7.1 Modeling & verifying solutions

Achievable AP Calculus AB

7. Differential equations

Our AP Calculus AB course is currently in development and is a work-in-progress.

Modeling & verifying solutions

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What you’ll learn:

How to write differential equations from verbal descriptions
Recognize relationships involving proportionality
How to verify solutions to differential equations

A differential equation relates a function to one or more of its derivatives. In other words, it connects a quantity to its rate(s) of change. Differential equations are useful for modeling real-world situations where the way something changes depends on its current value.

Modeling situations

Modeling questions ask you to translate a verbal description into a differential equation. Watch for key phrases such as “proportional,” “inversely proportional,” “rate of change,” and references to growth, decay, or cooling.

Here are some common phrases and translations, with $k$ as the constant of proportionality.

Phrase	Math translation
Rate of change of $y$ with respect to $x$	$\frac{d y}{d x}$
Proportional to $A$	$k A$
Inversely proportional to $A$	$\frac{k}{A}$
Proportional to product of $A$ and $B$	$k A B$
Proportional to square root of $t$	$k t$
Proportional to square of $v$ and inversely to $m$	$\frac{k v ^{2}}{m}$
Changing linearly	$k$

Examples

A population grows at a rate proportional to its current size. When the population is $800$ , it is growing at a rate of $120$ people per hour. Write a differential equation for this situation.

Because the rate of change of the population is directly proportional to the population itself, we define:

$P (t) =$ population at time $t$
$\frac{d P}{d t} =$ rate of change of the population with respect to time
$k =$ constant of proportionality

“Rate proportional to current size” translates to:

$\frac{d P}{d t} = k P$

This equation says that when $P$ increases, the growth rate $\frac{d P}{d t}$ increases in the same proportion.

Now use the given data to find $k$ . When $P = 800$ , the rate is $\frac{d P}{d t} = 120$ people per hour:

$120 = k (800)$

$k = 0.15$

So the differential equation that models this situation is

$\frac{d P}{d t} = 0.15 P$

The rate at which a person improves a skill is inversely proportional to their current skill level $L$ . When the skill level is $4$ , the improvement rate is $0.5$ units per day.

Solution

(spoiler)

“Inversely proportional to $L$ ” means proportional to $\frac{1}{L}$ , so the model is

$\frac{d L}{d t} = \frac{k}{L}$

where $k$ is the constant of proportionality.

This tells us:

The rate of change of the skill is proportional to the reciprocal of the current skill level $L$ .
When $L$ is small, $\frac{1}{L}$ is large, so improvement happens faster.
As $L$ increases, $\frac{1}{L}$ decreases, so the rate of improvement slows (the skill level itself is still increasing).

Now solve for $k$ using the given information. When $L = 4$ , the improvement rate is $\frac{d L}{d t} = 0.5$ :

$0.5 = \frac{k}{4}$

$k = 2$

So the differential equation is

$\frac{d L}{d t} = \frac{2}{L}$

A child’s shoe size increases linearly over time.

Solution

(spoiler)

Let shoe size be $S (t)$ . If $S (t)$ increases linearly, its graph is a straight line with constant slope. A constant slope means the derivative is constant.

So the differential equation is:

$\frac{d S}{d t} = k$

where $k$ is a constant representing the rate of increase in shoe size per unit time.

Verifying solutions

A solution to a differential equation is a function that makes the equation true when you substitute the function and its derivatives into the equation.

To verify a proposed solution:

Differentiate the given function as needed.
Substitute into the differential equation.
Check whether both sides match.

The function

$y = e^{3 x}$

is a solution to which of the following equations?

a) $y^{''} - 3 y^{'} = 6 e^{3 x}$
b) $y^{''} - 6 y^{'} + 9 y = 0$
c) $y^{''} - 9 y = 0$
d) $y^{''} + 3 y^{'} = 12 e^{3 x}$

Solution

(spoiler)

Differentiate $y = e^{3 x}$ :

$y = e^{3 x}$

$y^{'} = 3 e^{3 x}$

$y^{''} = 9 e^{3 x}$

Now test each choice.

For choice a):

$y^{''} - 3 y^{'} = 6 e^{3 x}$

$9 e^{3 x} - 3 (3 e^{3 x}) = 6 e^{3 x}$

$0 \neq = 6 e^{3 x}$

So choice a) is not correct.

Try choice b):

$y^{''} - 6 y^{'} + 9 y = 0$

$9 e^{3 x} - 6 (3 e^{3 x}) + 9 (e^{3 x}) = 0$

$0 = 0$

Choice b) is correct.

