7 Fig. 7 illustrates the growth of a population with time. The proportion of the ultimate (long term) population is denoted by \(x\), and the time in years by \(t\). When \(t = 0 , x = 0.5\), and as \(t\) increases, \(x\) approaches 1 .
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\caption{Fig. 7}
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One model for this situation is given by the differential equation
$$\frac { \mathrm { d } x } { \mathrm {~d} t } = x ( 1 - x )$$
- Verify that \(x = \frac { 1 } { 1 + \mathrm { e } ^ { - t } }\) satisfies this differential equation, including the initial condition.
- Find how long it will take, according to this model, for the population to reach three-quarters of its ultimate value.
An alternative model for this situation is given by the differential equation
$$\frac { \mathrm { d } x } { \mathrm {~d} t } = x ^ { 2 } ( 1 - x ) ,$$
with \(x = 0.5\) when \(t = 0\) as before.
- Find constants \(A , B\) and \(C\) such that \(\frac { 1 } { x ^ { 2 } ( 1 - x ) } = \frac { A } { x ^ { 2 } } + \frac { B } { x } + \frac { C } { 1 - x }\).
- Hence show that \(t = 2 + \ln \left( \frac { x } { 1 - x } \right) - \frac { 1 } { x }\).
- Find how long it will take, according to this model, for the population to reach three-quarters of its ultimate value.