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Derive an equation for population growth with a nonconstant population growth factor

by RoRi

From this previous exercise (Section 8.7, Exercise #13) we have the population growth law

    \[ x = \frac{M}{1+e^{-a(t-t_1)}}. \]

Find a formula that generalizes this equation in the case that the value of k is a function of time rather than a constant. Express this result in terms of the time t_0 at which x = \frac{M}{2}.

From our work in Exercise #13 (linked above) we know x^{-1} = v where v is the unique solution of

    \[ v' + k(t) M v = k(t). \]

Let

    \[ P(t) = k(t) M, \quad Q(t) = k(t), \qquad v(t_0) = \frac{2}{M}. \]

Then we apply Theorem 8.3 (page 310 of Apostol) with,

    \[ A(t) = M \int_{t_0}^t k(u) \, du. \]

Therefore,

    \begin{align*} v &= \left( \frac{2}{M} \right) e^{-M \int_{t_0}^t k(u) \, du} + e^{-M \int_{t_0}^t k(u) \, du} \cdot \int_{t_0}^t k(s) e^{M \int_{t_0}^s k(u) \, du} \, ds \\[9pt] &= \left( \frac{2}{M} \right) \exp \left( -M \int_{t_0}^t k(u) \, du \right) + \frac{1}{M} - \frac{1}{M} \exp\left( -M \int_{t_0}^t k(u) \, du \right) \\[9pt] &= \left( \frac{1}{M} \right) \exp \left( -M \int_{t_0}^t k(u) \, du \right) + \frac{1}{M} \\[9pt] &= \frac{\exp \left( M \int_{t_0}^t k(u) \, du \right) + 1}{M \exp \left( \int_{t_0}^t k(u) \, du \right)}. \end{align*}

Thus,

    \begin{align*} x &= \frac{M \exp \left( \int_{t_0}^t k(u) \, du \right)}{1 + \exp \left( M \int_{t_0}^t k(u) \, du \right)} \\[9pt] &= M \left( 1 + \exp \left( -M \int_{t_0}^t k(u) \, du \right) \right)^{-1}. \end{align*}

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