Modeling & verifying solutions | Differential equations

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 one that relates a function and its derivatives. These equations can describe how the rate of change of a quantity depends on the quantity itself and model real-world situations.

Modeling situations

Modeling questions involve translating word problems into differential equations. Pay attention to key words such as proportional," “inversely proportional,” “rate of change,” or 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

1. 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.

Since the rate of change of the population depends directly on the population itself, if we let

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

then the differential equation for this situation is

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

This tells us that as population size $P$ increases, the rate of growth/derivative $\frac{d P}{d t}$ increases proportionally.

We can find $k$ with the given information: when the population $P = 800$ , the rate of change $\frac{d P}{d t} = 120$ people per hour. So

$120 = k (800)$

$k = 0.15$

The differential equation that models this situation is

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

2. 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)

This situation can be modeled by the differential equation

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

where $k$ is again 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 the skill improves quickly
As $L$ increases, $\frac{1}{L}$ decreases, and the rate of improvement slows (not that the skill level decreases).

To find $k$ : when the skill level $L = 4$ , the improvement rate $\frac{d L}{d t} = 0.5$ units per day. Then

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

$k = 2$

So the differential equation that models this situation is

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

3. A child’s shoe size increases linearly over time.

Solution

(spoiler)

Let shoe size be $S (t)$ . Imagine the graph of $S (t)$ - if it increases linearly, then it’s a straight line with a positive slope. Then the rate of change of shoe size with respect to time (derivative) is a constant. So the differential equation modeling this is:

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

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

Verifying solutions

A solution to a differential equation is a function whose derivatives satisfy the differential equation.

For these types of questions, just differentiate the given function and substitute the derivatives into the differential equation to check if the equation is true.

1. 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)

Find the derivatives of $y = e^{3 x}$ and plug into each equation to check which one matches.

$y = e^{3 x}$

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

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

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 the correct answer.

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 the correct option.

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

Solution

(spoiler)

Given $y = k sin (x)$ , its derivatives are

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

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

Then substitute into the differential equation $y^{''} + 4 y = sin (x)$ and solve for $k$ .

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

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

$3 k = 1$

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

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

1. 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.

Since the rate of change of the population depends directly on the population itself, if we let

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

then the differential equation for this situation is

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

This tells us that as population size $P$ increases, the rate of growth/derivative $\frac{d P}{d t}$ increases proportionally.

We can find $k$ with the given information: when the population $P = 800$ , the rate of change $\frac{d P}{d t} = 120$ people per hour. So

$120 = k (800)$

$k = 0.15$

The differential equation that models this situation is

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

2. 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)

This situation can be modeled by the differential equation

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

where $k$ is again 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 the skill improves quickly
As $L$ increases, $\frac{1}{L}$ decreases, and the rate of improvement slows (not that the skill level decreases).

To find $k$ : when the skill level $L = 4$ , the improvement rate $\frac{d L}{d t} = 0.5$ units per day. Then

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

$k = 2$

So the differential equation that models this situation is

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

3. A child’s shoe size increases linearly over time.

Solution

(spoiler)

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

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

Verifying solutions

A solution to a differential equation is a function whose derivatives satisfy the differential equation.

For these types of questions, just differentiate the given function and substitute the derivatives into the differential equation to check if the equation is true.

1. 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)

Find the derivatives of $y = e^{3 x}$ and plug into each equation to check which one matches.

$y = e^{3 x}$

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

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

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 the correct answer.

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 the correct option.

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

Solution

(spoiler)

Given $y = k sin (x)$ , its derivatives are

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

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

Then substitute into the differential equation $y^{''} + 4 y = sin (x)$ and solve for $k$ .

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

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

$3 k = 1$

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