What is the general solution of y′=cos(x−y)y' = \cos(x-y)y′=cos(x−y) using the substitution v=x−yv = x-yv=x−y?
tan(x−y2+C)=x\tan(\frac{x-y}{2} + C) = xtan(2x−y+C)=x
sin(x−y)=x+C\sin(x-y) = x + Csin(x−y)=x+C
x−y=arctan(x)+Cx-y = \arctan(x) + Cx−y=arctan(x)+C
tan(x−y)=x+C\tan(x-y) = x + Ctan(x−y)=x+C