7.8 - Exponential Models with Differential Equations

Review: Exponential Functions

An exponential function has the form

y = a \cdot b^t

where:

$a$ is the initial value (the starting amount at $t = 0$ )

$b$ is the growth or decay factor

$b > 1$ means exponential growth
$0 < b < 1$ means exponential decay

$t$ is time (or whatever the independent variable is)

From Exponential Functions to Differential Equations

Take

y = a \cdot b^t

where

a

and

b

are constants, and differentiate with respect to

t

\frac{dy}{dt} = a \cdot b^t \cdot \ln(b)

Here’s the key observation:

a \cdot b^t

is just

y

. So substitute:

\frac{dy}{dt} = y \cdot \ln(b)

And

\ln(b)

is a constant, so call it

k

\frac{dy}{dt} = ky

That’s the differential equation form of exponential growth and decay.

The Exponential Model

\frac{dy}{dt} = ky

This says: the rate of change of a quantity is proportional to how much of it there is.

$k > 0$ means growth

$k < 0$ means decay

Recognition tip: any time the derivative equals a constant times the function itself, you’re looking at an exponential model.

Reading it in context: in

\frac{dP}{dt} = 0.03P

, with

P

in people and

t

in years,

\frac{dP}{dt}

is how fast the population is changing, in people per year. The constant

k = 0.03

says the growth rate is always

3\%

of the current population, per year. So when

P = 2000

, the population is growing at

0.03(2000) = 60

people per year.

Setting Up the Equations

Example 1: Population Growth

A bacteria population increases at a rate proportional to its current population.

Let

P

be the population:

\frac{dP}{dt} = kP

with

k

positive.

Example 2: Radioactive Decay

A radioactive substance decays at a rate proportional to the amount remaining.

Let

A

be the amount:

\frac{dA}{dt} = kA

with

k

negative.

Example 3: Investment Growth

Money grows at a rate proportional to the current balance.

Let

M

be the balance:

\frac{dM}{dt} = kM

with

k

positive.

Solving It: The General Solution

To solve

\frac{dy}{dt} = ky

, separate variables.

Step 1: Separate.

\frac{dy}{y} = k\,dt

Step 2: Integrate both sides.

\int \frac{dy}{y} = \int k\,dt \quad \Rightarrow \quad \ln|y| = kt + C_1

Step 3: Exponentiate.

|y| = e^{C_1} \cdot e^{kt}

Step 4: Simplify. Dropping the absolute value adds a

\pm

. Since

\pm e^{C_1}

is just another constant, call it

C

y = Ce^{kt}

where

C

is the initial value (at

t = 0

y = C

k

is the growth or decay constant, and

t

is time.

The Shortcut Formula

For any differential equation

\frac{dy}{dt} = ky

with initial condition

y(0) = y_0

y = y_0 e^{kt}

Once you recognize the pattern, you can skip separation of variables entirely and jump straight to this.

Example 1: Using the Shortcut

Solve

\frac{dy}{dt} = 4y

with

y(0) = 7

This matches the form with

k = 4

and

y_0 = 7

, so:

y = 7e^{4t}

Done.

Example 2: Decay Model

Solve

\frac{dN}{dt} = -0.05N

with

N(0) = 200

Exponential decay with

k = -0.05

N = 200e^{-0.05t}

The negative exponent tells you it’s decay.

Word Problem Strategy

Find the initial value (the amount at $t = 0$ )

Write the general form: $y = y_0 e^{kt}$

Use the extra information to find $k$

Answer the question with your complete equation

Example 3: Animal Growth

Problem: A puppy weighs 5 pounds at birth. After 4 weeks it weighs 8 pounds. Its weight increases at a rate proportional to its current weight. How much will it weigh at 10 weeks?

Step 1: Differential equation. Let

W

be weight in pounds.

\frac{dW}{dt} = kW

Step 2: General solution. Initial weight is

W(0) = 5

, so

W = 5e^{kt}

Step 3: Find $k$ . At

t = 4

W = 8

8 = 5e^{4k} \quad \Rightarrow \quad \frac{8}{5} = e^{4k} \quad \Rightarrow \quad \ln\left(\frac{8}{5}\right) = 4k

k = \frac{\ln(8/5)}{4} = \frac{0.4700}{4} \approx 0.1175

Important: store this value in your calculator, don’t round it yet. On a TI-84, press STO→ then ALPHA then a letter like A.

Step 4: Complete model.

W = 5e^{0.1175t}

Step 5: Weight at 10 weeks.

W(10) = 5e^{1.175} \approx 16.19 \text{ pounds}

Calculator Strategy: Storing Values

Why bother? If you round

k = 0.1175

0.118

, you introduce rounding error that compounds through later calculations.

How to store on a TI-84:

Calculate $k$

Press STO→

Press ALPHA, then a letter (like A)

Press ENTER

Recall it later with ALPHA and that letter

Now you get full precision without typing all those decimals.


ln(8/5)/4 → STO → A
5e^(A*10)

Doubling Time Problems

These tell you how long a quantity takes to double, but usually don’t give you the starting amount. That’s fine, because the initial amount cancels out.

