Let us for convenience choose $\alpha=\beta=1$. Then we get:

$$\begin{aligned} \frac{\dd u}{\dd t}&= \frac{1}{1+v}-u\\ \\\frac{\dd v}{\dd t}&= \frac{u}{1+u}-v\end{aligned}$$ We can determine an equilibrium, in this case even exactly by figuring out where the two derivatives in the differential equation system are simultaneously equal to zero. In the theory of dynamical systems, this is called a singular point.

For an equilibrium $(u,v)$ we should have

$$u=\frac{1}{1+v}\quad\text{and}\quad v=\frac{u}{1+u}$$ Substitution of the formula for $v$ in the first equation gives $$u = \frac{1}{1+\dfrac{u}{1+u}}$$ or by multiplying the numerator and denominator of the right-hand side with $u+1$ $$u=\frac{u+1}{2u+1}$$ So: $$u(2u+1)=u+1$$ Thus: $$2u^2=1$$ The only positive solution is $$u=\tfrac{1}{2}\sqrt{2}$$ Substitution in the formula for $v$ and simplification gives $$v=\sqrt{2}-1$$ So there is an equilibrium at $$(u,v)=\tfrac{1}{2}\sqrt{2}, \sqrt{2}-1)$$

It now turns out that solutions always go in the course of time toward this equilibrium; it is a so-called attracting equilibrium. You see this best in a so-called phase portrait. This is a diagram in which one often draws a number of solution curves at one time and draws the vector field corresponding to the 2-dimensional system of differential equations. At each point $(u,v)$ in the phase portrait we can namely calculate a vector $\cv{u'\\v'}$ and draw the vector in the plane. In our example, the phase portrait with one red solution curve that starts in $(u,v)=(2, 0.2)$ looks as follows: