So I know that a one form $\alpha$ is exact if $\alpha= df$ for some smooth $f$.
In other words, if you can take a derivative of a function and get the 1-form, the 1-form is exact?
So, how would one go about proving something like
$$a = 2x dx + dy$$
is exact?
If we assume $df$ is $2x dx + dy$ then wouldn't $f$ be $x^2 + y$? Since the derivative of $x^2$ will be $2x$ and the derivative of $y$ would be $1$?
All I have about 1-forms being closed is the following information:
$\alpha = adx + bdy$ with $da/dy = db/dx$ is closed. The definition of closed was introduced in the homework and I have no idea how to actually do one of these (since I have not seen any examples), could someone maybe point me to a source and or video describing this?
I assume you know some basic things about differential forms, like the exterior derivative ($\mathrm{d}$) and the wedge product ($\wedge$). Otherwise, I really have no clue how one can prove such a statement (because it is about differential forms...)
We have the fact that if a $1$-form is exact, then it is closed, as if $\alpha = \mathrm{d}f$, then $\mathrm{d}\alpha = \mathrm{d}^2 f = 0$ (recall that $\mathrm{d}^2=0$). It is in fact true for $k$-forms : if a $k$-form $\beta$ is the exterior derivative of a $(k-1)$-form, then $\mathrm{d}\beta=0$. The converse is, in general, false.
Here is a partial converse: the Poincaré lemma states that, if $M$ is a contractible manifold, then every closed exact form is exact.
In your example, I assume you are working on $\mathbb{R}^2$. Thus, let $f(x,y) = x^2 + y$. Then \begin{align} \mathrm{d}f &= \dfrac{\partial f}{\partial x} \mathrm{d}x + \dfrac{\partial f}{\partial y} \mathrm{d}y \\ &= 2x\mathrm{d}x + \mathrm{d}y \end{align} so $2x\mathrm{d}x + \mathrm{d}y$ is exact. But we could have looked at $\mathrm{d}\left(2x\mathrm{d}x + \mathrm{d}y \right)$: we would have noticed that it is zero, and thus, the Poincaré lemma would have said that there exists $g$ such that $\mathrm{d}g = 2x\mathrm{d}x+\mathrm{d}y$, because $\mathbb{R}^2$ is contractible. There exist infinitly many $g$ such that $\mathrm{d}g = 2x\mathrm{d}x + \mathrm{d}y$ (they all differ by some constant function).
In particular, in $\mathbb{R}^2$, the Poincaré lemma says that a $1$-form is closed if and only if it is exact. Let $\alpha = a(x,y) \mathrm{d}x + b(x,y)\mathrm{d}y$ be a $1$-form. Then \begin{align} \mathrm{d}\alpha = 0 & \iff \mathrm{d}\left(a(x,y) \mathrm{d}x + b(x,y)\mathrm{d}y \right)= 0 \\ &\iff \left((\partial_x a)\mathrm{d}x + (\partial_y a) \mathrm{d}y\right)\wedge \mathrm{d}x + \left((\partial_x b)\mathrm{d}x + (\partial_y b)\mathrm{d}y \right)\wedge \mathrm{d}y = 0 \\ &\iff (\partial_ya)\mathrm{d}y\wedge \mathrm{d}x + (\partial_xb) \mathrm{d}x\wedge\mathrm{d}y = 0 \\ & \iff \left(\partial_x b - \partial_y a \right)\mathrm{d}x\wedge \mathrm{d}y = 0 \\ &\iff \partial_y a = \partial_x b \end{align} And $\alpha$ is exact if and only if $\alpha$ is closed, if and only if $\partial_ya = \partial_x b$.