How to solve $\frac{dy}{dx}=\cos(x-y)$? How to solve $\dfrac{dy}{dx}=\cos(x-y)$ ? How do I separate x and y here ?
Please advise.
 A: Set $u=x-y$ then 
$$\frac{du}{dx}=1-\frac{dy}{dx}$$
and the original differential equation could be rewritten as
$$1-\frac{du}{dx}=\cos(u)\Rightarrow \frac{du}{dx}=1-\cos(u)$$
Using direct integration
$$\int\frac{1}{1-\cos(u)}\,du=\int\,dx\Leftrightarrow -\cot(\frac{u}{2})=x+c$$
In other words
$$u=2\cot^{-1}(-x-c)$$
Substituting $u=y-x$ yields
$$y=x+2\cot^{-1}(-x-c)$$
A: A substitution might be best.
$$z=x-y,z'=1-y'$$
$$1-z'=\cos z,z'=1-\cos z$$
$$x=\int\frac{dz}{1-\cos z}$$
From here, probably multiply top and bottom by $1+\cos z$.
A: with $$u=x-y$$ we get $$y'=1-u'$$ nand our equation will be
$$1-u'=\cos(u)$$
A: using $v=x-y$
$\frac{dv}{dx}=1-\frac{dy}{dx}$
Therefore:
$$\frac{dy}{dx}=\cos{(x-y)}v\equiv1-\frac{dv}{dx}=\cos{v}$$ By rearranging, we have:
$$\frac{dv}{1-\cos{v}}=dx$$
therefore:$$\int\frac{dv}{1-\cos{v}}=x+c$$
$$\int\frac{1+\cos{v}}{(1-\cos{v})(1+\cos{v})}dv=x+c$$
$$\int\frac{1+\cos{v}}{\sin^2{v}}dv=x+c$$
$$\int \csc^2v+\csc{v}\cot{v}dv=x+c$$
$$-(\cot{v}+\csc{v})=x+c$$  but $v=x-y$
therefore
$$-(\cot{(x-y)}+\csc{(x-y)})=x+c$$
