What's the difference between a differential equation and an initial condition?

A differential equation describes the relationship between a function and its derivatives — how fast something changes. An initial condition is a single known value of the function, like y(0) = 5. The equation alone has a whole family of solutions; the initial condition pins down one specific member.

How do I know when to include the constant k?

Whenever a problem says a rate is "proportional to" something, you must introduce a constant of proportionality k. "Directly proportional to x" means multiply by k (kx); "inversely proportional to x" means divide (k/x). Never leave k out — it is part of the model.

When verifying a solution, is it enough to check one point?

No. A function is a solution only if it satisfies the differential equation as an identity — true for every x in the domain. You substitute the function and its derivative, simplify both sides, and confirm they are algebraically equal, not just equal at one input.

Do I need to solve the differential equation to verify a solution?

No, and that is the point. Verification only requires differentiating the given candidate function and substituting into the equation. Actually solving the equation (for example by separation of variables) is a separate skill covered later in the unit.

AP-CALCBC-7.1-7.2

U7.1 Modeling and Verifying Differential Equations

Learn to turn verbal rate-of-change statements into differential equations and verify solutions by substitution — AP Calculus BC Unit 7.1-7.2.

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What this lesson covers

Every differential equation begins as a story about how something changes. A population grows, a cup of coffee cools, a savings account earns interest — each sentence hides a derivative. In this lesson you will learn two closely linked skills: translating a verbal description of a rate of change into a differential equation, and verifying that a proposed function actually solves a given differential equation.

These are foundational moves for all of Unit 7. Before you ever draw a slope field, run Euler's method, or separate variables, you must be able to read the English, spot the derivative, and write the equation. Verification then gives you a way to check any candidate solution — including ones you or the AP graders produce later.

Translating Words into Derivatives

A differential equation relates a function to its derivative. The phrase to watch for is rate of change: the rate of change of a quantity

y

with respect to time is

\frac{dy}{dt}

. Your job is to build an equation of the form

\frac{dy}{dt} = (\text{expression})

from a sentence.

Work phrase by phrase. The subject of the sentence names the quantity (

y

P

T

, etc.). The words after "rate of change" or "changes" become the left side. Everything describing what the rate equals becomes the right side.

English phrase	Symbol
rate of change of $y$	$\frac{dy}{dt}$
proportional to $y$	$ky$
proportional to the square of $y$	$ky^2$
inversely proportional to $y$	$\frac{k}{y}$
proportional to the product of $y$ and $(L-y)$	$ky(L-y)$
difference between $y$ and a constant $A$	$(y-A)$

The word "proportional" always introduces an unknown constant

k

. Do not drop it. "Directly proportional to" means multiply by

k

; "inversely proportional to" means divide. Watch for compound descriptions such as "proportional to the product of the amount present and the amount remaining," which produces the logistic form

ky(L-y)

you will study in U7.9.

Common Modeling Patterns

Certain phrasings recur so often on the exam that recognizing them saves time. "The rate of change is proportional to the amount present" gives

\frac{dy}{dt} = ky

, the exponential pattern. "The rate of change is proportional to the difference between the temperature and the surrounding temperature" gives Newton's cooling form

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

.

Pay attention to signs. If a quantity is decreasing — a substance decays, an object cools toward a lower temperature — the constant of proportionality will be negative, or the difference will be written so the sign works out. For cooling,

T > T_s

means

T - T_s > 0

, and since temperature is falling,

k

must be negative so that

\frac{dT}{dt} < 0

.

A frequent misconception is confusing the quantity with its rate. "The population is 500" is an initial condition,

P(0) = 500

, not part of the differential equation. "The population grows at 3% per year" describes the rate:

\frac{dP}{dt} = 0.03P

. Keep the equation (the relationship among derivatives) separate from the initial condition (a single value). AP problems often hand you both and expect you to label each correctly.

Verifying a Function Is a Solution

To verify that a function

y = f(x)

solves a differential equation, you substitute the function and its derivatives into the equation and check that both sides are identically equal — equal for all

x

in the domain, not just at one point.

The procedure has three steps. First, differentiate the candidate function to get every derivative that appears in the equation. Second, substitute those expressions into the differential equation. Third, simplify both sides and confirm they match. If the equation also comes with an initial condition, plug in the given input to check that too.

Step	Action
1	Compute $y'$ (and $y''$ if needed)
2	Substitute $y$ and $y'$ into the DE
3	Simplify; confirm LHS $=$ RHS for all $x$

A common error is checking only that the point satisfies the equation at a single value. Verification of a solution requires an identity. Another pitfall: when the equation is written in a form like

\frac{dy}{dx} = xy

, you must replace

y

with the actual function everywhere it appears on the right side, then compare to the derivative you computed on the left.

How the AP Exam Tests This

On the multiple-choice section, expect prompts such as "Which of the following differential equations models the situation?" where you match a sentence to an equation, and "Which function is a solution to

\frac{dy}{dx} = \dots

?" where you must differentiate each option mentally or by quick work.

On free-response questions, modeling usually appears as the opening line of a longer problem: "The rate at which water leaks from a tank is proportional to the square root of the volume." You write

\frac{dV}{dt} = k\sqrt{V}

and then proceed to later parts (slope fields, separation of variables, Euler's method). A correct equation here earns setup points and sets up every part that follows, so accuracy is high-value.

