=2sinx cosx-y^(2)cos x satisfying y((pi)/(2))=1 is given by (A) y=sin^(2)x (B) y^(2)=sin x (c) y^(2)=cos x+1 (D) y^(2)sin x=4cos^(2)x

Solution of 2y sin x(dy)/(dx)=sin2x-y^(2)cos x,x=(pi)/(2),y=1 is given by

Let y = y (x) be the solution of the differential equation cos x (dy)/(dx) + 2y sin x = sin 2x , x in (0, pi/2) . If y(pi//3) = 0, " then " y(pi//4) is equal to :

The solution of (dy)/(dx) + x sin 2y = x^(3) cos^(2) y is