Justifying the "Physicist's method" for ODEs using differential forms I need some help in untangling and solving the following exercise:

Let the curve $c:[a,b] \to \mathbb{R}^2, t \mapsto (t, y(t))$ be a
  solution for the ODE  $$ y'(x) = f(x, y(x)). $$ Justify the
  "Physicist's method" (no offense intended) of rearranging the equation
  $\frac{dy}{dx} = f(x,y)$ through formal multiplication of $dx$ to $$
dy = f(x,y)dx, $$ by showing that both differential forms agree in
  every point $c(t)$ on the tangent space.

As far as I understand the situation, we are considering the two differential forms
$$\begin{equation}
dy: \mathbb{R}^2 \to {\bigwedge}^1(T_p\mathbb{R}^2)\\
f(x,y)dx: \mathbb{R}^2 \to {\bigwedge}^1(T_p\mathbb{R}^2)\\
\end{equation}
$$
Here $dy$ is a constant differential form, in the sense that  $dy(p)(x,y) = y$ independent of $p=(v,w) \in \mathbb{R}^2$. However, in general $f(x,y)dx$ is not a constant differential form.
Now, if we choose a point on the curve $c$, say $p = c(t_0) \in c([a,b])$ then we have
$$
f(p)dx = f(c(t_0))dx = y'(t_0)dx,
$$
since $c$ is a solution to the ODE given above. I don't know how to proceed from this point. How can we argue that this equals $dy$?
 A: I think I may be the 'physicist' in question, but I'll give it a go. 
$dy$ is a one-form on $\mathbb{R}^2$, so: $dy_p: T_p \mathbb{R}^2\to \mathbb{R}$, i.e. it takes 1-simplexes defined in space isomorphic to $T_p \mathbb{R}^2$ and gives real values (via integration). In what is to follow $p\in\mathcal{D}$ is a fixed point and $\mathcal{D}\subseteq\mathbb{R}^2$ is the subspace where the solution of $y'=f$ is defined. Same defnitions apply to $\left(fdx\right)_p$
Next, since linear functionals can be linearly combined, we can define $dq_p=\left(dy-fdx\right)_p$. We want to prove that this new functional is zero.
Now, consider integral along the sufficiently short line-segment, 1-simplex, $\sigma_p=(p_0,p_1)$, that contains $p$. Let $\phi_p:[0,1]\to\mathcal{D}$ be the push-forward such that (with slight abuse of notation) $\phi_p\left([0,1]\right)=\sigma_p$. Then:
$\int_{\sigma_p} dq_p=\int_{\phi_p[0,1]} dq_p=\int_0^1\, \phi_p^*dq_p=\int_0^1 \left(\frac{d\bar{y}}{ds}-f\left(\phi\left(s\right)\right)\frac{d\bar{x}}{ds}\right)ds $
where $\bar{y}\left(s\right)=y\left(\phi\left(s\right)\right)$ and the same for $\bar{x}$. The good thing is that now we are no longer dealing with forms, so the simple chain rule applies:
$f\left(\phi\left(s\right)\right)\frac{d\bar{x}}{ds}=\frac{dy}{dx}\bigg|_{\phi\left(s\right)}\frac{d}{ds}x\left(\phi\left(s\right)\right)=\frac{d}{ds}\bar{y}$
so:
$\int_{\sigma_p} dq_p=0$
Thus we have a functional that gives zero for any 'vector' (simplex) we apply it to (via integration). It must be that $dq_p=0$ (how else would you define a zero-functional), which proves what you wanted.
A: The equality in $dy = f\, dx$ is very misleading, because strictly speaking it's not true. To see why, note that $dy$ is a differential $1$-form defined on $\Bbb{R}^2$, which means for each $p \in \Bbb{R}^2$, $dy_p : T_p \Bbb{R}^2 \to \Bbb{R}$ is linear. Similarly, $dx$ is also a differential $1$-form on $\Bbb{R}^2$. Let's for the sake of concreteness say $f: \Bbb{R}^2 \to \Bbb{R}$ is defined on all of $\Bbb{R}^2$, so that $f \, dx$ is still a $1$-form on $\Bbb{R}^2$.
So, if we just write $dy = f \, dx$, this means that the $1$-form on the LHS must equal the $1$-form on the RHS. But this is just not the case, because it amounts to saying that $dy$ and $dx$ are linearly-dependent over the module $C^{\infty}(\Bbb{R}^2)$. Just to really drive this point home, let's fix a point $p \in \Bbb{R}^2$, then, if that equality were true, it would mean $dy_p = f(p)\, dx_p$, where the equality is as elements in $T_p^*(\Bbb{R}^2)$ (the dual of the tangent space; i.e the cotangent space). But this is of course absurd, because if you evaluate both sides on the tangent vector $\dfrac{\partial}{\partial y}\bigg|_{p} \in T_p\Bbb{R}^2$, you'll get the absurd equality $1 = 0$. Yet again, the statement $dy = f \, dx$ is kind of like saying the row vector $(0 , 1)$ equals $\lambda \cdot (1,0)$ for some $\lambda \in \Bbb{R}$... which is plain wrong.
Now that I've hopefully convinced you that the equation taken literally is false, how do we interpret it? Well, the last sentence of your question gives a clue it says 

