Solve y′=cos(x−y)y' = \cos(x-y)y′=cos(x−y) using v=x−yv = x-yv=x−y. What is the resulting separable equation?
dv1−cosv=dx\frac{dv}{1-\cos v} = dx1−cosvdv=dx
dv1+cosv=dx\frac{dv}{1+\cos v} = dx1+cosvdv=dx
dv1−sinv=dx\frac{dv}{1-\sin v} = dx1−sinvdv=dx
dvcosv=dx\frac{dv}{\cos v} = dxcosvdv=dx