Problem source

If there are $x$ micrograms of bacteria in a nutrient medium, the population of bacteria will grow at the rate $(2K-x)x$ micrograms per hour. Show that, if $x=K$ when $t=0$, the population at time $t$ is given by 
\[
x(t)=K+K\frac{1-\mathrm{e}^{-2Kt}}{1+\mathrm{e}^{-2Kt}}.
\]
Sketch, for $t\geqslant0$, the graph of $x$ against $t$. What happens to $x(t)$ as $t\rightarrow\infty$?
Now suppose that the situation is as described in the first paragraph, except that we remove the bacteria from the nutrient medium at a rate $L$ micrograms per hour where $K^{2}>L$. We set $\alpha=\sqrt{K^{2}-L}.$
Write down the new differential equation for $x$. By considering a new variable $y=x-K+\alpha,$ or otherwise, show that, if $x(0)=K$ then $x(t)\rightarrow K+\alpha$ as $t\rightarrow\infty$.

Solution source

\begin{align*}
&& \dot{x} &= (2K-x)x \\
\Rightarrow && \int \d t &= \int \frac{1}{(2K-x)x} \d x \\
&&&= \int \frac1{2K}\left ( \frac{1}{2K-x} + \frac{1}{x} \right) \d x \\
&&&= \frac{1}{2K} \left (\ln x - \ln (2K-x) \right) \\
\Rightarrow && 2Kt+C &= \ln \frac{x}{2K-x} \\
t = 0, x = K: && C &= \ln \frac{K}{2K-K} = 0 \\
\Rightarrow && e^{2Kt} &= \frac{x}{2K-x} \\
\Rightarrow && e^{-2Kt} &= \frac{2K}{x} -1 \\
\Rightarrow && x &= \frac{2K}{1+e^{-2Kt}} \\
&&&= K + K \frac{1-e^{-2Kt}}{1+e^{-2Kt}}
\end{align*}

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){0.5+0.5*(1-exp(-(#1)))/(1+exp(-(#1)))};
    \def\xl{-1.5}; 
    \def\xu{5};
    \def\yl{-0.01}; \def\yu{1.5};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the reusable styles to keep code clean
    \tikzset{
        x=\xscale cm, y=\yscale cm,
        axis/.style={thick, draw=black!80, -{Stealth[scale=1.2]}},
        grid/.style={thin, dashed, gray!30},
        curveA/.style={very thick, color=cyan!70!black, smooth},
        curveB/.style={very thick, color=orange!90!black, smooth},
        dot/.style={circle, fill=black, inner sep=1.2pt},
        labelbox/.style={fill=white, inner sep=2pt, rounded corners=2pt} % Protects text from lines
    }
    
    % Draw background grid
    \draw[grid] (\xl,\yl) grid (\xu,\yu);
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right, black] {$t$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above, black] {$x$};
    
    % Define the bounding region with clip
    \begin{scope}
        \clip (\xl,\yl) rectangle (\xu,\yu);

        
        \draw[curveA, domain=0:6, samples=150] 
            plot ({\x},{\functionf(\x)});
        \draw[curveB, dashed] (0,1) -- (\xu, 1);

        \filldraw (0, 0.5) circle (1.5pt) node[left] {$K$};
        \filldraw (0, 1) circle (1.5pt) node[left] {$2K$};
    \end{scope}
    
    % Annotate Function Names
    

    \end{tikzpicture}
\end{center}

As $t \to \infty$ $x(t) \to 2K$

We now have
\[ \dot{x} = (2K-x)x - L\]
Suppose $y = x - K + \alpha$, where $\alpha = \sqrt{K^2-L}$ then,

\begin{align*}
&& \dot{y} &= \dot{x} \\
&&&= (2K-x)x - L \\
&&&= (2K - (y+K-\alpha))(y+K-\alpha) - L \\
&&&= (K+\alpha - y)y + (K+\alpha-y)(K-\alpha) - L \\
&&&= (K+\alpha-y)y + K^2-\alpha^2 -y(K-\alpha) - L \\
&&&= (K+\alpha-y)y -y(K-\alpha) \\
&&&= (2\alpha-y)y
\end{align*}

Note that when $t = 0, x = K, y = \alpha$ so we have the original equation but with $x \to y$ and $\alpha \to K$, in particular as $t \to \infty$ $y \to 2\alpha$ and $x \to 2\alpha - \alpha + K = K +\alpha$