"... by showing that both differential forms agree in every point c(t) on the tangent space."

But the tangent space of what? $\Bbb{R}^2$? Clearly not, as I've just shown above. What is actually meant is that these two differential forms agree at every point $c(t) \in \Bbb{R}^2$, when restricted to the (one-dimensional) subspace $T_{c(t)}\left(\text{image}(c) \right) \subset T_{c(t)} \Bbb{R}^2$. But what is the tangent space to the image of $c$?  It shouldn't be too hard to convince yourself that if you write $c(t) = (t, c_2(t))$ then the tangent space to the image equals the linear span of the (non-zero) vector
\begin{align}
\xi_{c(t)} :=\dfrac{\partial}{\partial x}\bigg|_{c(t)} + c_2'(t) \dfrac{\partial}{\partial y}\bigg|_{c(t)} \in T_{c(t)} \Bbb{R}^2
\end{align}
(i.e $c(t) = (t, c_2(t))$ implies $c'(t) = (1, c_2'(t))$ so the tangent space is just the
span of this vector). 
So, we have to show that for all $t \in [a,b]$ and for all $\zeta_{c(t)} \in T_{c(t)} \left( \text{image}(c)\right)$,
\begin{align}
dy_{c(t)}(\zeta_{c(t)}) &= f(c(t)) \cdot dx_{c(t)}(\zeta_{c(t)})
\end{align}
But notice that since the tangent space to the image is one-dimensional, it suffices to verify equality when evaluated on the basis vector $\xi_{c(t)}$ defined above; i.e it's enough to prove
\begin{align}
dy_{c(t)}(\xi_{c(t)}) &= f(c(t)) \cdot dx_{c(t)}(\xi_{c(t)}).
\end{align}
This is straight forward:
\begin{align}
dy_{c(t)}(\xi_{c(t)}) &= dy_{c(t)}\left( 
\dfrac{\partial}{\partial x}\bigg|_{c(t)} + c_2'(t) \dfrac{\partial}{\partial y}\bigg|_{c(t)}\right) \\
&= c_2'(t) \\
&= f(c(t)) \tag{$c$ solves the ODE} \\
&= f(c(t)) \cdot 1 \\
&= f(c(t))\cdot  dx_{c(t)}(\xi_{c(t)}).
\end{align}
So, this completes the proof.

Note that another way of stating the equality is that $c^*(dy) = c^*(f \, dx)$; i.e when you pull-back the two $1$-forms on $\Bbb{R}^2$ via the curve $c$, you get two $1$-forms , but now defined on $[a,b]$; and it is these forms which are equal.
