Show that with the transition rate formula\({\bf{r}}\left( {\bf{N}} \right){\bf{ = s}}\left( {{\bf{2N -- 1}}} \right)\), equation (22) takes the form of the equation for the logistic model (Section 3.2, equation

(14)). Solve (22) for this case.

Here equation (22) takes the form of the equation for the logistic model i.e\(\frac{{dN}}{{dt}} = cN - \int\limits_1^N {r(u)du} \).

Substitute the transition rule formula\(r\left( u \right) = s\left( {2u - 1} \right)\), then:

\(\begin{aligned}{c}\frac{{dN}}{{dt}} = cN - \int_1^N {s(2u - 1)du} \\ = cN - s{N^2} + s + sN - s\\ = - s{N^2} + (c + s)N\end{aligned}\)

If \(A = s\) and \({p_1} = \frac{c}{s} + 1\)then

\(\frac{{dN}}{{dt}} = - AN\left( {N - {p_1}} \right)\)

Variable separating then

\(\begin{aligned}{c}\frac{{dN}}{{ - s{N^2} + (c + s)N}} &= dt\\\smallint \frac{{dN}}{{ - s{N^2} + \left( {c + s} \right)N}} &= \smallint dt\\\frac{1}{{c + s}}\smallint \frac{{dN}}{N} + \frac{1}{{s + c}}\smallint \frac{s}{{c + s - sN}}dN &= t\\\ln \left( {\frac{{kN}}{{c + s - sN}}} \right) &= (c + s)t\end{aligned}\)

\(\begin{aligned}{c}\left( {\frac{{kN}}{{c + s - sN}}} \right) &= {e^{(c + s)t}}\\N = \frac{{(c + s){e^{(c + s)t}}}}{{k + s{e^{(c + s)t}}}}\\N &= \frac{{c + s}}{{k{e^{ - (c + s)t}} + s}}\end{aligned}\)

\(\begin{aligned}{c}N\left( 0 \right) &= {N_o}\\N\left( 0 \right) &= \frac{{c + s}}{{k + s}} &= {N_o}\\k &= \frac{{c + s - s{N_o}}}{{{N_o}}}\end{aligned}\)

Therefore, the value is\(N\left( t \right) = \frac{{{N_o}\left( {c + s} \right)}}{{s{N_o} + \left( {c + s - s{N_o}} \right){e^{ - \left( {c + s} \right)t}}}}\).