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 is basically an equation that involves one or more derivatives of a function. It relates a function to its rate(s) of change.

Some of the problems in this unit combine concepts from earlier chapters you’ve already studied, such as linear approximations, extrema, and integration, but presented in a different way. The key is to identify which concept each problem is testing.

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 $x$	$\frac{d y}{d x}$
Is proportional to $A$	$k A$
Is inversely proportional to $A$	$\frac{k}{A}$
Is proportional to the product of $A$ and $B$	$k A B$
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.

Start by defining a few variables. Let:

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

The phrase “grows at a rate proportional to its current size” means that the population’s rate of change at any time is proportional to the population size. This 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.

Next, use the given information to find the value of $k$ .

When $P = 800$ , the rate is $\frac{d P}{d t} = 120$ people per hour. Substituting these values in:

$120 = k (800)$

$k = 0.15$

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. What is a differential equation that models this situation?

Solution

(spoiler)

If the current skill level is defined as $L$ , then the rate of improvement is $\frac{d L}{d t}$ .

The phrase “inversely proportional to $L$ ” means $\frac{k}{L}$ . So:

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

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 (but the skill level itself is still increasing).

Now solve for $k$ by substituting in the given information. When $L = 4$ , the improvement rate is $0.5$ :

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

$k = 2$

So the differential equation that models the situation is:

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

A child’s shoe size increases linearly over time.

Solution

(spoiler)

Let’s define the shoe size as $S (t)$ .

If $S (t)$ increases linearly, its graph is a straight line. A straight line has a constant slope, or a derivative function that is just a number.

This number can be represented by a constant of proportionality $k$ .

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 satisfies the equation when you substitute it in with its derivatives.

You can think of the “solution” as the “original function” in the equation.

For example, $y = 2 x^{2}$ is a solution to the differential equation

$\frac{d y}{d x} = 4 x$

because if you take the derivative of $y$ and substitute it into the equation, both sides are equal.

General steps when a problem gives a function, or solution, and asks whether it satisfies a differential equation:

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 by plugging in the expressions for $y, y^{'},$ and $y^{''}$ .

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.

Choice (b):

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

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

$0 = 0$

Since both sides of the equation are equal, choice (b) is the correct differential equation.

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 the differential equation $y^{''} + 4 y = sin (x)$ :

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

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

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

A differential equation is basically an equation that involves one or more derivatives of a function. It relates a function to its rate(s) of change.

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 $x$	$\frac{d y}{d x}$
Is proportional to $A$	$k A$
Is inversely proportional to $A$	$\frac{k}{A}$
Is proportional to the product of $A$ and $B$	$k A B$
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.

Start by defining a few variables. Let:

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

The phrase “grows at a rate proportional to its current size” means that the population’s rate of change at any time is proportional to the population size. This 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.

Next, use the given information to find the value of $k$ .

When $P = 800$ , the rate is $\frac{d P}{d t} = 120$ people per hour. Substituting these values in:

$120 = k (800)$

$k = 0.15$

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. What is a differential equation that models this situation?

Solution

(spoiler)

If the current skill level is defined as $L$ , then the rate of improvement is $\frac{d L}{d t}$ .

The phrase “inversely proportional to $L$ ” means $\frac{k}{L}$ . So:

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

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 (but the skill level itself is still increasing).

Now solve for $k$ by substituting in the given information. When $L = 4$ , the improvement rate is $0.5$ :

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

$k = 2$

So the differential equation that models the situation is:

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

A child’s shoe size increases linearly over time.

Solution

(spoiler)

Let’s define the shoe size as $S (t)$ .

If $S (t)$ increases linearly, its graph is a straight line. A straight line has a constant slope, or a derivative function that is just a number.

This number can be represented by a constant of proportionality $k$ .

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 satisfies the equation when you substitute it in with its derivatives.

You can think of the “solution” as the “original function” in the equation.

For example, $y = 2 x^{2}$ is a solution to the differential equation

$\frac{d y}{d x} = 4 x$

because if you take the derivative of $y$ and substitute it into the equation, both sides are equal.

General steps when a problem gives a function, or solution, and asks whether it satisfies a differential equation:

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 by plugging in the expressions for $y, y^{'},$ and $y^{''}$ .

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.

Choice (b):

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

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

$0 = 0$

Since both sides of the equation are equal, choice (b) is the correct differential equation.