Graders reward showing the substitution explicitly when verifying. Write out

y'

, substitute, and state the conclusion in words: "Since both sides equal

2xe^{x^2}

, the function satisfies the equation." Do not skip to "yes." Also, remember that a differential equation generally has a family of solutions differing by a constant; verifying one particular member does not require finding the whole family — that comes in U7.6.

Key terms

Differential equation.: An equation relating a function to one or more of its derivatives, such as $\frac{dy}{dx} = ky$ .
Rate of change.: The derivative of a quantity with respect to an independent variable, typically $\frac{dy}{dt}$ for change over time.
Proportional.: Related by multiplication by a constant $k$ ; "y is proportional to x" means $y = kx$ , introducing an unknown constant.
Constant of proportionality.: The unknown constant $k$ that appears whenever a rate is described as proportional to something.
Initial condition.: A given value of the function at a specific input, such as $y(0) = 5$ , used to identify a particular solution.
Solution to a DE.: A function that, together with its derivatives, satisfies the differential equation as an identity over its domain.
Verification.: The process of substituting a candidate function and its derivatives into a differential equation to confirm both sides are equal.
Newton's Law of Cooling.: The model $\frac{dT}{dt} = k(T - T_s)$ , where an object's temperature changes proportionally to its difference from the surroundings.

Worked example

A tank contains a chemical whose amount

A

(in grams) changes over time

t

(in hours). The rate of change of the amount is proportional to the difference between 200 grams and the current amount. Initially the tank holds 50 grams. (a) Write a differential equation modeling the situation. (b) Verify that

A(t) = 200 - 150e^{-kt}

is a solution and satisfies the initial condition.

Part (a): The subject is the amount

A

, so the left side is

\frac{dA}{dt}

. "Proportional to the difference between 200 and the current amount" becomes

k(200 - A)

. The model is

\frac{dA}{dt} = k(200 - A).

The initial condition is

A(0) = 50

.

Part (b): First differentiate the candidate

A(t) = 200 - 150e^{-kt}

. Using the chain rule,

\frac{dA}{dt} = -150 \cdot (-k)e^{-kt} = 150k e^{-kt}.

Now compute the right side of the DE by substituting

A

k(200 - A) = k\big(200 - (200 - 150e^{-kt})\big) = k(150e^{-kt}) = 150k e^{-kt}.

Both sides equal

150k e^{-kt}

, so the function satisfies the differential equation for all

t

.

Finally check the initial condition:

A(0) = 200 - 150e^{0} = 200 - 150 = 50

. This matches, so the function is a valid particular solution to the modeled problem.

Practice questions

The rate at which a rumor spreads through a school of 800 students is proportional to the product of the number of students who have heard it,

N

, and the number who have not. Which differential equation models this situation?

$\frac{dN}{dt} = kN$
$\frac{dN}{dt} = k(800 - N)$
$\frac{dN}{dt} = kN(800 - N)$
$\frac{dN}{dt} = \frac{k}{N(800 - N)}$

Answer: $\frac{dN}{dt} = kN(800 - N)$

"Product of the number who have heard it and the number who have not" means multiply

N

(800 - N)

. "Proportional to" attaches the constant

k

. The result

\frac{dN}{dt} = kN(800 - N)

is the logistic pattern. The first choice ignores the unheard group, the second ignores

N

itself, and the fourth wrongly uses inverse proportionality.

Show whether

y = 3e^{x^2}

is a solution to the differential equation

\frac{dy}{dx} = 2xy

Answer: Yes, it is a solution.

Differentiate:

\frac{dy}{dx} = 3 \cdot 2x e^{x^2} = 6x e^{x^2}

. Now substitute

y = 3e^{x^2}

into the right side:

2xy = 2x(3e^{x^2}) = 6x e^{x^2}

. Both sides equal

6x e^{x^2}

for all

x

, so the function satisfies the equation and is a solution. The key is checking the identity everywhere, not at a single point.

A cup of coffee at 90°C cools in a room kept at 20°C. Its temperature

T

decreases at a rate proportional to the difference between its temperature and room temperature. Write a differential equation and an initial condition for

T(t)

, and state the sign of the proportionality constant.

Answer: $\frac{dT}{dt} = k(T - 20)$ , with $T(0) = 90$ and $k < 0$ .

The rate is

\frac{dT}{dt}

, proportional to the difference

(T - 20)

, giving

\frac{dT}{dt} = k(T - 20)

. Since the coffee is hotter than the room,

T - 20 > 0

, but the temperature is falling so

\frac{dT}{dt} < 0

; therefore

k

must be negative. The initial temperature gives

T(0) = 90

FAQ

What's the difference between a differential equation and an initial condition?: A differential equation describes the relationship between a function and its derivatives — how fast something changes. An initial condition is a single known value of the function, like $y(0) = 5$ . The equation alone has a whole family of solutions; the initial condition pins down one specific member.
How do I know when to include the constant k?: Whenever a problem says a rate is "proportional to" something, you must introduce a constant of proportionality $k$ . "Directly proportional to $x$ " means multiply by $k$ ( $kx$ ); "inversely proportional to $x$ " means divide ( $\frac{k}{x}$ ). Never leave $k$ out — it is part of the model.
When verifying a solution, is it enough to check one point?: No. A function is a solution only if it satisfies the differential equation as an identity — true for every $x$ in the domain. You substitute the function and its derivative, simplify both sides, and confirm they are algebraically equal, not just equal at one input.
Do I need to solve the differential equation to verify a solution?: No, and that is the point. Verification only requires differentiating the given candidate function and substituting into the equation. Actually solving the equation (for example by separation of variables) is a separate skill covered later in the unit.

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