For what value of $k$ does $y = k sin (x)$ satisfy the differential equation $y^{''} + 4 y = sin (x)$ ?

Solution

(spoiler)

Start with $y = k sin (x)$ and differentiate:

$y^{'} = k cos (x)$

$y^{''} = - k sin (x)$

Substitute into $y^{''} + 4 y = sin (x)$ :

$- k sin (x) + 4 k sin (x) = sin (x)$

$3 k sin (x) = sin (x)$

Since this must hold for all $x$ , the coefficients of $sin (x)$ must match:

$3 k = 1$

$k = \frac{1}{3}$

Modeling situations

Differential equations relate a function to its derivatives
Key phrases for translation:
- “Proportional to”: $k A$
- “Inversely proportional to”: $\frac{k}{A}$
- “Rate of change”: $\frac{d y}{d x}$
Recognize and translate verbal descriptions into equations using proportionality

Examples

Population growth proportional to current size: $\frac{d P}{d t} = k P$
- Find $k$ using given data: $k = \frac{120}{800} = 0.15$
- Final equation: $\frac{d P}{d t} = 0.15 P$
Skill improvement inversely proportional to skill level: $\frac{d L}{d t} = \frac{k}{L}$
- Find $k$ using data: $k = 2$
- Final equation: $\frac{d L}{d t} = \frac{2}{L}$
Linear increase (e.g., shoe size): $\frac{d S}{d t} = k$
- Constant rate of change

Verifying solutions

A solution satisfies the differential equation when substituted
Steps:
- Differentiate the proposed function as needed
- Substitute function and derivatives into the equation
- Check if both sides are equal

Example verifications

$y = e^{3 x}$ solves $y^{''} - 6 y^{'} + 9 y = 0$
- $y^{'} = 3 e^{3 x}$ , $y^{''} = 9 e^{3 x}$
- Substitution yields $0 = 0$
$y = k sin (x)$ solves $y^{''} + 4 y = sin (x)$ when $k = \frac{1}{3}$
- $y^{''} = - k sin (x)$
- $3 k sin (x) = sin (x) ⟹ k = \frac{1}{3}$

Modeling & verifying solutions

What you’ll learn:

How to write differential equations from verbal descriptions
Recognize relationships involving proportionality
How to verify solutions to differential equations

Modeling situations

Here are some common phrases and translations, with $k$ as the constant of proportionality.

Phrase	Math translation
Rate of change of $y$ with respect to $x$	$\frac{d y}{d x}$
Proportional to $A$	$k A$
Inversely proportional to $A$	$\frac{k}{A}$
Proportional to product of $A$ and $B$	$k A B$
Proportional to square root of $t$	$k t$
Proportional to square of $v$ and inversely to $m$	$\frac{k v ^{2}}{m}$
Changing linearly	$k$

Examples

A population grows at a rate proportional to its current size. When the population is $800$ , it is growing at a rate of $120$ people per hour. Write a differential equation for this situation.

Because the rate of change of the population is directly proportional to the population itself, we define:

$P (t) =$ population at time $t$
$\frac{d P}{d t} =$ rate of change of the population with respect to time
$k =$ constant of proportionality

“Rate proportional to current size” translates to:

$\frac{d P}{d t} = k P$

This equation says that when $P$ increases, the growth rate $\frac{d P}{d t}$ increases in the same proportion.

Now use the given data to find $k$ . When $P = 800$ , the rate is $\frac{d P}{d t} = 120$ people per hour:

$120 = k (800)$

$k = 0.15$

So the differential equation that models this situation is

$\frac{d P}{d t} = 0.15 P$

The rate at which a person improves a skill is inversely proportional to their current skill level $L$ . When the skill level is $4$ , the improvement rate is $0.5$ units per day.

Solution

(spoiler)

“Inversely proportional to $L$ ” means proportional to $\frac{1}{L}$ , so the model is

$\frac{d L}{d t} = \frac{k}{L}$

where $k$ is the constant of proportionality.

This tells us:

The rate of change of the skill is proportional to the reciprocal of the current skill level $L$ .
When $L$ is small, $\frac{1}{L}$ is large, so improvement happens faster.
As $L$ increases, $\frac{1}{L}$ decreases, so the rate of improvement slows (the skill level itself is still increasing).

