Consider the following very simple scenario.
Suppose that, after $t$ hours, the number of people at a small theme park is $p$. $p$ and $t$ are continuous variables.
On average in each hour, $0.5p$ new entrants come to the theme park, whilst the number of people who leave is directly proportional to $t^3$.
Formulating a DE is easy.
$\frac{dp}{dt}$ = $0.5p - kt^3$, where $k$ is a positive constant.
But what if $p$ is now given in terms of thousands? What will $\frac{dp}{dt}$ look like?
I believe the answer will be:
$\frac{dp}{dt}$ = $0.5p - \frac{kt^3}{1000}$, where $k$ is a positive constant.
This can be further simplified as $\frac{dp}{dt} = 0.5p - mt^3$, where $m$ is a positive constant and $m = \frac{k}{1000}$.
The "unit" of the "rate of change", "rate of increase" and "rate of decrease" must remain the same.
And if the rate of increase is in the "unit" of thousands, then the rate of decrease must be in thousands too, so that the rate of change, $\frac{dp}{dt}$ is in thousands.
Am I right to say this?
For this question, I was given the final answer of $\frac{dp}{dt} = 0.5p - rt^3$.
Of course, $r$ is just a dummy variable, but as arbitrary constants can be simplified e.g. $m = \frac{k}{1000}$, I'm not sure if this $r$ has been divided by $1000$ or not - it's unclear. I was wondering if I could get some opinions on this. Basic question still is, is my answer, $\frac{dp}{dt} = 0.5p - mt^3$, where $m$ is a positive constant and $m = \frac{k}{1000}$ correct?
(Side note: I know "unit" is not the formal and precise word here. Could anyone give me the precise language I need to describe this "rate" thing?)
