Non-dimensionalization problem I am trying to non-dimensionalize this problem but I am getting stuck and would appreciate some guidance.
This is the problem:
$\displaystyle x^\prime=rx\left(1-\frac{x}{k}\right)-\alpha xy$
$\displaystyle y^\prime=\rho y\left(1-\frac{y}{L}\right)-\beta xy$
This is what I've done: 
Let $x(t)=A \hat{x}(s)$ 
$y(t)=B \hat{y}(s)$
$t=\tau s$
$\displaystyle \frac{dx}{dt}=A \frac{d\hat{x}}{ds}\cdot \frac{ds}{dt}=\frac{A}{\tau} \frac{d\hat{x}}{ds}$
$\displaystyle \frac{dy}{dt}=B \frac{d\hat{y}}{ds}\cdot \frac{ds}{dt}=\frac{B}{\tau} \frac{d\hat{y}}{ds}$
$\displaystyle \frac{d}{dt}(t)=\frac{d}{dt}(\tau s)=\tau \frac{ds}{dt} \implies \frac{ds}{dt}=\frac{1}{\tau}$ 
$\displaystyle \frac{A}{\tau}\frac{d\hat{x}}{ds}=rA \hat{x}(s)\left(1-\frac{A\hat{x}(s)}{k}\right) - \alpha A \hat{x}(s)B\hat{y}(s)$
$\displaystyle \frac{B}{\tau}\frac{d\hat{y}}{ds}=\rho B \hat{y}(s)\left(1-\frac{B\hat{y}(s)}{L}\right) - \beta A \hat{x}(s)B\hat{y}(s)$
$\displaystyle \frac{d\hat{x}}{ds}=r \tau \hat{x}(s) \left(1-\frac{A \hat{x}(s)}{k}\right)-\alpha \tau B \hat{x}(s)\hat{y}(s)$
$\displaystyle \frac{d\hat{y}}{ds}=\rho \tau \hat{y}(s) \left(1-\frac{B \hat{y}(s)}{L}\right)-\beta A \tau \hat{x}(s)\hat{y}(s)$
I am getting stuck on the substitutions for $A, B, \text{and} \ \tau.$ I've tried a handful of combinations but they all leave us with $6$ parameters still. I would appreciate some help. 
 A: Past this point, the procedure depends on your intuition for what the constants will be. You want to go through and choose 3 combinations of your parameters to be 1, in such a way that the values of $A$, $B$, and $\tau$ are specified uniquely. (For example, we can't specify both $r \tau$ and $\rho \tau$ because then $\tau$ would need two values at once.)
In this problem, knowing at a glance that the problem is predator-prey-like, two choices seem to be rather intuitive: $A=k$, $B=L$. This is because $x$ and $y$ appear with these coefficients in the original problem, and because it keeps $(\hat{x},\hat{y})$ in $[0,1] \times [0,1]$ if they start out there, which is convenient for analysis. Anyway, then we have
\begin{align}
\frac{d \hat{x}}{ds} & = & r \tau \hat{x}(s) \left ( 1 - \hat{x}(s) \right ) - \alpha \tau L \hat{x}(s) \hat{y}(s) \\
\frac{d \hat{y}}{ds} & = & \rho \tau \hat{y}(s) \left ( 1 - \hat{y}(s) \right ) - \beta \tau k \hat{x}(s) \hat{y}(s)
\end{align}
Knowing that this is a predator-prey-like model motivates what to do next as well. $r$ and $\rho$ are essentially growth factors; they describe the time scale on which $x$s and $y$s respectively would grow if they were not affecting one another. So it makes sense to set the time scale according to one or the other (again, we cannot do both). Taking $r \tau = 1$ gives
\begin{align}
\frac{d \hat{x}}{ds} & = & \hat{x}(s) \left ( 1 - \hat{x}(s) \right ) - \frac{\alpha L}{r} \hat{x}(s) \hat{y}(s) \\
\frac{d \hat{y}}{ds} & = & \frac{\rho}{r} \hat{y}(s) \left ( 1 - \hat{y}(s) \right ) - \frac{\beta k}{r} \hat{x}(s) \hat{y}(s)
\end{align}
So the remaining dimensionless parameters, which become the independent variables, are $\frac{\rho}{r}$, $\frac{\alpha L}{r}$, and $\frac{\beta k}{r}$.
This isn't the only way to do this; essentially any consistent choice would have worked. But this one is convenient, because it puts everything on "natural" scales. $\hat{x}$ and $\hat{y}$ will be in $[0,1]$, and the time scale is in terms of the growth time of the $x$s. It also makes everything we can't easily understand, namely the effect of the two consumption constants and the effect of the difference in the growth rates, into the independent parameters. The rest of it we can easily understand (the logistic growth equation is easy to solve).