Now solve for $k$ using the given information. When $L = 4$ , the improvement rate is $\frac{d L}{d t} = 0.5$ :

$0.5 = \frac{k}{4}$

$k = 2$

So the differential equation is

$\frac{d L}{d t} = \frac{2}{L}$

A child’s shoe size increases linearly over time.

Solution

(spoiler)

Let shoe size be $S (t)$ . If $S (t)$ increases linearly, its graph is a straight line with constant slope. A constant slope means the derivative is constant.

So the differential equation is:

$\frac{d S}{d t} = k$

where $k$ is a constant representing the rate of increase in shoe size per unit time.

Verifying solutions

A solution to a differential equation is a function that makes the equation true when you substitute the function and its derivatives into the equation.

To verify a proposed solution:

Differentiate the given function as needed.
Substitute into the differential equation.
Check whether both sides match.

The function

$y = e^{3 x}$

is a solution to which of the following equations?

a) $y^{''} - 3 y^{'} = 6 e^{3 x}$
b) $y^{''} - 6 y^{'} + 9 y = 0$
c) $y^{''} - 9 y = 0$
d) $y^{''} + 3 y^{'} = 12 e^{3 x}$

Solution

(spoiler)

Differentiate $y = e^{3 x}$ :

$y = e^{3 x}$

$y^{'} = 3 e^{3 x}$

$y^{''} = 9 e^{3 x}$

Now test each choice.

For choice a):

$y^{''} - 3 y^{'} = 6 e^{3 x}$

$9 e^{3 x} - 3 (3 e^{3 x}) = 6 e^{3 x}$

$0 \neq = 6 e^{3 x}$

So choice a) is not correct.

Try choice b):

$y^{''} - 6 y^{'} + 9 y = 0$

$9 e^{3 x} - 6 (3 e^{3 x}) + 9 (e^{3 x}) = 0$

$0 = 0$

Choice b) is correct.

For what value of $k$ does $y = k sin (x)$ satisfy the differential equation $y^{''} + 4 y = sin (x)$ ?

Solution

(spoiler)

Start with $y = k sin (x)$ and differentiate:

$y^{'} = k cos (x)$

$y^{''} = - k sin (x)$

Substitute into $y^{''} + 4 y = sin (x)$ :

$- k sin (x) + 4 k sin (x) = sin (x)$

$3 k sin (x) = sin (x)$

Since this must hold for all $x$ , the coefficients of $sin (x)$ must match:

$3 k = 1$

$k = \frac{1}{3}$

Achievable AP Calculus AB

7. Differential equations

Our AP Calculus AB course is currently in development and is a work-in-progress.

Modeling & verifying solutions

5 min read

Font

Discuss

Feedback

What you’ll learn:

How to write differential equations from verbal descriptions
Recognize relationships involving proportionality
How to verify solutions to differential equations

Modeling situations

Here are some common phrases and translations, with $k$ as the constant of proportionality.

Phrase	Math translation
Rate of change of $y$ with respect to $x$	$\frac{d y}{d x}$
Proportional to $A$	$k A$
Inversely proportional to $A$	$\frac{k}{A}$
Proportional to product of $A$ and $B$	$k A B$
Proportional to square root of $t$	$k t$
Proportional to square of $v$ and inversely to $m$	$\frac{k v ^{2}}{m}$
Changing linearly	$k$

Examples

A population grows at a rate proportional to its current size. When the population is $800$ , it is growing at a rate of $120$ people per hour. Write a differential equation for this situation.

Because the rate of change of the population is directly proportional to the population itself, we define:

$P (t) =$ population at time $t$
$\frac{d P}{d t} =$ rate of change of the population with respect to time
$k =$ constant of proportionality

“Rate proportional to current size” translates to:

$\frac{d P}{d t} = k P$

This equation says that when $P$ increases, the growth rate $\frac{d P}{d t}$ increases in the same proportion.

Now use the given data to find $k$ . When $P = 800$ , the rate is $\frac{d P}{d t} = 120$ people per hour:

$120 = k (800)$

$k = 0.15$

So the differential equation that models this situation is

$\frac{d P}{d t} = 0.15 P$

The rate at which a person improves a skill is inversely proportional to their current skill level $L$ . When the skill level is $4$ , the improvement rate is $0.5$ units per day.

Solution

(spoiler)

“Inversely proportional to $L$ ” means proportional to $\frac{1}{L}$ , so the model is

$\frac{d L}{d t} = \frac{k}{L}$

where $k$ is the constant of proportionality.

