Population Models 2

Problem 2.

Denote $x(t)$ the amount of fish in a lake. Assume the exponential growth model, $x'(t)=K x(t)$, $x(0)=x_0$. How do we set the fishing quota $H>0$ if we want $x(t)$ to be positive for $t>0$?

Show Solution

The model is $$x'(t)=Kx(t)-H.$$ The equilibrium solution is $x(t)=\frac{H}{K}$.
Separating the variables $$\frac{1}{Kx-H}\,dx=1\,dt.$$ Integrating both sides we obtain

$$\begin{align*} &\Step{1B}{\frac{1}{K}\ln|Kx-H|=t+c,}\\ &\Step{2B}{\ln|Kx-H|=Kt+c,} \\ &\Step{3B}{Kx-H=ce^{Kt}.} \\ \end{align*}$$

Solving for $x$ $$x(t)=ce^{Kt}+\frac{H}{K}.$$ From the initial condition $$c=x_0-\frac{H}{K}.$$ So we obtain $$x(t)=\frac{H}{K}+\left( x_0-\frac{H}{K}\right) e^{Kt}.$$ Thus we have to choose $H\leq x_0 K$. $\qquad\blacksquare$