I have been thinking about the differentials that we use in derivatives and integrals.
For example, I have an equation:
$${\int{w}{dr}} = \text{other stuff}$$
The context for this strange equation was: $$\begin{align*} q'(t) &= a - \frac{b}{r(t)},\\ r'(t) &= c - d r(t) - e q(t)\\ r''(t)&= d r'(t) - e q'(t)\\ r''(t)&= d r' - e a + \frac{eb}{r}\\ w&=r'\\ w' &= d w - ea + \frac{eb}{r}\\ \frac{dw}{dr} \frac{dr}{dt}&= d w - ea + \frac{eb}{r}\\ w \frac{dw}{dr} &= dw - ea + \frac{eb}{r} \end{align*}$$
Integrate both sides with respect to $r$: $$\frac{w^2}{2} = \int{d*w dr} - ea r + eb log(r)$$ Shift the integral to the left hand side: $$\int d * w dr = - \frac{w^2}{2} - ea * r - eb * log(r)$$
Divide by $d$: $$\int w dr = \frac{w^2}{2d} - \frac{ear}{d}-\frac{eb}{d} log(r)$$ And I get to that $w\, dr$.
So if I can replace $w$ with ${\frac{dr}{dt}}$, and differentials can be multiplied like normal, I would get ${\frac{dr^2}{dt}}$.
Then I have the idea of multiplying by ${\frac{dt}{dt}}$. Can I go ahead and do this? That would give me $${\int{\frac{dr^2}{dt^2} dt}}$$ Which looks like ${w^2}$ to me. But can I do that?