This tells us:

The rate of change of the skill is proportional to the reciprocal of the current skill level $L$ .
When $L$ is small, $\frac{1}{L}$ is large, so improvement happens faster.
As $L$ increases, $\frac{1}{L}$ decreases, so the rate of improvement slows (the skill level itself is still increasing).

Now solve for $k$ using the given information. When $L = 4$ , the improvement rate is $\frac{d L}{d t} = 0.5$ :

$0.5 = \frac{k}{4}$

$k = 2$

So the differential equation is

$\frac{d L}{d t} = \frac{2}{L}$

A child’s shoe size increases linearly over time.

Solution

(spoiler)

Let shoe size be $S (t)$ . If $S (t)$ increases linearly, its graph is a straight line with constant slope. A constant slope means the derivative is constant.

So the differential equation is:

$\frac{d S}{d t} = k$

where $k$ is a constant representing the rate of increase in shoe size per unit time.

Verifying solutions

A solution to a differential equation is a function that makes the equation true when you substitute the function and its derivatives into the equation.

To verify a proposed solution:

Differentiate the given function as needed.
Substitute into the differential equation.
Check whether both sides match.

The function

$y = e^{3 x}$

is a solution to which of the following equations?

a) $y^{''} - 3 y^{'} = 6 e^{3 x}$
b) $y^{''} - 6 y^{'} + 9 y = 0$
c) $y^{''} - 9 y = 0$
d) $y^{''} + 3 y^{'} = 12 e^{3 x}$

Solution

(spoiler)

Differentiate $y = e^{3 x}$ :

$y = e^{3 x}$

$y^{'} = 3 e^{3 x}$

$y^{''} = 9 e^{3 x}$

Now test each choice.

For choice a):

$y^{''} - 3 y^{'} = 6 e^{3 x}$

$9 e^{3 x} - 3 (3 e^{3 x}) = 6 e^{3 x}$

$0 \neq = 6 e^{3 x}$

So choice a) is not correct.

Try choice b):

$y^{''} - 6 y^{'} + 9 y = 0$

$9 e^{3 x} - 6 (3 e^{3 x}) + 9 (e^{3 x}) = 0$

$0 = 0$

Choice b) is correct.

For what value of $k$ does $y = k sin (x)$ satisfy the differential equation $y^{''} + 4 y = sin (x)$ ?

Solution

(spoiler)

Start with $y = k sin (x)$ and differentiate:

$y^{'} = k cos (x)$

$y^{''} = - k sin (x)$

Substitute into $y^{''} + 4 y = sin (x)$ :

$- k sin (x) + 4 k sin (x) = sin (x)$

$3 k sin (x) = sin (x)$

Since this must hold for all $x$ , the coefficients of $sin (x)$ must match:

$3 k = 1$

$k = \frac{1}{3}$

Modeling situations

Differential equations relate a function to its derivatives
Key phrases for translation:
- “Proportional to”: $k A$
- “Inversely proportional to”: $\frac{k}{A}$
- “Rate of change”: $\frac{d y}{d x}$
Recognize and translate verbal descriptions into equations using proportionality

Examples

Population growth proportional to current size: $\frac{d P}{d t} = k P$
- Find $k$ using given data: $k = \frac{120}{800} = 0.15$
- Final equation: $\frac{d P}{d t} = 0.15 P$
Skill improvement inversely proportional to skill level: $\frac{d L}{d t} = \frac{k}{L}$
- Find $k$ using data: $k = 2$
- Final equation: $\frac{d L}{d t} = \frac{2}{L}$
Linear increase (e.g., shoe size): $\frac{d S}{d t} = k$
- Constant rate of change

Verifying solutions

A solution satisfies the differential equation when substituted
Steps:
- Differentiate the proposed function as needed
- Substitute function and derivatives into the equation
- Check if both sides are equal

Example verifications

$y = e^{3 x}$ solves $y^{''} - 6 y^{'} + 9 y = 0$
- $y^{'} = 3 e^{3 x}$ , $y^{''} = 9 e^{3 x}$
- Substitution yields $0 = 0$
$y = k sin (x)$ solves $y^{''} + 4 y = sin (x)$ when $k = \frac{1}{3}$
- $y^{''} = - k sin (x)$
- $3 k sin (x) = sin (x) ⟹ k = \frac{1}{3}$

Modeling & verifying solutions

What you’ll learn:

How to write differential equations from verbal descriptions
Recognize relationships involving proportionality
How to verify solutions to differential equations

Modeling situations

Here are some common phrases and translations, with $k$ as the constant of proportionality.