Example 4: Population Doubling

Problem: A population doubles every 12 years. Find

k

Let

P_0

be the initial population. After 12 years the population is

2P_0

2P_0 = P_0 e^{12k}

Divide by

P_0

2 = e^{12k} \quad \Rightarrow \quad \ln(2) = 12k

k = \frac{\ln(2)}{12} \approx \frac{0.6931}{12} \approx 0.0578

Half-Life Problems

Same idea, but the quantity drops to half its original amount.

Example 5: Radioactive Decay

Problem: An isotope has a half-life of 20 years. Find

k

Let

A_0

be the initial amount. After 20 years it’s

\frac{1}{2}A_0

\frac{1}{2}A_0 = A_0 e^{20k} \quad \Rightarrow \quad \frac{1}{2} = e^{20k}

k = \frac{\ln(1/2)}{20} = \frac{-0.6931}{20} \approx -0.0347

k

is negative, as it should be for decay.

Quick Reference Formulas

Doubling time. If a quantity doubles every

T

units of time:

2 = e^{kT} \quad \Rightarrow \quad k = \frac{\ln(2)}{T}

Half-life. If a quantity has half-life

T

\frac{1}{2} = e^{kT} \quad \Rightarrow \quad k = \frac{\ln(1/2)}{T} = -\frac{\ln(2)}{T}

Example 6: Complete Half-Life Problem

Problem: Carbon-14 has a half-life of 5,730 years. A fossil contains 30% of its original carbon-14. How old is it?

Step 1: Find $k$ .

k = \frac{\ln(1/2)}{5730} \approx -0.000121

Step 2: Write the decay model.

A = A_0 e^{-0.000121t}

Step 3: Solve for $t$ when $A = 0.30A_0$ .

0.30 = e^{-0.000121t} \quad \Rightarrow \quad \ln(0.30) = -0.000121t

t = \frac{\ln(0.30)}{k} \approx 9{,}953 \text{ years}

Use the stored value of

k

here. With the rounded

k = -0.000121

, you’d get about

9{,}950

years, which is off by almost

3

years. That’s the rounding error the calculator strategy above warns about.

Example 7: Motion Along a Line

Motion problems are differential equations too. Velocity is the rate of change of position, and acceleration is the rate of change of velocity, so finding a position or velocity function means solving a differential equation with initial conditions.

Problem: A boat is moving in a straight line when its engine shuts off. From then on, its velocity

v

(in feet per second) decreases at a rate proportional to its velocity. When the engine shuts off (

t = 0

v = 20

, and

2

seconds later,

v = 15

. Find

v(t)

, and find how far the boat travels in the first

4

seconds.

Step 1: Differential equation. The rate of change of velocity is proportional to velocity, so this is the exponential model:

\frac{dv}{dt} = kv

Step 2: Solve. With

v(0) = 20

, the shortcut gives

v = 20e^{kt}

Step 3: Find

k

. At

t = 2

v = 15

15 = 20e^{2k} \quad \Rightarrow \quad k = \frac{\ln(3/4)}{2} \approx -0.1438

k

is negative, as it should be, since the boat is slowing down. So

v(t) = 20e^{kt}

with that stored

k

Step 4: Distance. Position

x

satisfies another differential equation,

\frac{dx}{dt} = v(t)

, with

x(0) = 0

at the moment the engine shuts off. The velocity is always positive, so the distance traveled is the change in position:

x(4) = 0 + \int_0^4 20e^{kt}\,dt = \frac{20}{k}\left(e^{4k} - 1\right) \approx 60.83 \text{ feet}

If a problem gives acceleration as a function of

t

instead (like

\frac{dv}{dt} = 6t - 2

), there’s nothing to separate. Antidifferentiate to get

v

, use the initial velocity to find

C

, then do it again to get position. Topic 8.2 works through one of those.

Summary of Problem Types

Type	Differential Equation	Condition	Result
Basic exponential	$\frac{dy}{dt} = ky$	$y(0) = y_0$	$y = y_0e^{kt}$
Find $k$ from two points	$\frac{dy}{dt} = ky$	$y(t_1) = y_1$	Substitute and solve
Doubling time	$\frac{dy}{dt} = ky$	Doubles in time $T$	$k = \frac{\ln(2)}{T}$
Half-life	$\frac{dy}{dt} = ky$	Halves in time $T$	$k = -\frac{\ln(2)}{T}$

Common Mistakes

Rounding $k$ too early. Store the full value in your calculator instead.

Forgetting the negative sign in decay problems. Decay should give $k < 0$ .

Confusing the initial value with some other data point. The initial value is always at $t = 0$ .

Mixing time units. Keep everything in years, or everything in days, but don’t mix.

Key Takeaways

$\frac{dy}{dt} = ky$ means “the rate of change is proportional to the current amount”

Motion along a line is a differential equation too: $\frac{dv}{dt}$ gives velocity, and $\frac{dx}{dt} = v$ gives position, each with its own initial condition

$k > 0$ is growth, $k < 0$ is decay

Shortcut: $y = y_0 e^{kt}$ , so you can skip separation of variables once you spot the pattern

$y_0$ is the amount at $t = 0$

Find $k$ using a second data point, doubling time, or half-life

Store intermediate values to avoid rounding errors

Doubling time: $k = \frac{\ln(2)}{T}$

Half-life: $k = -\frac{\ln(2)}{T}$

Only round the final answer

Keep time units consistent

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