\(\blacksquare \) The most basic model is call Malthusian model given by \(\frac{dp}{dt}=ap\left ( t\right ) \) which says that rate of change of population is proportional to current population size. \(a\) is constant. The solution is \(p\left ( t\right ) =p_{0}e^{a\left ( t-t_{0}\right ) }\). Where \(p\left ( t\right ) \) is population at time \(t\) and \(p_{0}\) is initial population at time \(t_{0}\). This model is OK when population is small. A better model is called logistic model given by \begin{align*} \frac{dp}{dt} & =ap\left ( t\right ) -bp^{2}\left ( t\right ) \\ p\left ( t_{0}\right ) & =p_{0} \end{align*}
Where \(b\) is the competition factor. Also constant and positive. It is much smaller than \(a\). The solution to the above is\begin{equation} p\left ( t\right ) =\frac{ap_{0}}{bp_{0}+\left ( a-bp_{0}\right ) e^{-a\left ( t-t_{0}\right ) }}\tag{1} \end{equation} In this model, we are normally given \(p_{0}\) and given \(\left ( t-t_{0}\right ) \) and given what is called the limiting value \(\frac{a}{b}\) which is \(\lim _{t\rightarrow \infty }p\left ( t\right ) \). Then asked to find population \(p\left ( t\right ) \) after sometime. This will be \(\left ( t-t_{0}\right ) \). We need to find \(a\). Once we find \(a\), then we find \(b\) from the limiting value. The trick is to find \(a\). To do this, we first use (1) from the information given. The problem will always say that the population doubles every so many years, or the population increases at rate of some percentage per year and so on. Use this to find \(a\) from (1). Now we know \(b\). Then use (1) again now to find the population at some future time as the problem says. See HW1, last problem for an example.
\(\blacksquare \) If a problem says substance decays exponentially, this means \(M\left ( t\right ) =M_{0}e^{-Ct}\). where \(C>0\). Need to find \(C\) from other problem information. Typically problem gives half life to do this. For example, see problem section 1.8, problem 3. It says:
substance \(x\) decays exponentially, and only half of the given quantity remains after 2 years. How long it takes for \(5\) lb decay to \(1\) lb? Solution is\[ M=M_{0}e^{-Ct}\] After 2 years, \(M=\frac{M_{0}}{2}\), hence \(\frac{M_{0}}{2}=M_{0}e^{-2C}\). Hence \(\frac{1}{2}=e^{-2C}\) or \(\ln \left ( \frac{1}{2}\right ) =-2C\), hence \(C=-\frac{1}{2}\ln \left ( \frac{1}{2}\right ) =\frac{1}{2}\ln \left ( 2\right ) \). Now we know \(C\), we can finish the solution.\begin{align*} M & =M_{0}e^{-\frac{1}{2}\ln \left ( 2\right ) t}\\ 1 & =5e^{-\frac{1}{2}\ln \left ( 2\right ) t}\\ \frac{1}{5} & =e^{-\frac{1}{2}\ln \left ( 2\right ) t}\\ \ln \left ( \frac{1}{5}\right ) & =-\frac{1}{2}\ln \left ( \frac{1}{5}\right ) t\\ t & =-2\frac{\ln \left ( \frac{1}{5}\right ) }{\ln \left ( 2\right ) }\\ & =2\frac{\ln 5}{\ln 2}\\ & =4.643\text{ years} \end{align*}
If it says it grows exponentially, then \(M=M_{0}e^{Ct}\) instead.