Math Core

The simplest and most useful differential equation says that a quantity changes at a rate proportional to its own size. It describes bacteria multiplying, money compounding continuously, drugs clearing from the bloodstream and radioactive atoms decaying. With separation of variables you can solve it once and for all.

Solving $\dfrac{dy}{dt} = ky$

Suppose $\dfrac{dy}{dt} = ky$ with $y(0) = y_0$ , and $y > 0$ . Separate and integrate:

$\frac{1}{y}\,dy = k\,dt \quad\Longrightarrow\quad \ln y = kt + C.$

At $t = 0$ , $\ln y_0 = C$ . So $\ln y = kt + \ln y_0$ , and exponentiating gives $y = e^{kt}e^{\ln y_0} = y_0e^{kt}$ .

Exponential models

The only solutions of $\dfrac{dy}{dt} = ky$ are the exponential functions

$y = y_0e^{kt},$

where $y_0 = y(0)$ is the initial amount.

If $k > 0$ , $y$ shows exponential growth.
If $k < 0$ , $y$ shows exponential decay.

This is why AP questions can say "the rate of change of $y$ is proportional to $y$ " and expect you to write $y = y_0e^{kt}$ immediately. You may quote this result without re-deriving it, though you should still be able to separate variables if asked to show the work.

Finding $k$ from data

Usually you're not told $k$ . Instead you get two data points: the initial amount and the amount at some later time. Substitute and solve for $k$ with a natural log.

Worked example: Bacteria growth

A bacteria population $P$ grows at a rate proportional to its size. There are 200 bacteria at time $t = 0$ and 1600 after 3 hours. Find $P(t)$ and the population after 5 hours.

Solution. The model is $P = 200e^{kt}$ . At $t = 3$ :

$1600 = 200e^{3k} \quad\Longrightarrow\quad e^{3k} = 8 \quad\Longrightarrow\quad k = \frac{\ln 8}{3} = \frac{3\ln 2}{3} = \ln 2.$

So $P(t) = 200e^{(\ln 2)t} = 200\cdot 2^t$ : the population doubles every hour. After 5 hours, $P(5) = 200 \cdot 2^5 = 6400$ bacteria.

Half-life and doubling time

For decay, the half-life is the time it takes for half of the quantity to remain. For growth, the doubling time is the time it takes to double. Both are constants: they don't depend on how much you start with.

If the half-life is $h$ , then $\tfrac12 y_0 = y_0e^{kh}$ , so $e^{kh} = \tfrac12$ and

$k = \frac{\ln(1/2)}{h} = -\frac{\ln 2}{h}.$

Similarly, a doubling time $d$ gives $k = \dfrac{\ln 2}{d}$ . You can also write the model directly as $y = y_0\left(\tfrac12\right)^{t/h}$ or $y = y_0 \cdot 2^{t/d}$ , which is often quicker.

Worked example: Radioactive decay

A radioactive isotope has a half-life of 12 years. How much of an 80-gram sample remains after 30 years? Round to the nearest hundredth of a gram.

Solution. Thirty years is $\dfrac{30}{12} = 2.5$ half-lives, so

$y(30) = 80\left(\tfrac12\right)^{2.5} = \frac{80}{2^{2.5}} = \frac{80}{4\sqrt2} = \frac{20}{\sqrt2} = 10\sqrt2 \approx 14.14 \text{ grams}.$

Equivalently, $k = -\dfrac{\ln 2}{12}$ and $y(30) = 80e^{-30\ln 2/12} \approx 14.14$ grams.

Newton's law of cooling

In the modeling lesson you met $\dfrac{dT}{dt} = k(T - A)$ , where $A$ is the constant surrounding temperature. It is not exactly $\dfrac{dy}{dt} = ky$ , but a substitution makes it one. Let $y = T - A$ , the temperature difference. Since $A$ is constant, $\dfrac{dy}{dt} = \dfrac{dT}{dt} = k(T - A) = ky$ . So the difference decays exponentially:

$T - A = (T_0 - A)e^{kt} \quad\Longrightarrow\quad T = A + (T_0 - A)e^{kt}.$

Worked example: Cooling soup

A pot of soup at $190^\circ$ F is set on a counter in a $70^\circ$ F kitchen. Ten minutes later the soup is $150^\circ$ F. Assuming Newton's law of cooling, when will the soup reach $100^\circ$ F? Round to the nearest tenth of a minute.

Solution. The model is $T = 70 + 120e^{kt}$ . At $t = 10$ :

$150 = 70 + 120e^{10k} \quad\Longrightarrow\quad e^{10k} = \frac{80}{120} = \frac23 \quad\Longrightarrow\quad k = \frac{\ln(2/3)}{10}.$

Now set $T = 100$ : $30 = 120e^{kt}$ , so $e^{kt} = \dfrac14$ and

$t = \frac{\ln(1/4)}{k} = \frac{10\ln(1/4)}{\ln(2/3)} \approx 34.2 \text{ minutes}.$

T = 70 + 120e^(kt) with k = ln(2/3)/10 ≈ −0.0405. The temperature approaches the room temperature of 70°F (dashed).Open in grapher →

Common mistake

Newton's law of cooling is not $T = T_0e^{kt}$ . That formula would make the soup cool toward $0^\circ$ , not toward room temperature. It's the difference $T - A$ that decays exponentially, so the model always has the form $T = A + (T_0 - A)e^{kt}$ .

Tip

Keep $k$ exact (as a logarithm) until the last step. Rounding $k$ early, say to $-0.04$ , can shift a final answer enough to miss the AP's three-decimal accuracy requirement.

Practice

Practice 1

Find the solution $y(t)$ of $\dfrac{dy}{dt} = -0.3y$ with $y(0) = 50$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 2

A quantity $y$ satisfies $\dfrac{dy}{dt} = ky$ , and it doubles every 3 years. What is $k$ ?

Practice 3

A culture starts with 1000 cells, and the number of cells doubles every 4 hours. Assuming exponential growth, how many cells are present after 10 hours? Round to the nearest whole number.