I'm working on a nonlinear predator prey model and am struggling finding my equilibrium points. I've done this three times and gotten three different answers. I'd appreciate if someone could check these to see what they get.
The system is
$$X_{t+1}=X_t+rX_t\left(1-\frac{X_t}{k}\right)-sX_tY_t$$ $$Y_{t+1}=Y_t-dY_t+\epsilon X_tY_t$$
All of the parameters are positive. It is easy to show that $(0,0)$ is an equilibrium point. In my work, I have found $(\frac{d}{\epsilon},\frac{r(k-x)}{sk})$ to be one when I did the work once, but another time I found $(k,\frac{r(k-x)}{sk})$, as well as other variations of these. If someone could verify these with me, it would be greatly appreciated. I'm probably making some kind of algebra mistake.
If $(X,Y)$ is an equilibrium point then \begin{align*} 0 &= rX - rX^2/k -sXY\\ 0 &= -dY +\epsilon XY \end{align*}
If $X=0$, then $Y=0$ is an equilibrium point (and no other $Y$ is). If $Y=0$, then $X^2 = kX$ and we have a second equilibrium point $(k,0)$.
Otherwise, if neither $X$ nor $Y$ are zero, \begin{align*} 0 &= r - rX/k -sY\\ 0 &= -d +\epsilon X. \end{align*} So $X = \frac{d}{\epsilon}$ and $Y = \frac{r}{s} - \frac{rd}{\epsilon k s}.$
So the three equilibrium points are $$(0,0),\ (k,0),\ \left(\frac{d}{\epsilon}, \frac{r}{s} - \frac{rd}{\epsilon k s}\right).$$