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 $t>0$?
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
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 $H\leq x_0 K$. $\qquad\blacksquare$
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 $H\leq x_0 K$. $\qquad\blacksquare$
