Population

Problem 1.

The model is \[ y'(t)=Ky(t)(M-y(t)), \] where $K,M>0$ are constants. Determine the general solution if $y(0)=y_0>0$.

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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*}

Expand the following: \begin{align} (x+1)^2 &\Step{1B}{= (x+1)(x+1)}\\ &\Step{2B}{= x(x+1) + 1(x+1)}\\ &\Step{3B}{= (x^2+x) + (x+1)}\\ &\Step{4B}{= x^2 + (x + x) + 1}\\ &\Step{5B}{= x^2+2x+1}\\ \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

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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

\begin{align*} &\Step{1C}{\frac{1}{K}\ln|Kx-H|=t+c,}\\ &\Step{2C}{\ln|Kx-H|=Kt+c,} \\ &\Step{3C}{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 $H\leq x_0 K$.
In this case \[ \lim_{t\to\infty}x(t)=\frac{H}{K}\qquad\blacksquare. \]

Problem 3.

Heating

Problem 1.

If the temperature of the bread is $120^oC$, of the air is $30^oC$ and $K=0.0366$, determine the temperature of the bread $60$ minutes later.

We use the Newton's law of cooling,
\[
T'(t)=K(M-T(t)),
\] where $T(t)$ is the temperature of the bread, $M$ is the temperature of the air and $K\gt 0$ is a constant. Denote $T_0:=T(0)$. The constant function solution of the equation is
\[
T(t)=\underset{\sim}{M}.
\] This is a solution only in case of $T(0)=M$.
Now determine the non-constant solution.
From the differential equation
\[
\frac{1}{K(M-T)}\,dT=1\,dt.
\] Integrating both sides we obtain

\begin{align*} &\cssId{Step1}{-\frac{1}{K}\ln\left| M-T \right| =t+c,}\\ &\cssId{Step2}{\ln\left| M-T \right| =-Kt+c,} \\ &\cssId{Step3}{M-T =ce^{-Kt},} \\ &\cssId{Step4}{T(t) =M+ce^{-Kt}.}\\ \end{align*}

From the initial condition
\[
c=T_0-M.
\] Using this value of $c$ we obtain
\[
\bbox[lightblue,5px,border:2px solid red]{\color{#800000}{ \bf{T(t)=M+(T_0-M)e^{-Kt}.}}}
\] So
\[
T(60)=30+90e^{-2.1960}=40.0123.
\] It is worth to note that
\[
\lim_{t\to\infty}T(t)=M,
\] independently of the initial value $T_0$. $\blacksquare$

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