I have a nonlinear system of differential equations for functions $x(t)$ and $y(t)$ with parameters $\epsilon$ and $\lambda$
$$\frac{\text dx}{\text dt}=(1-x)\cdot[1-\epsilon x-2\lambda xy(1-x)]$$ $$\frac{\text dy}{\text dt}=-\lambda xy^2(1-x)$$
I'm trying to look at the steady state of this system. Looking at $\frac{\text dy}{\text dt}=0$ we have three cases: $x=0$, and $x=1$, and $y=0$
For $x=0$ we have $\frac{\text dx}{\text dt}=1$, so this cannot be a steady state.
For $y=0$ we have equilibrium at $x=1/\epsilon$ or $x=1$
For $x=1$, $\frac{\text dx}{\text dt}=0$ no matter what the other parameters or $y$ are equal to, and this is what I have a question on. From solving this numerically it looks like when $\epsilon<1$ we end up in this third case, and the equilibrium of $y$ depends on both $\epsilon$ and $\lambda$. However, since we have equilibrium when $x=1$ no matter what $y$ is, I am not sure how to determine the equilibrium value for $y$ given the parameters $\epsilon$ and $\lambda$.
How do we determine the steady state of $y$ here in terms of $\epsilon$ and $\lambda$ when $\epsilon<1$ and when $x=1$?
EDIT: I realize this also depends on the initial conditions. I am using $x(0)=0$ and $y(0)=1$. Because of this it seems like this is not a simple task...