Find the general solution to y′=cos(x−y)y' = \cos(x-y)y′=cos(x−y) using v=x−yv = x-yv=x−y.
x=ln∣sec(v)+tan(v)∣+Cx = \ln|\sec(v) + \tan(v)| + Cx=ln∣sec(v)+tan(v)∣+C
x=tan(v/2)+Cx = \tan(v/2) + Cx=tan(v/2)+C
x=2tan−1(v)+Cx = 2\tan^{-1}(v) + Cx=2tan−1(v)+C
x=sec(v)+Cx = \sec(v) + Cx=sec(v)+C