Unit 7Checking if a function satisfies a differential equation

Concept 4.119Built for May 2027 AP Exam

Checking if a function satisfies a differential equation

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

A differential equation is a condition on a function, and a candidate either meets it or does not. To decide, differentiate the candidate, substitute it into the right side, and compare two finished expressions. If they are the same function, the candidate is a solution; if the comparison fails anywhere on the interval, it is not. The check is mechanical, which is exactly why it is worth points: no integration, no guessing, just differentiation and algebra. Two details decide most cases. The comparison has to hold on an interval rather than at a single point, and a missing chain-rule factor is the most frequent reason a correct answer gets marked wrong. When the candidate is given implicitly, differentiate both sides with respect to the independent variable and carry the unknown derivative as a symbol.

On the AP Exam

AP Calculus AB tests this skill through the representations and reasoning described below.

What you’ll see: Rarely as a free-response part of its own. It appears as a multiple-choice item — “which of these functions satisfies the equation” or “show that $y = \ldots$ is a solution” — and inside the first line of a longer free-response part, where writing down the verification earns a point before any solving begins.
What you’ll do: Differentiate the candidate, substitute it into the side of the equation that does not already contain a derivative, simplify each side on its own line, and state that the two agree.
Memory hook: Differentiate, substitute, compare. Three verbs, three lines, no integration anywhere.
Use it on the exam: For MCQs: Do not try to solve the differential equation. Differentiate each offered function and substitute; the one that produces a true statement is the answer, and the check costs about twenty seconds per option. For FRQs: Write the two sides on separate lines before claiming they match. A reader cannot award the substitution point to work in which the two expressions were never visibly compared.

Concept Review

What “satisfies” Actually Means

Figure 1 — Verification is a comparison of two computed expressions. The candidate goes in, two sides come out, and the claim is that they are the same function.

A first-order differential equation has the form $\frac{dy}{dx} = F(x, y)$ . A function $y = f(x)$ satisfies the equation on an interval $I$ when substituting $f$ for $y$ and $f'$ for $\frac{dy}{dx}$ produces a statement that is true for every $x$ in $I$ :

$\text{for all } x \in I: \qquad f'(x) = F\left(x, f(x)\right)$

The phrase that does the work is for all $x$ in $I$ . An equation that holds at one convenient value has not been verified, and the exam writes distractors that exploit exactly that gap.

Verification is a comparison of two computed expressions. The candidate goes in, two sides come out, and the claim is that they are the same function.

The Three-Line Check

Line	What you write	What it proves
1	$y = 3e^{2x} \Rightarrow y' = 6e^{2x}$	the derivative of the candidate, chain rule included
2	right side $= 2y = 2\left(3e^{2x}\right) = 6e^{2x}$	the equation's own expression, with the candidate substituted
3	$6e^{2x} = 6e^{2x}$ for all $x$	the two sides are the same function

Nothing else is required. The verification is finished the moment line 3 is written, and every additional step risks introducing an error into work that was already complete.

Implicit Candidates

When the candidate arrives as a relation rather than a function, differentiate both sides with respect to the independent variable and keep $\frac{dy}{dx}$ as a symbol. For $y^{2} = x^{3} + 7$ :

$2y\frac{dy}{dx} = 3x^{2} \qquad\Longrightarrow\qquad \frac{dy}{dx} = \frac{3x^{2}}{2y}$

So the relation satisfies $\frac{dy}{dx} = \frac{3x^{2}}{2y}$ on any interval where $y \neq 0$ . The implicit route is the only practical one here, because solving for $y$ first introduces a $\pm$ sign and a restricted domain that the original relation does not have.

Families and Parameters

A one-parameter family can satisfy an equation for every value of the parameter, and showing that is often the whole task. For $y = Ce^{2x}$ :

$y' = 2Ce^{2x} = 2\left(Ce^{2x}\right) = 2y$

The parameter cancels because differentiation is linear: multiplying a solution by a constant produces another solution. That is why the general solution of $\frac{dy}{dx} = 2y$ is written with a $C$ rather than with a single function, and why a verification that works for $C = 3$ does not automatically prove the family claim — you must keep the $C$ symbolic until the last line.

Where the Check Fails: One Point Versus an Interval

The candidate $y = 2x + 1$ passes a check against $\frac{dy}{dx} = 2x$ at $x = 1$ , because both sides equal $2$ there. It fails at every other value, since $y' = 2$ is constant while $2x$ is not. A single-point match is a coincidence of two graphs crossing, and crossing once is not the same as coinciding.

The interval matters for a second reason: a solution can be valid on one piece of its domain and meaningless on another. The relation $y^{2} = 2x + C$ satisfies $\frac{dy}{dx} = \frac{1}{y}$ only where $y \neq 0$ , so a complete answer names that exclusion. On the exam the exclusion is usually worth a point of its own when the question says “on what interval.”

