Population Models 1

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 $y(0)=y_0>0$.

Show Solution

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

partial fraction decomposition

$$\frac{1}{Ky(M-y)}=\frac{1}{KM(M-y)}+\frac{1}{KMy}.$$ 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