What is the process of nondimensionalizing an equation? Question: I need to scale time by $\frac{1}{I}$ and species by $P$
for the following equation
$\frac{dS}{dt}=I(1-\frac{S}{P})-\frac{ES}{P}$
where 
P - Size of the source pool of species on the mainland
S - number of species on the island
E - Extinction Rate for species
I - Immigration Rate for species
t - time
My Attempt:
So I need to introduce new variables, called m and n, to make this equation dimensionless.
$m = \frac{S}{S^*}, n = \frac{t}{t^*}$
$mS^*=S, nt^*=t$
So substituting these values into the equation, we have
$\frac{d(mS^*)}{d(nt^*)}=I(1-\frac{(mS^*)}{P})-\frac{E(mS^*)}{P}$
Diving $\frac{S^*}{t^*} $ throughout the equation, we have
$\frac{dm}{dn}=\frac{t^*I}{S^*}-\frac{Imt^*}{P}-\frac{Emt^*}{P}$
Ok. This is where the easy part ends. So, my $t^*$ is by itself and that's a unit of time, so since the time is already given why not let $t^*=\frac{1}{I}$
$\frac{dm}{dn}=\frac{(\frac{1}{I})I}{S^*}-\frac{Im(\frac{1}{I})}{P}-\frac{Em(\frac{1}{I})}{P}$
Except I have distributed by accident, so my equation is really 
$\frac{dm}{dn}=I(\frac{1-mS^*}{P})\frac{t^*}{S^*}-\frac{Emt^*}{P}$
Now let  $t^*=\frac{1}{I}$
$\frac{dm}{dn}=I(\frac{1-mS^*}{P})\frac{(\frac{1}{I})}{S^*}-\frac{Em(\frac{1}{I})}{P}$
$\frac{dm}{dn}=\frac{1-mS^*}{S^*}-\frac{Em}{IP}$
And then let $S =P$ because S is species and P is species so the units match.
$\frac{dm}{dn}=(1-\frac{mP}{P})-\frac{Em}{IP}$
$\frac{dm}{dn}=(\frac{1-\frac{mP}{P}}P)-\frac{Em}{IP}$
$\frac{dm}{dn}=(\frac{1-m}{P})-\frac{Em}{IP}$
If we factor out $\frac{1}{P}$, we have 
$\frac{dm}{dn}=\frac{1}{P}((1-m)-\frac{Em}{I})$
which is almost close, but apparently a better substitution would be to let $t^*=\frac{I}{P}$ so that more parameters would cancel out. So that would mean let my $t^*$ be $\frac{unit-of-time}{species}$
Would it be ok to just let $t^*=\frac{1}{I}$ because that was given in the first place. And all the units will need to match first before I begin. I noticed that when I distributed the I, I have
$\frac{dS}{dt}=(I-\frac{IS}{P})-\frac{ES}{P}$
So basically that's 
$\frac{d[species]}{d[time]}=([time]-\frac{[time][species]}{[species]})-\frac{[species][species]}{[species]}$
Don't I need a $\frac{1}{I}$ or $\frac{1}{time}$ for the last part of the equation where there are species all over the place? ANd $\frac{species}{(time)^2}$ is needed for the only time in the first section? If that's the case then the middle part needs a $\frac{species}{(time)^2}$ before I continue. 
 A: The instructions for nondimensionalizing an equation are as follows:


*

*Identify all the independent and dependent variables.

*Replace each of them with a quantity scaled relative to a characteristic unit of measure to be determined.

*Divide through by the coefficient of the highest order polynomial or derivative term.

*Choose judiciously the definition of the characteristic unit for each variable so that the coefficients of as many terms as possible become 1.

*Rewrite the system of equations in terms of their new dimensionless quantities.


So given
$$
\frac{dS}{dT}=I\left(1-\frac SP\right)-\frac{ES}P\tag{1}
$$
Choose $S=\alpha s$, $T=\beta t$, $P=\gamma p$, $E=\delta e$ and $I=\epsilon i$. With some work, you can get
$$
\frac{ds}{dt}=\beta\epsilon i\left(\frac1\alpha-\frac{s}{p\gamma}\right)-\frac{\beta\delta}{\gamma}\frac{es}{p}
$$
Since $S$ and $P$ are both populations of species, we can probably expect that $\alpha=\gamma$. Similarly, $\delta=\epsilon$ makes sense since $I$ and $E$ are rates of changes of the populations. Thus, we get
$$
\frac{ds}{dt}=\frac{\beta\epsilon}\alpha i\left(1-\frac{s}{p}\right)-\frac{\beta\epsilon}{\alpha}\frac{es}{p}\tag{2}
$$
If you choose that $\beta\epsilon=\alpha$ (so you only need to define two of these three scales, since the third is given by the two), then (2) becomes
$$
\frac{ds}{dt}=i\left(1-\frac{s}{p}\right)-\frac{es}{p}\tag{1*}
$$
which, unless I've made a mistake somewhere, appears to be a dimensionless form of (1).
