Math Core

Many real situations are easier to describe by how fast something changes than by a formula for the thing itself. A population grows faster when there are more organisms; hot coffee cools faster when it is much hotter than the room. A differential equation captures a rule like that in symbols, and learning to write and read one is the first step toward solving it.

What a differential equation is

Definition

Differential equation

A differential equation is an equation that involves an unknown function and one or more of its derivatives. For example,

$\frac{dy}{dt} = 0.4y, \qquad \frac{dy}{dx} = x - y, \qquad y'' + 9y = 0.$

The order of a differential equation is the highest derivative that appears. The first two above are first order; the third is second order.

Notice what is different from the equations you solved in algebra. In $x^2 - 5x + 6 = 0$ , the unknown is a number. In $\dfrac{dy}{dt} = 0.4y$ , the unknown is a whole function $y(t)$ , and the equation tells you how that function's rate of change relates to its current value.

In AP Calculus AB, almost every differential equation you meet is first order and written in the form

$\frac{dy}{dx} = (\text{some expression in } x \text{ and } y).$

Translating words into a differential equation

Word problems describe rates in sentences. The key vocabulary is the word proportional.

" $y$ is proportional to $x$ " means $y = kx$ for some constant $k$ .
" $y$ is inversely proportional to $x$ " means $y = \dfrac{k}{x}$ .
"The rate of change of $y$ with respect to $t$ " means $\dfrac{dy}{dt}$ .

Put those together and you can translate almost any statement.

Statement	Differential equation
The rate of change of $P$ with respect to $t$ is proportional to $P$ .	$\dfrac{dP}{dt} = kP$
The rate of change of $h$ is proportional to the square root of $h$ .	$\dfrac{dh}{dt} = k\sqrt{h}$
The rate of change of $A$ is inversely proportional to $A$ .	$\dfrac{dA}{dt} = \dfrac{k}{A}$
The rate of change of $T$ is proportional to the difference between $T$ and 70.	$\dfrac{dT}{dt} = k(T - 70)$
The rate of change of $N$ is proportional to the product of $N$ and $1000 - N$ .	$\dfrac{dN}{dt} = kN(1000 - N)$

The constant $k$ is called the constant of proportionality. Its sign carries meaning: if the quantity is growing, the rate is positive; if it is shrinking, the rate is negative. Many textbooks write a decreasing model with an explicit minus sign, like $\dfrac{dh}{dt} = -k\sqrt{h}$ with $k > 0$ , so the sign is visible at a glance.

Reading a model

A differential equation $\dfrac{dy}{dt} = f(t, y)$ is a rule that gives the rate of change of $y$ at every moment from the current values of $t$ and $y$ .

Plug in the current values to get the instantaneous rate, with units of ( $y$ -units) per ( $t$ -unit).
$\dfrac{dy}{dt} > 0$ means $y$ is increasing at that moment; $\dfrac{dy}{dt} < 0$ means it is decreasing.

Worked example: Translating a statement

A town's population $P$ grows at a rate proportional to the population. When $P = 20{,}000$ people, the population is growing by 600 people per year. Write a differential equation for $P$ , including the value of the constant.

Solution. "Rate proportional to the population" gives $\dfrac{dP}{dt} = kP$ . Use the given moment to find $k$ :

$600 = k(20{,}000) \quad\Longrightarrow\quad k = 0.03.$

So $\dfrac{dP}{dt} = 0.03P$ , with $t$ in years. The model says the town grows by 3% of its current size per year, at every instant.

Newton's law of cooling

A classic model says that an object cools (or warms) at a rate proportional to the difference between its temperature and the temperature of its surroundings. If $T$ is the object's temperature and $A$ is the constant ambient (room) temperature, then

$\frac{dT}{dt} = -k(T - A), \qquad k > 0.$

Check that the signs make sense. If the object is hotter than the room, $T - A > 0$ , so $\dfrac{dT}{dt} < 0$ and it cools. If the object is colder than the room, $T - A < 0$ , so $\dfrac{dT}{dt} > 0$ and it warms up. And the bigger the gap, the faster the change.