Phrase	Math translation
Rate of change of $y$ with respect to $x$	$\frac{d y}{d x}$
Proportional to $A$	$k A$
Inversely proportional to $A$	$\frac{k}{A}$
Proportional to product of $A$ and $B$	$k A B$
Proportional to square root of $t$	$k t$
Proportional to square of $v$ and inversely to $m$	$\frac{k v ^{2}}{m}$
Changing linearly	$k$

Examples

A population grows at a rate proportional to its current size. When the population is $800$ , it is growing at a rate of $120$ people per hour. Write a differential equation for this situation.

Because the rate of change of the population is directly proportional to the population itself, we define:

$P (t) =$ population at time $t$
$\frac{d P}{d t} =$ rate of change of the population with respect to time
$k =$ constant of proportionality

“Rate proportional to current size” translates to:

$\frac{d P}{d t} = k P$

This equation says that when $P$ increases, the growth rate $\frac{d P}{d t}$ increases in the same proportion.

Now use the given data to find $k$ . When $P = 800$ , the rate is $\frac{d P}{d t} = 120$ people per hour:

$120 = k (800)$

$k = 0.15$

So the differential equation that models this situation is

$\frac{d P}{d t} = 0.15 P$

The rate at which a person improves a skill is inversely proportional to their current skill level $L$ . When the skill level is $4$ , the improvement rate is $0.5$ units per day.

Solution

(spoiler)

“Inversely proportional to $L$ ” means proportional to $\frac{1}{L}$ , so the model is

$\frac{d L}{d t} = \frac{k}{L}$

where $k$ is the constant of proportionality.

This tells us:

The rate of change of the skill is proportional to the reciprocal of the current skill level $L$ .
When $L$ is small, $\frac{1}{L}$ is large, so improvement happens faster.
As $L$ increases, $\frac{1}{L}$ decreases, so the rate of improvement slows (the skill level itself is still increasing).

Now solve for $k$ using the given information. When $L = 4$ , the improvement rate is $\frac{d L}{d t} = 0.5$ :

$0.5 = \frac{k}{4}$

$k = 2$

So the differential equation is

$\frac{d L}{d t} = \frac{2}{L}$

A child’s shoe size increases linearly over time.

Solution

(spoiler)

Let shoe size be $S (t)$ . If $S (t)$ increases linearly, its graph is a straight line with constant slope. A constant slope means the derivative is constant.

So the differential equation is:

$\frac{d S}{d t} = k$

where $k$ is a constant representing the rate of increase in shoe size per unit time.

Verifying solutions

A solution to a differential equation is a function that makes the equation true when you substitute the function and its derivatives into the equation.

To verify a proposed solution:

Differentiate the given function as needed.
Substitute into the differential equation.
Check whether both sides match.

The function

$y = e^{3 x}$

is a solution to which of the following equations?

a) $y^{''} - 3 y^{'} = 6 e^{3 x}$
b) $y^{''} - 6 y^{'} + 9 y = 0$
c) $y^{''} - 9 y = 0$
d) $y^{''} + 3 y^{'} = 12 e^{3 x}$

Solution

(spoiler)

Differentiate $y = e^{3 x}$ :

$y = e^{3 x}$

$y^{'} = 3 e^{3 x}$

$y^{''} = 9 e^{3 x}$

Now test each choice.

For choice a):

$y^{''} - 3 y^{'} = 6 e^{3 x}$

$9 e^{3 x} - 3 (3 e^{3 x}) = 6 e^{3 x}$

$0 \neq = 6 e^{3 x}$

So choice a) is not correct.

Try choice b):

$y^{''} - 6 y^{'} + 9 y = 0$

$9 e^{3 x} - 6 (3 e^{3 x}) + 9 (e^{3 x}) = 0$

$0 = 0$

Choice b) is correct.

For what value of $k$ does $y = k sin (x)$ satisfy the differential equation $y^{''} + 4 y = sin (x)$ ?

Solution

(spoiler)

Start with $y = k sin (x)$ and differentiate:

$y^{'} = k cos (x)$

$y^{''} = - k sin (x)$

Substitute into $y^{''} + 4 y = sin (x)$ :

$- k sin (x) + 4 k sin (x) = sin (x)$

$3 k sin (x) = sin (x)$

Since this must hold for all $x$ , the coefficients of $sin (x)$ must match:

$3 k = 1$

$k = \frac{1}{3}$