Common Misconception Check

Myth 1: “If substituting gives a true statement, the work is done.” Truth: It is done only if the statement is true for every $x$ in an interval. One value proves nothing.
Myth 2: “I can check the answer by integrating the equation instead.” Truth: Integration is a different task with a different failure mode. Verification never requires an antiderivative.
Myth 3: “A candidate that fails must be algebraically wrong.” Truth: Most failures are differentiation errors in the candidate, not errors in the equation. Recompute the derivative before concluding anything.

Topic-Specific Study Method

Differentiate First, Then Substitute

Borrowed idea: Differentiate the candidate before you substitute it anywhere. That single step exposes a missing chain-rule factor at once. Then put the candidate into the right side and simplify each side separately, so the comparison is between two finished expressions. They must agree as functions on an interval, not at a single point. A candidate that satisfies the equation at one convenient value has proved nothing, because the derivative condition is a statement about a whole interval.

Exam disguise	What you see on the page	What your brain must translate	Required execution
Explicit candidate	$y = 3e^{2x}$ , equation given	differentiate, then substitute	three lines, no integration
Implicit candidate	$y^{2} = x^{3} + 7$	differentiate both sides, keep $y'$	solve for $y'$ , note $y \neq 0$
Parameterised family	$y = Ce^{2x}$ for all $C$	keep $C$ symbolic to the end	the parameter must cancel
“Show that” phrasing	a claim, not a question	the answer is the verification itself	write the comparison explicitly
Distractor built on one point	options that agree at a single $x$	test a second value	reject any option that only matches once

① Differentiate on its own line. Combining the derivative with the substitution in one line is where chain-rule factors disappear.

② Simplify each side independently before comparing. Comparing a partly simplified left side with a partly simplified right side invites a false mismatch.

③ End with a sentence, not a symbol. “The two sides agree for all $x$ , so the function is a solution” is the sentence that carries the point.

Training & Error Diagnosis

Core Archetype (Simplest Form): Show that $y = x^{2}e^{3x}$ is a solution of $\frac{dy}{dx} = 3y + 2xe^{3x}$ .

Standard Execution:

Differentiate. By the product rule, $y' = 2xe^{3x} + 3x^{2}e^{3x}$ .

Substitute. The right side is $3\left(x^{2}e^{3x}\right) + 2xe^{3x}$ .

Compare. Both equal $3x^{2}e^{3x} + 2xe^{3x}$ , so the candidate is a solution for all real $x$ .

State. The identity holds on $(-\infty, \infty)$ .

Exam Variation (Context / Parameter Shift): The relation $y^{2} = 2x + C$ is claimed to satisfy $\frac{dy}{dx} = \frac{1}{y}$ . Verify the claim and state where it holds.

Solution: Differentiating both sides with respect to $x$ gives $2y\frac{dy}{dx} = 2$ , so $\frac{dy}{dx} = \frac{1}{y}$ . The algebra is valid for every $C$ , but the division requires $y \neq 0$ , and $y = 0$ occurs exactly when $2x + C = 0$ , that is at $x = -\frac{C}{2}$ . So the relation satisfies the equation on the two intervals cut off by that point, and a complete answer names the exclusion rather than claiming the whole real line.

Common Traps & Failure Signals:

Chain rule dropped (Signal: $y' = 3e^{2x}$ for $y = 3e^{2x}$ ): the inner rate of $2$ must appear.
Single-point check (Signal: one value of $x$ substituted, then a conclusion): the claim is about an interval.
Comparison never written (Signal: substitution with no equality line): the reader cannot award the point.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Chain Rule Dropped: You differentiated $3e^{2x}$ as $3e^{2x}$ instead of $6e^{2x}$ , which is the single most common way an otherwise correct verification collapses into a false negative on the exam.
Point Substituted: One convenient value is not an interval.
One Side Only: You never computed the right side, so nothing was compared.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; No released free-response part since 2017 scores verification as a step of its own; the skill appears in the multiple-choice section and as an unstated premise inside the free-response parts that follow it
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; Every free-response set that mentions a differential equation states the equation as given rather than asking the student to confirm a candidate
FRQ Years Appeared: None in Released FRQs ---

Practice Examples

MCQ 1

Test Each Option

Which of these functions satisfies the differential equation $\frac{dy}{dx} = 2xy$ ?

MCQ 2

An Implicit Candidate

The relation $y^{2} = x^{3} + 7$ defines $y$ implicitly. Which differential equation does it satisfy?

MCQ 3

Reading the Parameter Off the Equation

For which value of the constant $k$ is $y = e^{kt}$ a solution of $\frac{dy}{dt} = -5y$ ?

