7.1 Intro to differential equations

Achievable AP Calculus AB

7. Differential equations

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Intro to differential equations

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What you’ll learn

How to model verbal descriptions into differential equations
Identify relationships involving proportionality
How to verify solutions to differential equations

A differential equation is an equation that relates an unknown function to one or more of its derivatives.

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 the keyword “proportional to,” which means one quantity is always a constant multiple of another. We represent the multiplier with $k$ , known as the constant of proportionality.

“Rate is proportional to $y$ ”:
- Rate $= k y$
“Rate is inversely proportional to $y$ ”:
- Rate $= \frac{k}{y}$
“ $y$ is changing at a constant rate of…”
- Rate $= k$
- A constant rate doesn’t depend on the quantity size or time and is just a fixed number.

Example 1: Directly proportional

A population of bacteria $P (t)$ grows at a rate directly proportional to the square root of the population size at time $t$ hours. If the population is $100$ when it is growing at a rate of $50$ bacteria per hour, write a differential equation that models this situation.

Solution

(spoiler)

Translate the verbal phrase - a “rate (of growth) directly proportional to the square root of population size” translates to

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

Next, use the initial condition given to solve for $k$ . When $P = 100$ bacteria, it is growing at a rate of $\frac{d P}{d t} = 50$ bacteria per hour. Substitute these into the equation:

$5050 k = k 100 = 10 k = 5$

Therefore the differential equation that models this situation is

$\frac{d P}{d t} = 5 P$

Example 2: Inversely proportional

The rate at which a person improves a particular skill is inversely proportional to their current skill level $L (t)$ . When the $L = 4$ , the rate of improvement is $0.5$ units per day. Write the differential equation that models this behavior.

Solution

(spoiler)

The current skill level is defined as $L (t)$ , so the rate of improvement of the skill is $\frac{d L}{d t}$ .

Since the rate is inversely proportional to the current skill level,

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

This tells us:

The improvement rate 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 $0.5$ :

$0.5 k = \frac{k}{4} = 2$

Therefore, the differential equation that models the situation is

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

Example 3: Constant rate

Water is leaking out of a tank such that the volume of the water, $V (t)$ , decreases at a constant rate of $3$ gallons per minute. Write a differential equation that models this situation.

Solution

(spoiler)

You’ve already in fact practiced the notation for differential equations in the related rates chapter!

Since the volume of water is decreasing, its rate of change with respect to time must be negative, and the differential equation that models this situation is

$\frac{d V}{d t} = - 3$

Verifying solutions

Recall from section 6.7 on indefinite integrals that a solution (either a general or particular one) is essentially the “original function” that has a given derivative.

On the AP exam, verifying a solution means checking whether a given function satisfies a differential equation. To do this:

Differentiate the given function as needed.
Substitute into the differential equation.
Simplify to check if both sides match.

Example 1: Matching equations

The function

$y = e^{3 x}$

is a solution to which of the following differential 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)

Since the answers choices involve $y^{''}$ , the second derivative, differentiate $y = e^{3 x}$ twice:

$y = e^{3 x}$

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

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

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.

Example 2: Finding an unknown constant

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:

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

Differential equations

Equation relating an unknown function to one or more of its derivatives
“Solution” = a function that satisfies the equation when substituted in

Modeling verbal descriptions

“Rate of change of y with respect to x” → $\frac{d y}{d x}$
Key proportionality translations (k = constant of proportionality):
- Proportional to A → $k A$ ; inversely proportional to A → $\frac{k}{A}$
- Proportional to product of A and B → $k A B$ ; changing linearly → $k$
Find k by substituting given numerical values into the equation

Common model types

Growth/decay proportional to current size → $\frac{d P}{d t} = k P$
Inversely proportional rate → $\frac{d L}{d t} = \frac{k}{L}$ (faster change when L is small)
Linear change → $\frac{d S}{d t} = k$ (constant derivative)

Verifying solutions

Differentiate the given function as needed (find y′, y″, etc.)
Substitute y and its derivatives into the equation
Confirm both sides are equal; if solving for k, match coefficients

Intro to differential equations

What you’ll learn

How to model verbal descriptions into differential equations
Identify relationships involving proportionality
How to verify solutions to differential equations

A differential equation is an equation that relates an unknown function to one or more of its derivatives.

Modeling situations

Modeling questions ask you to translate a verbal description into a differential equation.

Watch for the keyword “proportional to,” which means one quantity is always a constant multiple of another. We represent the multiplier with $k$ , known as the constant of proportionality.

“Rate is proportional to $y$ ”:
- Rate $= k y$
“Rate is inversely proportional to $y$ ”:
- Rate $= \frac{k}{y}$
“ $y$ is changing at a constant rate of…”
- Rate $= k$
- A constant rate doesn’t depend on the quantity size or time and is just a fixed number.

Example 1: Directly proportional

A population of bacteria $P (t)$ grows at a rate directly proportional to the square root of the population size at time $t$ hours. If the population is $100$ when it is growing at a rate of $50$ bacteria per hour, write a differential equation that models this situation.

Solution

(spoiler)

Translate the verbal phrase - a “rate (of growth) directly proportional to the square root of population size” translates to

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

Next, use the initial condition given to solve for $k$ . When $P = 100$ bacteria, it is growing at a rate of $\frac{d P}{d t} = 50$ bacteria per hour. Substitute these into the equation:

$5050 k = k 100 = 10 k = 5$

Therefore the differential equation that models this situation is

$\frac{d P}{d t} = 5 P$

Example 2: Inversely proportional

The rate at which a person improves a particular skill is inversely proportional to their current skill level $L (t)$ . When the $L = 4$ , the rate of improvement is $0.5$ units per day. Write the differential equation that models this behavior.

Solution

(spoiler)

The current skill level is defined as $L (t)$ , so the rate of improvement of the skill is $\frac{d L}{d t}$ .

Since the rate is inversely proportional to the current skill level,

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

This tells us:

The improvement rate 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 $0.5$ :

$0.5 k = \frac{k}{4} = 2$

Therefore, the differential equation that models the situation is

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

Example 3: Constant rate

Water is leaking out of a tank such that the volume of the water, $V (t)$ , decreases at a constant rate of $3$ gallons per minute. Write a differential equation that models this situation.

Solution

(spoiler)

You’ve already in fact practiced the notation for differential equations in the related rates chapter!

Since the volume of water is decreasing, its rate of change with respect to time must be negative, and the differential equation that models this situation is

$\frac{d V}{d t} = - 3$

Verifying solutions

Recall from section 6.7 on indefinite integrals that a solution (either a general or particular one) is essentially the “original function” that has a given derivative.

On the AP exam, verifying a solution means checking whether a given function satisfies a differential equation. To do this:

Differentiate the given function as needed.
Substitute into the differential equation.
Simplify to check if both sides match.

Example 1: Matching equations

The function

$y = e^{3 x}$

is a solution to which of the following differential 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)

Since the answers choices involve $y^{''}$ , the second derivative, differentiate $y = e^{3 x}$ twice:

$y = e^{3 x}$

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

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

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.

Example 2: Finding an unknown constant

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 now in "early access" - get 50% off for a limited time.