Differential Equation Modeling
Proportional Growth and Decay
$\frac{dx}{dt} = kx(t)$
Growth if $k > 0$, decay if $k < 0$.
Example
A population grows proportionally. At $t = 3$ hours, the population size is 400. At $t = 10$ hours, the population size is 2000. What was the initial population size at $t = 0$?
$\frac{dP}{dt} = kP(t) \qquad P(3) = 400 \qquad P(10) = 2000 \qquad k > 0$
- Solve the differential equation to get $P(t)$:
\(\frac{1}{P(t)}dP = kdt\) \(\int \frac{1}{P(t)}dP = \int kdt\) \(ln|P(t)| = kt + c\) \(P(t) = \pm e^{kt + c}\) \(P(t) = c_1e^{kt} \qquad c_1 = \pm e^{c}\)
- Determine $k$ using the values given:
\(P(3) = c_1 e^{k(3)} = 400 \qquad P(10) = c_1 e^{k(10)} = 2000\) \(c_1 = \frac{400}{e^{k(3)}} \qquad c_1 = \frac{2000}{e^{k(10)}}\) \(\frac{400}{e^{k(3)}} = \frac{2000}{e^{k(10)}}\) \(400e^{10k} = 2000e^{3k}\) \(\frac{e^{10k}}{e^{3k}} = \frac{2000}{400}\) \(e^{7k} = 10\) \(7k = ln(10)\) \(k = \frac{1}{7}ln(10)\)
- And then determine $c_1$:
- Finally, determine $P(0)$:
\(P(0) = \frac{400}{e^{(\frac{1}{7}ln(10))(3)}}e^{\frac{1}{7}ln(10)(0)}\) \(P(0) = \frac{400}{e^{(\frac{1}{7}ln(10))(3)}}\) \(P(0) \approx 149\)
Newton’s Law of Cooling and Warming
\[\frac{dT(t)}{dt} = k(T(t) - T_a)\]Where $T(t)$ is the temperature of the object at time $t$ and $T_a$ is the ambient temperature.
Example
A thermometer is taken outside where it is 5 degrees. After 1 minute, it reads 55 degrees. After 5 minutes, it reads 30 degrees. What was the initial temperature of the thermometer?
\[\frac{dT(t)}{dt} = k(T(t) - 5) \qquad T(1) = 55 \qquad T(5) = 300\]Logistic Growth
\[\frac{dP}{dt} = P(a - bP) \qquad a > 0 \qquad b > 0\]Example
A population of 1000 students is exposed to a virus. The rate of spread is proportional to the number of infected and non-infected students. Determine the number of infected students after 6 days if after 4 days, 50 students were infected.
\[\frac{dP}{dt} = P(a - bP) \qquad P(0) = 1 \qquad P(4) = 50\]