The task at hand is to find an expression for the equilibrium solutions of the following system of ODEs: $$ \left\{ \begin{array}{ll} X'=&X(1-X)-YX/(0.2+X) \\ Y'=&0.5YX/(X+0.2)-kY \end{array} \right. $$ where $X(t)$ and $Y(t)$ represent a prey and predator population respectively, with constant $k$.
Attempted solution. I have attempted to begin by letting $X'=Y'=0$ and then trying to solve for specific variables. But I am unsure as to what variable we should be solving for.
Any help would be greatly appreciated! Thank you in advance.
The equilibrium points are the solutions for
$$ x(1-x)-\frac{x y}{x+0.2} = 0\\ \frac 12\frac{x y}{x+0.2}-k y = 0 $$
or
$$ (1,0),\ \ \left(\frac{0.4k}{1-2k}, \frac{0.04(5-12k)}{(2k-1)^2}\right), \ \ (0, 0) $$
Using the linear approximation we can qualify those points. Now with the jacobian
$$ J_k = \left( \begin{array}{cc} \frac{y x}{(x+0.2)^2}-2 x-\frac{y}{x+0.2}+1 & -\frac{x}{x+0.2} \\ \frac{y}{2 (x+0.2)}-\frac{x y}{2 (x+0.2)^2} & \frac{x}{2 (x+0.2)}-k \\ \end{array} \right) $$
calculated at the equilibrium points and then calculating their eigenvalues we have respectively
$$ (-1, 0.416667 - k),\ \ (\lambda_1(k), \lambda_2(k)),\ \ (1,-k) $$
$\lambda_1(k)$ and $\lambda_2(k)$ are involved expressions so we show their behavior using a graphic representation. Red for $\lambda_1(k)$ and blue for $\lambda_2(k)$. Attached a stream plot for $k = 1$. In black the locus described for the second equilibrium point with $k$. There is also the possibility of Hopf bifurcations.