`dy/dx=tan^2(x+y)`

Solve the following initial value problem: (dy)/(dx)=tan2x,y(0)=2

The solution of the differential equation 2x^2y"dy"/"dx"=tan(x^2y^2) -2xy^2 , where y=sqrt(pi/2) where x =1, is

The solution of the differential equation 2x^(2)y(dy)/(dx)=tan(x^(2)y^(2))-2xy^(2), given y(1)=(pi)/(2), is

Solution of the equation cos^(2)x(dy)/(dx)-(tan2x)y=cos^(4)x,|x|<(pi)/(4), wheny ((pi)/(6))=(3sqrt(3))/(8)is

The integrating factor of the differential equation (dy)/(dx)=y tan x-y^(2)sec x, is

If y=(cos x)^((cosx)^((cos x )...oo)) then prove that dy/dx=(y^2 tan x)/(y log (cosx)-1).

The solution of the differential equation 2x^(2)y(dy)/(dx) = tan(x^(2)y^(2))-2xy^(2) , given y(1) = sqrt(pi/2) , is

(dy)/(dx)=1+x tan(y-x)

x(dy)/(dx)=y-x(tan y)/(x)dydx