Example of exact form

Question

Consider the differential 1-form $\omega = ydx+dy$. I need to show that this is not exact, and find an example of a function $G(x,y)$ such that $G\omega=G(x,y)(ydx+dy)$ is an exact form.

I have done the first part of the question via contradiction, but I am stuck on the second part. I know that I need to find a function so that $G\omega=d\phi $ for some $\phi$ but I am not sure how to go about this. Is there something simple that I am missing?

Answer 1

Well, I think you want to find nonzero $G$ as $0$ is exact.

Another approach :

De Rham cohomology of $\mathbb{R}^2$ vanishes, thus every closed form is exact. Hence it suffices to show $dG\omega=(G_x dx+G_y dy)\wedge (ydx+dy)-G(dx\wedge dy)=(G_x-yG_y-G)dx\wedge dy=0$. Hence you need to find a function $G$ with $G_x-yG_y-G=0$.

Answer 2

I think that if you look at the corresponding vector field <1,y>, it has non-zero curl, and so it is not a gradient field, not conservative, and hence the differential form you have is not exact.