(This is kind-of a crazy ramble; it's kind-of not a questions and kind-of not a request for comments, but it's more of a request for comments.)
We all know what the second derivative of a real-valued function of a real-variable, but I'm going to look at it in kind-of a crazy way. When we have $f: \mathbb{R} \to \mathbb{R}$, $\mathbb{R}$ is kind-of a trivial Riemannian manifold, and the function's derivative is actually a 1-form, $df = f'(x)dx \in \Omega^1(\mathbb{R})$. Then, without actually realizing what we're doing, we take the Hodge star of $df$, $v(x) =\ ^*df = f'(x)$ and we take its differential $dv = f''(x)dx$, and then we take its Hodge star, $a(x) =\ ^*dv = f''(x)$.
To go from acceleration to position, we take the Hodge star of $a(x)$, $\omega = a(x)dx$, and, since $dx \wedge dx = 0$, $\omega$ is closed, and, supposing we're actually interested in a closed, bounded interval, or the whole real line, with no homology, we have that $\omega$ is exact, and hence has an anti-derivative $v(x)$. Then, we run the same gambit, take $v$'s Hodge star $\alpha$, note it is closed and hence exact, and take its anti-derivative, $f(x)$.
Now, suppose, for the sake of simplicity, space-time is a closed, pseudo-Riemannian (1,3) manifold $M^4$ (with no black holes or cusps or other icky phenomena). If we start with a real-valued function $f$ on $M$, we could take its differential $df$. We could then take the Hodge star $\theta =\ ^*df \in \Omega^3(M)$, take its differential $d\theta \in \Omega^4(M)$, and then take its Hodge star, $a(p)$. Multiplying this by some kind of mass and setting this equal to a scalar function $F(p)$ could give some version of Newton's Second Law.
Now, suppose we have a scalar function $a(p)$ on $M$. We could take its Hodge star, $\ ^*a \in \Omega^4(M)$; since $\Omega^5(M^4) = 0$, this is closed, and, supposing it is not a generator of $H^4(M)$, it would have an anti-derivative $\theta \in \Omega^3(M)$. Taking $\theta$'s Hodge star, we have $\alpha \in \Omega^1(M)$. If $d\alpha = 0 \in \Omega^2(M)$ and $\alpha$ is not a generator of $H^1(M)$, we could take an anti-derivative for $\alpha$, $f(p)$, a kind of "second integral" of $a(p)$.
We all know the d'Alembertian/Laplace-Beltrami operator, $\ ^*\Box^2 =\ ^*d^*d: \Omega^0(M) \to \Omega^0(M)$, has solutions to $\ ^*\Box^2f = 0$ with physical significance; it there any way this crazy way of looking at second-order DEs $m\ ^*\Box^2f = F$ could be useful?