The exterior derivative is defined as the unique $\mathbb{R}$-linear mapping, such that $df$ is a differential one-form for a zero-form $ f $, $d d\alpha = 0$ for any $\alpha$, and that it is an antiderivation.
By the word "unique" I question myself if there is an "inverse" of the exterior derivative. If it exists, how would solve for $\omega$ if we knew $\alpha$ in an equation of the type:
$ d\omega = \alpha $
On
There is a nice machinery of inverting (locally in a star-shaped subset, basically the Poincare lemma in practical formulation) covariant derivative using linear homotopy operator. The word 'linear' is here essential since then homotopy operator is nilpotent and has many useful properties that allows to do practical computations efficiently. The best source is the book: D. Edelen "Applied exterior calculus", Dover, however you can also look into the series of papers that elaborate it to various setups:
There is also quite extensive use of linear homotopy operator in the calculus of variations in a "functional forms" setup. See the nice and standard book by P. Olver "Applications of Lie Groups to Differential Equations", Springer. Some other sources related to variational calculus are summarized in:
https://arxiv.org/abs/2110.04346
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In your equation (by applying again $d$) we get $d\alpha=0$, so $\alpha$ is, and must be, closed. Otherwise you have an ill-posed problem. When you assume you are working on $U$ which is star-shaped - $\alpha$ is also exact. Then you can use linear homotopy operator $H$ (for definition see above sources), which can be used as a projector (precisely the projector is $dH$ and fulfills homotopy invariance formula: $dH+Hd=Id + s$, where $s$ is nonzero only for $0$-forms=functions on $U$, so usually you can assume that $s$ term is irrelevant) on exact forms, i.e., since $\alpha$ is already exact so the projector on exact forms does not change $\alpha$, i.e., $\alpha=dH\alpha$. So we get $d\omega=dH\alpha$, so $d(\omega-H\alpha)=0$. Again, $\omega -H\alpha$ is closed, and hence, exact on $U$, so $\omega = H\alpha+\beta$, where $\beta$ is an arbitrary exact/closed form (some kind of constant of integration). As you see in the setup of linear homotopy operator solving such equations on a star-shaped set is as easy as solving linear ODEs. The question about the existence of global solution on the whole manifold is more delicate and involves topology of underlying manifold.
A form is exact iff it's the exterior derivative of some form. Suppose $\alpha$ is not exact. Then no such $\omega$ exists. Thus the exterior derivative has no inverse in general.