7.1 - Modeling Situations with Differential Equations

Introduction to Differential Equations

A differential equation is an equation involving a rate of change. The general form looks like:

\frac{dy}{dx} = \text{(some expression)}

These equations are essential for modeling real-world situations where we know how quantities change rather than knowing the quantities themselves.

More precisely, a differential equation relates a function to its own derivatives, and often to the independent variable too. In

\frac{dy}{dx} = x^2y

, the unknown function is

y

, the independent variable is

x

, and the equation ties the slope

\frac{dy}{dx}

to both of them. So the answer to a differential equation is a function, not a number.

The order of a differential equation is its highest derivative.

\frac{dy}{dx} = x^2y

is first order, and an equation with

\frac{d^2y}{dx^2}

in it, like the acceleration in Example 3 below, is second order. Slope fields, separation of variables, and Euler’s method (BC) all work with first-order equations.

Proportionality Relationships

Before working with differential equations, you need to understand two types of proportional relationships.

Direct Proportionality

When variable

a

is directly proportional to variable

b

a = k \cdot b

where

k

is the constant of proportionality. The word “proportional” with no qualifier always means directly proportional.

Inverse Proportionality

When variable

a

is inversely proportional to variable

b

a = \frac{k}{b} \quad \text{or} \quad a = k \cdot \frac{1}{b}

You’ll also see this stated as “proportional to the reciprocal of

b

.”

Example 1: Direct Proportionality with Multiple Variables

Problem: The rate of change of

y

with respect to

x

is proportional to the product of

s

and the rate of change of

v

with respect to

x

Setting up the differential equation:

Since this is direct proportionality, with no modifier given:

\frac{dy}{dx} = k \cdot s \cdot \frac{dv}{dx}

The constant

k

is the proportionality constant, and the whole right side is the product of

s

and

\frac{dv}{dx}

Example 2: Inverse Proportionality with a Physical Application

Problem: The force

F

of a spring on a diving board is related to the distance

d

it is stretched. The rate of change of force with respect to distance is inversely proportional to the natural logarithm of the distance

d

Step 1: Write the general differential equation

\frac{dF}{dd} = \frac{k}{\ln(d)}

Step 2: Use the given information to find $k$

We’re told: - The rate of change of

F

5

units per centimeter - The distance stretched is

0.3

centimeters

Substitute:

5 = \frac{k}{\ln(0.3)}

Solve for

k

k = 5 \cdot \ln(0.3) = 5 \cdot (-1.204) \approx -6.020

Step 3: Write the specific differential equation

\frac{dF}{dd} = \frac{-6.020}{\ln(d)}

Example 3: Second Derivatives and Acceleration

Problem: A runner moves in a straight line along a track. Her position from the starting point is given by

p(t)

, where

t

is in minutes. During the first minute, her acceleration is proportional to the square root of time.

Analysis:

Acceleration is the second derivative of position, so using Leibniz notation:

\frac{d^2p}{dt^2} = k\sqrt{t}

Important distinctions: - “Square root of

t

” means

\sqrt{t}

, or

t^{1/2}

- “Square of

t

” means

t^2

- “Cube root of

t

” means

\sqrt[3]{t}

, or

t^{1/3}

Critical notation point: when you’re dealing with velocity or acceleration in differential equations, it’s clearest to use Leibniz notation (

\frac{d}{dx}

), not function notation like

p''(t)

, though the exam accepts both.

For second derivatives: - Position:

p(t)

- Velocity:

\frac{dp}{dt}

- Acceleration:

\frac{d^2p}{dt^2}

Example 4: Proportional to a Difference

Problem: A cup of hot chocolate cools in a room that stays at

68

degrees Fahrenheit. Its temperature

T

changes at a rate (in degrees per minute) that is proportional to the difference between

T

and the room temperature.

Setting up the differential equation:

“The difference between

T

and the room temperature” is

T - 68

, and the whole difference gets multiplied by

k

\frac{dT}{dt} = k(T - 68)

The hot chocolate is warmer than the room, so

T - 68 > 0

, and it’s cooling, so

\frac{dT}{dt} < 0

. That means

k

has to be negative. The word “difference” is your signal to put a subtraction inside parentheses.

Working with Differential Equations: General Strategy

Identify the rate of change being described

Determine the type of proportionality, direct or inverse

Set up the equation structure:

Direct: $k \cdot (\text{expression})$

Inverse: $\frac{k}{\text{expression}}$ , or $k \cdot \frac{1}{\text{expression}}$

If specific values are given, substitute them to solve for $k$

Write the complete differential equation

Key Takeaways

Differential equation: an equation containing a derivative, written as $\frac{dy}{dx} = \text{(expression)}$

Solutions are functions: a differential equation relates a function to its derivatives, so solving one means finding a function

Proportional to a difference: “proportional to the difference between $T$ and $68$ ” means $\frac{dT}{dt} = k(T - 68)$

Direct proportionality: $a = k \cdot b$ , and the word “proportional” alone means direct

Inverse proportionality: $a = \frac{k}{b}$ , also called “proportional to the reciprocal”

Use Leibniz notation for differential equations. The exam accepts prime notation too, but it’s clearer to write $\frac{dy}{dx}$ , not $y'$

Second derivatives in Leibniz notation: $\frac{d^2y}{dx^2}$ for acceleration problems

Finding $k$ : when you’re given specific values, substitute them into the equation to solve for the constant of proportionality

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