Worked example: Evaluating a rate

A cup of tea is poured at $190^\circ$ F in a room kept at $70^\circ$ F. Its temperature $T$ , in degrees Fahrenheit, satisfies $\dfrac{dT}{dt} = -0.1(T - 70)$ , where $t$ is in minutes.

(a) How fast is the tea cooling when it is poured?

(b) What is the temperature of the tea at the moment it is cooling at $3^\circ$ F per minute?

Solution. (a) At $T = 190$ :

$\frac{dT}{dt} = -0.1(190 - 70) = -12.$

The tea is cooling at $12^\circ$ F per minute (the rate of change is $-12$ degrees per minute).

(b) "Cooling at 3 degrees per minute" means $\dfrac{dT}{dt} = -3$ :

$-3 = -0.1(T - 70) \quad\Longrightarrow\quad T - 70 = 30 \quad\Longrightarrow\quad T = 100^\circ\text{F}.$

Using the equation to describe behavior

You can learn a lot from a differential equation before solving it. Since the right side tells you the sign of the derivative, it tells you when the quantity rises and falls.

Worked example: Where is the solution increasing?

A fish population $y$ (in hundreds) is modeled by $\dfrac{dy}{dt} = 0.2y(8 - y)$ for $y > 0$ . For which population values is the population increasing? What happens when $y = 8$ ?

Solution. For $y > 0$ , the factor $0.2y$ is positive, so the sign of $\dfrac{dy}{dt}$ matches the sign of $8 - y$ .

If $0 < y < 8$ , then $8 - y > 0$ , so $\dfrac{dy}{dt} > 0$ and the population is increasing.
If $y > 8$ , then $\dfrac{dy}{dt} < 0$ and the population is decreasing.
If $y = 8$ , then $\dfrac{dy}{dt} = 0$ : the population is not changing at all. A population of 800 fish stays at 800.

A value of $y$ that makes $\dfrac{dy}{dt} = 0$ for all $t$ , like $y = 8$ above, gives a constant solution called an equilibrium solution. You will see these again when you study slope fields.

Common mistake

Don't confuse the quantity with its rate. In $\dfrac{dT}{dt} = -0.1(T - 70)$ , the number 70 is a temperature, and plugging in $T = 190$ gives a rate ( $-12$ degrees per minute), not a temperature. Always ask whether the question wants $T$ or $\dfrac{dT}{dt}$ , and give units that match.

Tip

To check a translation, test the sign. If the story says the quantity is shrinking, make sure your equation produces a negative $\dfrac{dy}{dt}$ for realistic values of $y$ .

Practice

Practice 1

The rate of change of a quantity $A$ with respect to time $t$ is inversely proportional to the square of $A$ . Which differential equation models this situation, where $k$ is a constant?

Practice 2

A population satisfies $\dfrac{dP}{dt} = 0.03P$ , where $t$ is in years. At what rate, in organisms per year, is the population growing when $P = 5000$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

A metal rod heated to $350^\circ$ C is placed in a room kept at $22^\circ$ C. The rod's temperature $T$ changes at a rate proportional to the difference between $T$ and the room temperature. Which equation models $T$ , where $k$ is a positive constant?

Practice 4

The temperature $T$ of a bowl of soup, in degrees Celsius, satisfies $\dfrac{dT}{dt} = -0.2(T - 20)$ , where $t$ is in minutes. Find $\dfrac{dT}{dt}$ when $T = 85$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

A quantity $y > 0$ satisfies $\dfrac{dy}{dt} = y(6 - y)$ . For which values of $y$ is $y$ increasing?

Practice 6

Water drains from a tank so that the volume $V$ , in liters, satisfies $\dfrac{dV}{dt} = -2\sqrt{V}$ , where $t$ is in minutes. What is the volume at the moment the water is draining at 10 liters per minute?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

A quantity $y$ changes at a rate proportional to $y$ . When $y = 50$ , $\dfrac{dy}{dt} = 4$ . Find $\dfrac{dy}{dt}$ when $y = 80$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.