Problem 1.
The model is
\[
y'(t)=Ky(t)(M-y(t)),
\] where $K,M>0$ are constants. Determine the general solution if the initial value is $y(0)=y_0>0$.
The equilibrium solution is $y(t)=M$. The equation is separable and we obtain
\begin{equation}
\label{eq:diff}
\frac{1}{Ky(M-y)}\,dy=1\,dt .
\end{equation} By \[
\frac{1}{Ky(M-y)}=\frac{1}{KM(M-y)}+\frac{1}{KMy}. \] Integrating both sides in \eqref{eq:diff} we obtain
Solving for $y$ \[ y(t)=\frac{cMe^{KMt}}{1+ce^{KMt}}. \] From the initial condition we find \[ c=\frac{y_0}{M-y_0}. \] Substituting this value of $c$ we obtain \[ y(t)=\frac{y_0 M}{(M-y_0)e^{-KMt}+y_0}. \] It is worth to note that \[ \lim_{t\to\infty}y(t)=M. \qquad\blacksquare \]
\begin{equation}
\label{eq:diff}
\frac{1}{Ky(M-y)}\,dy=1\,dt .
\end{equation} By \[
\frac{1}{Ky(M-y)}=\frac{1}{KM(M-y)}+\frac{1}{KMy}. \] Integrating both sides in \eqref{eq:diff} we obtain
\begin{align*}
&\Step{1A}{\frac{1}{KM}\ln|y|-\frac{1}{KM}\ln|M-y|=t+c,}\\
&\Step{2A}{\ln\left| \frac{y}{M-y} \right| =KMt+c,} \\
&\Step{3A}{\frac{y}{M-y}=ce^{KMt}.} \\
\end{align*}
Solving for $y$ \[ y(t)=\frac{cMe^{KMt}}{1+ce^{KMt}}. \] From the initial condition we find \[ c=\frac{y_0}{M-y_0}. \] Substituting this value of $c$ we obtain \[ y(t)=\frac{y_0 M}{(M-y_0)e^{-KMt}+y_0}. \] It is worth to note that \[ \lim_{t\to\infty}y(t)=M. \qquad\blacksquare \]
Let $K:=0.1, M:=60, y_0:=3$. The graph of $y(t)$ is
Let $K:=0.1, M:=6, y_0:=25$. The graph of $y(t)$ is
