7.8 Exponential Models with Differential Equations

Exponential models come from the differential equation $\frac{dy}{dt}=ky$, which says the rate of change of a quantity is proportional to the size of the quantity. Solving it by separation of variables gives $y=y_0e^{kt}$, where $y_0$ is the starting amount and $k$ is the growth ($k>0$) or decay ($k<0$) constant. For AP Calculus, interpret $k$, $y_0$, and the units in context.

Why This Matters for the AP Calculus Exam

This topic connects the differential equation skills from earlier in Unit 7 (separation of variables and initial conditions) to real-world growth and decay. On the AP Calculus exam you may need to:

• Translate a verbal statement like "the rate of change is proportional to the amount" into $\frac{dy}{dt} = ky$.
• Solve a separable equation and apply an initial condition to find the particular solution.
• Use the model $y = y_0 e^{kt}$ to predict future values or solve for time.
• Interpret what the variables and the constant $k$ mean in context.

You can recognize the exponential form and use it directly without re-deriving it every time, but you should still be ready to show the separation-of-variables steps when a problem asks for them. Writing clear setup, correct notation, and labeled units makes your work easier to follow and check.

Key Takeaways

• The phrase "rate of change is proportional to the size of the quantity" always means $\frac{dy}{dt} = ky$.
• The general solution is $y = y_0 e^{kt}$, where $y_0 = y(0)$ is the initial value.
• A positive $k$ models growth; a negative $k$ models decay.
• Find $k$ by plugging a second known data point into $y = y_0 e^{kt}$ and using natural logs.
• Keep the constant of integration when you integrate, and separate variables before integrating.
• Track units on $t$ and $k$ so your final answer makes sense in context.

What Is a Differential Equation Here?

A differential equation relates a quantity to its own rate of change. In exponential models, you know how fast something is changing relative to how much is present, and you want the full function that describes the quantity over time.

The model you use is:

$$\frac{dy}{dt} = ky$$

• $\frac{dy}{dt}$ is the rate of change of the quantity over time (people, milligrams, bacteria, etc.).
• $k$ is the proportionality constant that sets the rate and direction of change.
  • $k > 0$ means the quantity is growing.
  • $k < 0$ means the quantity is decaying.

Solving the Differential Equation

You can derive $y = y_0 e^{kt}$ from $\frac{dy}{dt} = ky$ using separation of variables.

1. Start with the differential equation: $\frac{dy}{dt} = ky$.
2. Separate the variables: Move all $y$ terms to one side and all $t$ terms to the other: $\frac{1}{y}\,dy = k\,dt$.
3. Integrate both sides: The integral of $\frac{1}{y}\,dy$ is $\ln|y|$, and the integral of $k\,dt$ is $kt$. This gives $\ln|y| = kt + C$, where $C$ is the constant of integration.
4. Solve for $y$: Exponentiate both sides to undo the natural log: $e^{\ln|y|} = e^{kt+C}$, which simplifies to $|y| = e^C \cdot e^{kt}$.
5. Use the initial condition: Suppose $y = y_0$ when $t = 0$. Plugging in gives $|y_0| = e^C \cdot e^0$, so $e^C = |y_0|$. Then $|y| = |y_0| \cdot e^{kt}$.
6. Final exponential model: Assuming $y$ stays positive, drop the absolute values to get $y = y_0 e^{kt}$.

Here $y_0$ is the starting value, $e$ is Euler's number, and $kt$ combines the rate constant and time.

Worked Example: Population Growth

A small town has a population of 2,000 people. The population increases at a rate proportional to its current size. After 3 years, the population grows to 3,000 people. What will the population be after 10 years?

Identify the given information:

• Initial population $y_0 = 2000$
• Population after 3 years: $3000$
• Goal: find the population after 10 years.

Set up the model $y = y_0 e^{kt}$ with $y = 2000$ when $t = 0$.

Find $k$ using the point at $t = 3$, $y = 3000$:

$$3000 = 2000 \cdot e^{3k}$$

Divide both sides by 2000:

$$\frac{3000}{2000} = e^{3k}$$ $$1.5 = e^{3k}$$

Take the natural log of both sides:

$$\ln(1.5) = 3k$$ $$k = \frac{\ln(1.5)}{3} \approx 0.13516$$

Now use $t = 10$ in $y = 2000 \cdot e^{kt}$:

$$y = 2000 \cdot e^{0.13516 \times 10}$$ $$y = 2000 \cdot e^{1.3516} \approx 2000 \cdot 3.8637 \approx 7727.4$$

The population is expected to be about 7,727 people after 10 years.