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 the differential equation $y^{''} + 4 y = sin (x)$ :

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

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

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

A differential equation is basically an equation that involves one or more derivatives of a function. It relates a function to its rate(s) of change.

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 $x$	$\frac{d y}{d x}$
Is proportional to $A$	$k A$
Is inversely proportional to $A$	$\frac{k}{A}$
Is proportional to the product of $A$ and $B$	$k A B$
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.

Start by defining a few variables. Let:

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

The phrase “grows at a rate proportional to its current size” means that the population’s rate of change at any time is proportional to the population size. This 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.

Next, use the given information to find the value of $k$ .

When $P = 800$ , the rate is $\frac{d P}{d t} = 120$ people per hour. Substituting these values in:

$120 = k (800)$

$k = 0.15$

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. What is a differential equation that models this situation?

Solution

(spoiler)

If the current skill level is defined as $L$ , then the rate of improvement is $\frac{d L}{d t}$ .

The phrase “inversely proportional to $L$ ” means $\frac{k}{L}$ . So:

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

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 (but the skill level itself is still increasing).

Now solve for $k$ by substituting in the given information. When $L = 4$ , the improvement rate is $0.5$ :

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

$k = 2$

So the differential equation that models the situation is:

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

A child’s shoe size increases linearly over time.

Solution

(spoiler)

Let’s define the shoe size as $S (t)$ .

If $S (t)$ increases linearly, its graph is a straight line. A straight line has a constant slope, or a derivative function that is just a number.

This number can be represented by a constant of proportionality $k$ .

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 satisfies the equation when you substitute it in with its derivatives.

You can think of the “solution” as the “original function” in the equation.

For example, $y = 2 x^{2}$ is a solution to the differential equation

$\frac{d y}{d x} = 4 x$

because if you take the derivative of $y$ and substitute it into the equation, both sides are equal.

General steps when a problem gives a function, or solution, and asks whether it satisfies a differential equation:

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 by plugging in the expressions for $y, y^{'},$ and $y^{''}$ .

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.

Choice (b):

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

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

$0 = 0$

Since both sides of the equation are equal, choice (b) is the correct differential equation.

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 the differential equation $y^{''} + 4 y = sin (x)$ :

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

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

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

A differential equation is basically an equation that involves one or more derivatives of a function. It relates a function to its rate(s) of change.

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 $x$	$\frac{d y}{d x}$
Is proportional to $A$	$k A$
Is inversely proportional to $A$	$\frac{k}{A}$
Is proportional to the product of $A$ and $B$	$k A B$
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.

Start by defining a few variables. Let:

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

The phrase “grows at a rate proportional to its current size” means that the population’s rate of change at any time is proportional to the population size. This 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.

Next, use the given information to find the value of $k$ .

When $P = 800$ , the rate is $\frac{d P}{d t} = 120$ people per hour. Substituting these values in:

$120 = k (800)$

$k = 0.15$

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. What is a differential equation that models this situation?

Solution

(spoiler)

If the current skill level is defined as $L$ , then the rate of improvement is $\frac{d L}{d t}$ .

The phrase “inversely proportional to $L$ ” means $\frac{k}{L}$ . So:

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

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 (but the skill level itself is still increasing).

Now solve for $k$ by substituting in the given information. When $L = 4$ , the improvement rate is $0.5$ :

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

$k = 2$

So the differential equation that models the situation is:

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

A child’s shoe size increases linearly over time.

Solution

(spoiler)

Let’s define the shoe size as $S (t)$ .

If $S (t)$ increases linearly, its graph is a straight line. A straight line has a constant slope, or a derivative function that is just a number.

This number can be represented by a constant of proportionality $k$ .

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 satisfies the equation when you substitute it in with its derivatives.

You can think of the “solution” as the “original function” in the equation.

For example, $y = 2 x^{2}$ is a solution to the differential equation

$\frac{d y}{d x} = 4 x$

because if you take the derivative of $y$ and substitute it into the equation, both sides are equal.

General steps when a problem gives a function, or solution, and asks whether it satisfies a differential equation:

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 by plugging in the expressions for $y, y^{'},$ and $y^{''}$ .

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.

Choice (b):

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

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

$0 = 0$

Since both sides of the equation are equal, choice (b) is the correct differential equation.

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 the differential equation $y^{''} + 4 y = sin (x)$ :

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

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

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

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