Converting 'velocity with respect to distance' to 'distance with respect to time'

Question

If I have a formula for velocity with respect to distance, like:

$73 (km / s / megaparsec)$

And I want to convert it to a formula for velocity (or any of its derivatives or its integral) with respect to time, how would I go about it?

My approach thus far is as follows:

$v(s) = 73s(t)$

By the chain rule:

$\frac{ds'}{ds} = \frac{ds'}{dt} * \frac{dt}{ds}$

$\frac{ds'}{ds} = s''(t) * \frac{1}{s'(t)}$ (because of $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$)

I can solve this differential equation with the characteristic equation:

$s^2 - 73s = 0$

I arrive at:

$s(t) = e^{74 * t}$

And this doesn't work because substituting values for $t$ here does not agree with what we would expect given $v(s)$

Answer 1

Your value $73\text{ (km/hr)/megaparsec}$ has units of inverse seconds. It is one value of the Hubble constant relating the speed an object is receding from us to the distance to the object. You can naively take the inverse and get $4.227\cdot 10^{17}$ seconds or $13.39$ billion years. It is roughly the age of the universe in that if you reverse the expansion and do not let things accelerate under gravity this is the time until everything collapses. It is an observed constant at the present epoch. If you want to solve a differential equation for distance as a function of time, you need to consider the time rate of change of the Hubble constant. Ignoring gravitational slowing and dark energy expansion, you would expect the Hubble "constant" to decrease as $\frac 1t$ because when the time since the big bang doubles, things are twice as far away but moving at the same speed relative to us. The constant speed lets you say $s=vt$, where the $v$ is the current speed.

Answer 2

You are making it too complicated. We have:

$$\frac{ds}{dt} = v = 73s$$

which immediately gives us

$$s=Ae^{73t}$$

for some constant $A$. Your differential equation looks highly suspicious to me: what exactly does $\dfrac{ds'}{ds}$ mean?