How to Use This on the AP Calculus Exam

Problem Solving

Use this routine for any exponential growth or decay problem:

1. Recognize the phrase "proportional to the amount" and write $\frac{dy}{dt} = ky$.
2. Write the solution $y = y_0 e^{kt}$ and substitute the initial value $y_0$.
3. Plug in a second known point to solve for $k$ using natural logs.
4. Answer the actual question (a future value, or a time you solve for) and include units.

Free Response

If a problem asks you to find a general or particular solution, show the separation-of-variables steps: separate, integrate both sides, add the constant of integration, and use the initial condition to solve for the constant. Skipping the separation step or dropping the constant of integration limits the points you can earn.

Common Trap

Not every differential equation with a fraction leads to a log or exponential answer. Only equations in the form $\frac{dy}{dt} = ky$ (rate proportional to amount) give exponential solutions. Check the structure before assuming the answer is exponential.

Practice Problems

Try these before checking the solutions.

1. The rate at which a drug leaves the bloodstream is proportional to the amount in the bloodstream. A dose of 200 mg is given to a patient. After 3 hours, approximately 127.3 mg remain. The amount of the drug in milligrams after $t$ hours is $A(t)$. Write an equation for $A(t)$.
2. Bacteria in a culture increase at a rate proportional to the number present. If the number of bacteria doubles in five hours, in how many hours will the number quadruple?

Solution for Problem 1

Identify the given information:

• $y_0 = 200$
• After 3 hours ($t = 3$), $y = 127.3$
• Goal: find the equation that models the situation.

Use $y = y_0 e^{kt}$ with $y = 200$ when $t = 0$. Find $k$ using the point at $t = 3$, $y = 127.3$:

$$127.3 = 200 \cdot e^{3k}$$

Divide both sides by 200:

$$\frac{127.3}{200} = e^{3k}$$ $$0.6365 = e^{3k}$$

Take the natural log of both sides:

$$\ln(0.6365) = 3k$$ $$k = \frac{\ln(0.6365)}{3} \approx -0.151$$

Since $\frac{dy}{dt} = kA$ and $A(0) = 200$, the model is:

$$A(t) = 200 \cdot e^{-0.151t}$$

The negative $k$ confirms decay, which fits a drug leaving the bloodstream.

Solution for Problem 2

There are no exact numbers, only ratios, so pick convenient starting values.

• Let $y_0 = 100$ when $t = 0$.
• After 5 hours ($t = 5$), $y = 200$ (doubled).
• Goal: find how long until $y = 400$ (quadrupled).

Use $y = y_0 e^{kt}$ with $y = 100$ when $t = 0$. Find $k$ from the point at $t = 5$, $y = 200$:

$$200 = 100 \cdot e^{5k}$$ $$\frac{200}{100} = e^{5k}$$ $$2 = e^{5k}$$

Take the natural log of both sides:

$$\ln(2) = 5k$$ $$k = \frac{\ln(2)}{5} \approx 0.139$$

Now solve for the time when $y = 400$:

$$400 = 100 \cdot e^{0.139t}$$ $$4 = e^{0.139t}$$

Take the natural log of both sides:

$$\ln(4) = 0.139t$$ $$t = \frac{\ln(4)}{0.139} \approx 9.97 \ \text{hours}$$

Common Misconceptions

• "Any differential equation with a fraction gives a log or exponential answer." Only the form $\frac{dy}{dt} = ky$ (rate proportional to amount) produces exponential solutions. Look at the actual structure first.
• "You can skip separating variables." You need to move all $y$ terms to one side and all $t$ terms to the other before integrating. Skipping this leads to incorrect work and lost points.
• "The constant of integration does not matter." Forgetting $C$ when you integrate means you cannot correctly use the initial condition to find the particular solution.
• "$k$ is always positive." A positive $k$ models growth and a negative $k$ models decay. Solving for $k$ tells you which is happening; do not assume the sign.
• "$y_0$ is just any data point." $y_0$ is specifically the value of $y$ when $t = 0$. Use a different known point to solve for $k$.
• "Units do not matter." Track the units on time and on $k$ so your prediction makes sense in context.

Related AP Calculus Guides

• Unit 7 Overview: Differential Equations
• 7.2 Verifying Solutions for Differential Equations
• 7.3 Sketching Slope Fields
• 7.6 Finding General Solutions Using Separation of Variables
• 7.5 Approximating Solutions Using Euler’s Method
• 7.7 Finding Particular Solutions Using Initial Conditions and Separation of Variables