5.5: Understanding Differential Equations

We will now outline a process one can follow to make sense of a difficult or complicated differential equation:

1. Understand what each variable is measuring with correct units.
2. Write down what relationship the differential equation is describing in common, no-nonsense language.
3. Explain why those relationships seem to make sense.

Let’s do an example with the population equation.

Simple Population Growth

A population of snakes is declining according to the model \(S'(t) = -0.05 S(t)\), where \(S(t)\) is number of snakes, and \(t\) is measured in years. Follow the steps above to explain this differential equation.

1. As given in the problem, \(S(t)\) is number of snakes at time \(t\), \(t\) is measured in years. Since \(S'(t)\) is a rate of change of the snakes, it is snakes lost per year.
2. The population of snakes is decreasing, because of the negative sign, in proportion to the number of snakes. Hence, a good answer here is “The more snakes you have, the more snakes you lose.”
3. Of course we don’t know why the snakes are dying off, but it makes sense that the more snakes you have the more you lose, since there are more snakes with the potential to die.

The next example involves a system of differential equations which makes it a bit more complicated. A system of differential equations is just like a system of equations: you now have perhaps several unknown functions, and you want to find all the unknown functions so that all the equations are true at the same time.

Wolves and Deer

Let \(W(t)\) be the population of wolves and \(D(t)\) be the population of deer, with \(t\) measured in years. They satisfy the differential equation

\[\begin{align*} D'(t) & = -0.1(W(t) - 30) \\ W'(t) & = 0.1(D(t) - 300). \end{align*}\]

Explain this differential equation.

1. As given in the problem, \(D(t)\) is the number of deer at a given time, \(W(t)\) the number of wolves, and \(t\) is measured in years. \(D'(t)\) is how fast the deer population is changing in deer per year, and \(W'(t)\) the same thing for wolves measured in wolves per year.
2. We see that a wolf population larger than \(30\) makes the deer population go down. A deer population above 300 makes the wolf population go up.
3. These relationships make sense since if we have a lot of wolves, they hunt the deer and the population of deer goes down. If we have a lot of deer, there is a lot of food for the wolves, so the wolf population goes up.

Let’s look at a rocket equation.

Rocket Equation

A rocket that is blasting off has height \(h(t)\) in meters and has mass of fuel \(m(t)\) in kg (here, \(t\) is measured in seconds). These quantities follow the differential equations

\[\begin{align*} h''(t) & = \frac{-10000m'(t)}{50 + m(t)} - 9.8\\ m'(t) & = -0.1 \end{align*}\]

Use the steps to explain this differential equation.

1. As given in the problem, \(h(t)\) is the height measured in meters, \(m(t)\) is the mass of fuel in kg, and \(t\) is time measured in seconds. We also have \(h'(t)\) would be the velocity in the upward direction in m/s, and \(h''(t)\) is acceleration upward of the rocket in m/s\(^2\). \(m'(t)\) is the change in mass of rocket fuel in kg/s.
2. We see that acceleration, \(h''(t)\), is related to \(-10000m'(t)\). That means as we lose rocket fuel, we gain faster acceleration. We also have \(50 + m(t)\) in the bottom of the fraction — that means we gain acceleration more slowly since we are dividing by this quantity. We also lose 9.8 additional units of acceleration.

   The second equation is much simpler — it just says we are losing rocket fuel at a rate of 0.1 kg/s.

3. This is more complicated, but these relationships do make sense. For example, the 9.8 units of acceleration lost are due to gravity. It makes sense that we gain acceleration as we lose fuel, since we burn fuel to make the rocket go faster. Finally, dividing by \(50 + m(t)\) is to account for the fact that the heavier the rocket is, the slower it will accelerate.