91Ó°ÊÓ

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\(N\left( t \right) = \frac{{{N_o}\left( {c + s} \right)}}{{s{N_o} + \left( {c + s - s{N_o}} \right){e^{ - \left( {c + s} \right)t}}}}\).