Consider the system of DE $$\frac{dx}{dt}=-axy+b$$ $$\frac{dy}{dt}=axy-cy$$
Where a,b,c are positive constants.
Show that the system has a unique equilibrium point.
Show that any solution in the system that starts close enough to it's equilibrium point tends finally to the equilibrium point when $t$ tends to infinity.
I did 1.
The equilibrium point is $\overline x=(\frac{c}{a},\frac{b}{c})$, and to prove uniqueness I assumed there was another equilibrium point $x^* $ and after calculation I got $x^*=\overline x.$
And I'm not quite sure what should I do to solve 2.
Can someone help me please?
Look for example at this question. The second answer also contains a detailed link about the theory behind this. What it boils down to is to calculate the Jacobian $$J(x, y) = \begin{bmatrix} \dfrac{\partial x'}{\partial x} & \dfrac{\partial x'}{\partial y} \\ \dfrac{\partial y'}{\partial x} & \dfrac{\partial y'}{\partial y} \end{bmatrix} $$ You need to calculate the eigenvalues of this matrix. You can then show that the real part of the eigenvalues are both negative, so the equilibrium is stable.
In case I did not do ant mistakes, you get $$\lambda_{1.2}=\frac{-\frac{ab}{c}\pm\sqrt{\left(\frac{ab}{c}\right)^2-4ab}}{2}$$ If the quantity under the square root is positive, it will be less than $ab/c$, so both solutions are negative. If the quantity is negative, you get two complex solutions, but the real part is $-\frac{ab}{2c}\lt 0$, since $a,b,c\gt 0$.