It is easy to check that the 1-form $$\frac{xdx+ydy}{x^2+y^2}$$ is co-closed, just by using $★δk=d★k$. But since $$\frac{xdx+ydy}{x^2+y^2}=-1×★ \frac{-ydx+xdy}{x^2+y^2},$$ I thought that there are no 2-form $a$ that satisfies $$\frac{xdx+ydy}{x^2+y^2}=\delta a.$$ (As same reason that $\frac{-ydx+xdy}{x^2+y^2}$ is closed form but not exact form.) So am I right? Is the 1- form $$\frac{xdx+ydy}{x^2+y^2}$$ co-closed but not co-exact?
And additionally, is this related with homology or cohomology?