Prey-Predators systems 
Some solutions
We can, without much effort, determine some more solution-curves. Taking the initial condition \(x_{0}=0\) (absence of pouts) and \(y_{0}>0\), then the function \(x\) is identically null and, from the system of equations, we conclude that \(y^{'}=-Cy\) and, therefore, \(y\) is strictly decreasing; that is, if there are no pouts, the number of sharks decreases to extinction - which was expected. Similarly, if there are no sharks \(\left(y_{0}=0\right)\) and \(x_{0}>0\), \(y\) remains zero and the number of pouts grows exponentially – which was also predictable.
Let us draw in the first quadrant the solution curves \((x,y)\) that we have already determined: the equilibrium points, \(Z_{1}=\left(0,0\right)\) and \(Z_{2}=\left(\frac{C}{D},\frac{A}{B}\right)\), and the two half-lines \(r_{1}=\left\{ (0,y_{0}e^{-Ct}):t\geq0\right\} \) and \(r_{2}=\left\{ (x_{0}e^{At},0):t\geq0\right\} \).
From the uniqueness of the solution it follows that no solution with initial condition in the first quadrant can leave this region, otherwise it would have to intersect one of the four curves we have already identified. This means that if \(x_{0},y_{0}>0\), then \(x\) and \(y\) remain positive which means that the two species do not die out.
