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One of the simplest models for a differential equation is when the rate of growth or decay of a thing is proportional to the current quantity of the thing. The mass of a radioactive substance over time and the population of an organism over time are typical examples.
The rate of decay of a radioactive substance is proportional to the mass of the radioactive substance. The typical measurement is the half-life, which is the length of time that half of the radioactive substance will decay. The ODE is typically written as,
\begin{equation} Q' = -kQ \end{equation}where $Q(t)$ is the mass of radioactive substance at time $t$ and $k$ is the rate of radioactive decay. The convention is for $k$ to be a positive number, so the RHS is written as $-kQ$ meaning the rate of change, $Q'$, is always negative.
Draw a direction field for the radioactive decay of a substance assuming you start with $Q(0) > 0$ for a positive mass and decay rate $k=0.5$.
\begin{equation} Q' = -0.5Q \end{equation}Solve the initial value problem.
\begin{equation} Q' = -0.5Q \qquad Q(0) = 10 \end{equation}Determine the half-life for the following ODE for the radioactive decay of a substance.
\begin{equation} Q' = -3Q \qquad Q(0) = Q_{0} \end{equation}The rate of growth of a organism, such as bacteria in milk, can be modelled as proportional to the current amount of bacteria in the milk. The model works well as long as the amount of bacteria is small compared the the quantity of milk. The ODE is typically written as,
\begin{equation} P' = kP \end{equation}where $P(t)$ is the mass of radioactive substance at time $t$ and $k$ is the rate of population growth. The convention is for $k$ to be a positive number, so the RHS $P'$ is always positive because the population of bacteria in the milk is always increasing.
Draw a direction field for the population growth of bacteria with initial population $P(0) > 0$ and growth rate $k=10$.
\begin{equation} P' = 10P \end{equation}Solve the initial value problem.
\begin{equation} P' = 10P \qquad P(0) = 0.7 \end{equation}Solve for $k$ if $P(0)=3$ and $P(2)=8$.
\begin{equation} P' = kP \qquad P(0)=3 \quad P(2)=8 \end{equation}