The simplest population growth law is given by
![\[ \frac{dx}{dt} = kx, \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-085fbe17b0535db87744dc30a7c68371_l3.png "Rendered by QuickLaTeX.com")
where 
is the population and 
is a constant dependent on the type of population in question. A more involved growth law in which the population is subject to a maximum constraint 
, gives an equation for population growth of
![\[ \frac{dx}{dt} = kx(M-x), \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-23ce2e187f0e4b4799f4a49bef70a298_l3.png "Rendered by QuickLaTeX.com")
where is a constant, or possibly a function of time.
Express the population as a function of 
in each of these growth laws (with and both constant). Prove that the result in the second growth law can be expressed as:
![\[ x = \frac{M}{1 + e^{-a(t-t_1)}}, \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-09d1658cae0d6f36401b486d20ce05fa_l3.png "Rendered by QuickLaTeX.com")
where 
is a constant and 
is the time at which 
.
Proof. For the first growth law,
![\[ kx = \frac{dx}{dt} \quad \implies \quad x' - kx = 0. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-c212a6f0aafbad1d757e9ba5df3f5462_l3.png "Rendered by QuickLaTeX.com")
This has solutions of the form
![\[ x = x_0 e^{k(t-t_0)} \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-113518f2c8c93ce328033fa1d1b762f4_l3.png "Rendered by QuickLaTeX.com")
where 
at time 
.
For the second, more complicated, growth law we have
![\[ \frac{dx}{dt} = kx (M-x) \quad \implies \quad x' - kMx = -kx^2. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-0ae74f2946c1aae5391a1f4fcd6a2b18_l3.png "Rendered by QuickLaTeX.com")
This is a Bernoulli equation (as seen in this exercise, Section 8.5 Exercise #13) so we know 
where 
is the unique solution to
![\[ v' - P(t) v = - Q(t) \qquad \text{where} \quad P(t) = -kM, \quad Q(t) = -k. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-9735ee019c31628eb4b2aae6523a068d_l3.png "Rendered by QuickLaTeX.com")
Thus, we are looking for the unique solution of
![\[ v' + kMv = k \qquad \text{with} \qquad x(t_0) = x_0 \implies v(t_0) = \frac{1}{x_0}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-2654424d1285e8754314b0024c728fbc_l3.png "Rendered by QuickLaTeX.com")
Using Theorem 8.3 (page 310 of Apostol) for the solutions of first-order linear differential equations, we have
![\begin{align*} v &= \frac{1}{x_0} e^{-kM(t-t_0)} + e^{-kM(t-t_0)} \int_{t_0}^t ke^{km(u-t_0)} \, du \\[9pt] &= \frac{1}{x_0} e^{-kM(t-t_0)} + e^{-kM(t-t_0)} \left( \frac{k}{e^{kMt_0}} \cdot \frac{1}{kM} \cdot \left( e^{kMt} - e^{kMt_0} \right) \right) \\[9pt] &= \frac{1}{x_0} e^{-kM(t-t_0)} + \frac{1}{Me^{kMt}} \left( e^{kMt} - e^{kMt_0} \right) \\[9pt] &= \frac{1}{x_0} e^{-kM(t-t_0)} + \frac{1}{M} - \frac{1}{M} \cdot e^{-kM(t-t_0)} \\[9pt] &= \frac{1}{Mx_0} \left( e^{-kM(t-t_0)} (M- x_0) + x_0) \right). \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-90cd1399b82a9068bdd7828be089f396_l3.png "Rendered by QuickLaTeX.com")
Therefore,
![\[ x = Mx_0 \left( e^{-kM(t-t_0) (M-x_0) + x_0 \right)^{-1}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-07db9fd26d810340ae656cfed628a1e7_l3.png "Rendered by QuickLaTeX.com")
So, if we have 
, then 
and 
, so
![\[ x = \frac{M\left( \frac{M}{2} \right)}{e^{-\alpha(t-t_0)} \left(M - \frac{M}{2} \right) + \frac{M}{2}} = \frac{M}{e^{-\alpha(t-t_0)} + 1} \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-8ebdf173d56ebd73fd8e1d21f325be9b_l3.png "Rendered by QuickLaTeX.com")
where 
Rori- wondering if you could explain how you got T0=T1 at the end. Is this just the obvious choice to remove the exponential? Thanks.
Okay think I got it now- T1 was just used for the initial condition ,a, and M/2 ,a constant value, is set as x(a)=b.