Solve 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+C\tan(\frac{x-y}{2} + C) = x + Ctan(2x−y+C)=x+C
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
y=x−2arctan(x+C)y = x - 2 \arctan(x + C)y=x−2arctan(x+C)
y=sin(x)+Cy = \sin(x) + Cy=sin(x)+C