Problem:
Find the steady states and check for stability of the model \begin{align*} X_{t+1} &= rX_te^{r(1-X/k)-aY_t} \\ Y_{t+1} &= X_t(1-e^{-aY_t}) \end{align*}
Attempt: Calculating the steady states: \begin{align*} X(1-re^{r(1-X/k)-aY})=0 &\rightarrow X=0 \\ Y=X(1-e^{-aY}) &\rightarrow Y=0 \end{align*} Thus $(0,0)$ is a trivial steady state. We also have $$ X(1-re^{r(1-X/k)-aY})=0 \rightarrow \frac{1}{r}=e^{-(aY-r)} \rightarrow \log r = aY -r \rightarrow Y= \frac{\log r +r }{a}$$ and $$\frac{\log r +r }{a}= X \bigg( 1- \frac{1}{re^r}\bigg) \rightarrow X = \frac{re^r(\log r+r)}{a(re^r -1)}$$ Thus a non-trivial state is, with $r>1$ $$\bigg( \frac{re^r(\log r+r)}{a(re^r -1)},\frac{\log r +r }{a}\bigg).$$ Did I made a mistake here? Any help would be much appreciated!
You have with $X \ne 0$: $$ X\left(1-re^{r(1-X/k)-aY}\right)=0 \implies \frac{1}{r}=e^{-(aY-r)} $$ but shouldn't it be $$ X\left(1-re^{r(1-X/k)-aY}\right)=0 \implies \frac{1}{r}=e^{-(aY-r+rX/k)} $$
UPDATE Assuming $X\ne 0$, the second equation yields $$ X = \frac{Y}{1-e^{-aY}} = \frac{Ye^{aY}}{e^{aY}-1} $$ and we can plug this into the first one. We get: $$ \begin{split} \frac{1}{r} &= e^{-(aY-r+rX/k)} \\ \ln r &= aY-r+\frac{r}{k} X \\ r + \ln r &= aY + \frac{r}{k} \frac{Ye^{aY}}{e^{aY}-1} \\ \end{split} $$ which I believe is transcendental, so does not have an easily expressible analytic solution.