What is the general solution of the equation y′=cos(x+y)y' = \cos(x+y)y′=cos(x+y)?
ln∣sec(x+y)+tan(x+y)∣=x+C\ln|\sec(x+y) + \tan(x+y)| = x + Cln∣sec(x+y)+tan(x+y)∣=x+C
tan(x+y2)=x+C\tan(\frac{x+y}{2}) = x + Ctan(2x+y)=x+C
sin(x+y)=x+C\sin(x+y) = x + Csin(x+y)=x+C
cos(x+y)=x+C\cos(x+y) = x + Ccos(x+y)=x+C