Writing Differential Equations — AP® Calculus AB Notes

So far you have been handed a function and asked for its derivative. Many real questions run the other way. A biologist cannot write down the size of a bacteria culture for every hour, but she can say how fast it grows: the bigger the culture, the faster it grows. A cook cannot predict the temperature of soup minute by minute, but he knows hot soup cools faster than warm soup. Each of these is a statement about a rate. Written with a derivative, it becomes a differential equation. This page is about writing and reading them; later topics solve them.

What a differential equation is

A differential equation is an equation that contains the derivative of an unknown function. Here are three:

$\frac{dP}{dt} = 0.26P \qquad \frac{dT}{dt} = -0.05(T - 20) \qquad \frac{dy}{dx} = x + y$

Read the first one aloud: "the rate of change of $P$ with respect to $t$ equals 0.26 times $P$ ." It does not tell you what $P$ is at any time. It tells you how fast $P$ changes once you know $P$ . If $P = 500$ cells, then $P$ is growing at $0.26 \cdot 500 = 130$ cells per hour. If $P = 800$ , it is growing at 208 cells per hour.

That is the big shift. An ordinary equation such as $y = x^2$ gives values. A differential equation gives rates, and the rate usually depends on the current value.

Translating "proportional to"

Most word problems use one phrase: proportional to. " $A$ is proportional to $B$ " means $A = kB$ for some constant $k$ , called the constant of proportionality. When $A$ is a rate, it goes on the left as a derivative.

Words	Differential equation
The rate of change of $y$ is proportional to $y$ .	$\frac{dy}{dt} = ky$
The rate of change of $y$ is proportional to the square root of $y$ .	$\frac{dy}{dt} = k\sqrt{y}$
The rate of change of $y$ is inversely proportional to $y^2$ .	$\frac{dy}{dt} = \frac{k}{y^2}$
The temperature changes at a rate proportional to the difference between $T$ and 20.	$\frac{dT}{dt} = k(T - 20)$
The rate of change of $y$ with respect to $x$ is proportional to the product of $x$ and $y$ .	$\frac{dy}{dx} = kxy$

Two checks catch most mistakes. First, the sentence talks about a rate, so a derivative must appear. "Grows proportionally to its size" is $\frac{dP}{dt} = kP$ , never $P = kt$ . Second, look at the variable after "proportional to". If it says the population, the right side uses $P$ , not $t$ .

The sign of the rate

Words like "decreases" or "decays" fix a sign. "The amount of medicine decreases at a rate proportional to the amount present" is $\frac{dA}{dt} = -kA$ with $k > 0$ . You could also write $\frac{dA}{dt} = kA$ and say $k < 0$ ; both are correct if you state the sign.

The sign of the right side tells you where the function rises or falls, before you know any formula. For $\frac{dy}{dt} = 0.4(50 - y)$ , the rate is positive when $y < 50$ and negative when $y > 50$ . So $y$ climbs toward 50 from below and falls toward 50 from above. That is a lot of information from one line.

Finding the constant

Worked example. Coffee at 90 °C sits in a 20 °C room. It cools at a rate proportional to the difference between its temperature and the room's. At 90 °C it is cooling at 3.5 °C per minute. Write the differential equation, then find the rate when the coffee is at 55 °C.

Name the quantity: $T$ , the temperature in °C, at time $t$ in minutes. "Proportional to the difference" gives $\frac{dT}{dt} = k(T - 20)$ .

Use the known rate. Cooling means $\frac{dT}{dt} = -3.5$ when $T = 90$ : $-3.5 = k(90 - 20) = 70k$ , so $k = -0.05$ per minute.

Now predict: at $T = 55$ , $\frac{dT}{dt}$ $= -0.05(55 - 20)$ $= -1.75$ °C per minute. The coffee still cools, at half the speed, because it is half as far above room temperature.

The graph of T = 20 + 70e^(−0.05t) for t from 0 to 60 minutes falls from 90 °C toward the dashed room-temperature line at 20 °C. Tangent segments at T = 90, 55 and 30 have slopes −3.5, −1.75 and −0.5 °C per minute, each equal to −0.05 times (T − 20). — **Figure 1.** The temperature of the coffee in the worked example. Each tangent segment has slope $-0.05(T - 20)$ , so the graph flattens as $T$ nears 20 °C. LevlPrep original diagram.

Notice the units of $k$ . The left side is in °C per minute and $T - 20$ is in °C, so $k$ is in "per minute". In $\frac{dP}{dt} = 0.26P$ with $t$ in hours, $k = 0.26$ per hour.

Reading a model from data

A table can tell you which model fits. If a table lists $P$ and $\frac{dP}{dt}$ and the ratio $\frac{dP/dt}{P}$ is the same in every column, the rate is proportional to $P$ , and that ratio is $k$ . If the rate were the same in every column, the model would be $\frac{dP}{dt} = c$ , a constant rate, and the graph would be a line.

Common slips

Writing the quantity instead of its rate: $P = kP$ or $P = kt$ for "grows at a rate proportional to its size".
Using $T$ instead of $T - 20$ in a cooling problem: the coffee cools toward room temperature, not toward 0 °C.
Losing the sign: "decreases" needs a negative rate.
Mixing up "proportional" ( $ky$ ) and "inversely proportional" ( $\frac{k}{y}$ ).