FRQ 4

Verification With and Without an Explicit Formula

(a) Show that $y = 4e^{-3x} + 2$ is a solution of $\frac{dy}{dx} = -3(y - 2)$ for all real $x$ .

(b) Show that every relation of the form $y^{2} = 2x + C$ satisfies $\frac{dy}{dx} = \frac{1}{y}$ , and state the intervals on which the verification is valid.

(c) A student checks $y = 2x + 1$ against $\frac{dy}{dx} = 2x$ at $x = 1$ , obtains $2 = 2$ , and concludes that the function is a solution. Explain what is wrong with that conclusion.

Cheatsheet

Core Reference

	Action	Failure if skipped
1	differentiate the candidate on its own line	a chain factor goes missing
2	substitute the candidate into the non-derivative side	the comparison has nothing to compare
3	simplify each side separately	a false mismatch from partial algebra
4	state equality for all $x$ in an interval	the point is not awarded
5	name any excluded value (implicit case)	half the domain claim is lost

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Unit 7Checking if a function satisfies a differential equation

Concept 4.119Built for May 2027 AP Exam

Checking if a function satisfies a differential equation

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

AP Calculus AB tests this skill through the representations and reasoning described below.

What you’ll see: Rarely as a free-response part of its own. It appears as a multiple-choice item — “which of these functions satisfies the equation” or “show that $y = \ldots$ is a solution” — and inside the first line of a longer free-response part, where writing down the verification earns a point before any solving begins.
What you’ll do: Differentiate the candidate, substitute it into the side of the equation that does not already contain a derivative, simplify each side on its own line, and state that the two agree.
Memory hook: Differentiate, substitute, compare. Three verbs, three lines, no integration anywhere.
Use it on the exam: For MCQs: Do not try to solve the differential equation. Differentiate each offered function and substitute; the one that produces a true statement is the answer, and the check costs about twenty seconds per option. For FRQs: Write the two sides on separate lines before claiming they match. A reader cannot award the substitution point to work in which the two expressions were never visibly compared.

Concept Review

What “satisfies” Actually Means

Figure 1 — Verification is a comparison of two computed expressions. The candidate goes in, two sides come out, and the claim is that they are the same function.

$\text{for all } x \in I: \qquad f'(x) = F\left(x, f(x)\right)$

The phrase that does the work is for all $x$ in $I$ . An equation that holds at one convenient value has not been verified, and the exam writes distractors that exploit exactly that gap.

Verification is a comparison of two computed expressions. The candidate goes in, two sides come out, and the claim is that they are the same function.

The Three-Line Check

Line	What you write	What it proves
1	$y = 3e^{2x} \Rightarrow y' = 6e^{2x}$	the derivative of the candidate, chain rule included
2	right side $= 2y = 2\left(3e^{2x}\right) = 6e^{2x}$	the equation's own expression, with the candidate substituted
3	$6e^{2x} = 6e^{2x}$ for all $x$	the two sides are the same function

Nothing else is required. The verification is finished the moment line 3 is written, and every additional step risks introducing an error into work that was already complete.

Implicit Candidates

When the candidate arrives as a relation rather than a function, differentiate both sides with respect to the independent variable and keep $\frac{dy}{dx}$ as a symbol. For $y^{2} = x^{3} + 7$ :

$2y\frac{dy}{dx} = 3x^{2} \qquad\Longrightarrow\qquad \frac{dy}{dx} = \frac{3x^{2}}{2y}$

Families and Parameters

A one-parameter family can satisfy an equation for every value of the parameter, and showing that is often the whole task. For $y = Ce^{2x}$ :

$y' = 2Ce^{2x} = 2\left(Ce^{2x}\right) = 2y$

Where the Check Fails: One Point Versus an Interval

Common Misconception Check

Myth 1: “If substituting gives a true statement, the work is done.” Truth: It is done only if the statement is true for every $x$ in an interval. One value proves nothing.
Myth 2: “I can check the answer by integrating the equation instead.” Truth: Integration is a different task with a different failure mode. Verification never requires an antiderivative.
Myth 3: “A candidate that fails must be algebraically wrong.” Truth: Most failures are differentiation errors in the candidate, not errors in the equation. Recompute the derivative before concluding anything.

Topic-Specific Study Method

Differentiate First, Then Substitute

Exam disguise	What you see on the page	What your brain must translate	Required execution
Explicit candidate	$y = 3e^{2x}$ , equation given	differentiate, then substitute	three lines, no integration
Implicit candidate	$y^{2} = x^{3} + 7$	differentiate both sides, keep $y'$	solve for $y'$ , note $y \neq 0$
Parameterised family	$y = Ce^{2x}$ for all $C$	keep $C$ symbolic to the end	the parameter must cancel
“Show that” phrasing	a claim, not a question	the answer is the verification itself	write the comparison explicitly
Distractor built on one point	options that agree at a single $x$	test a second value	reject any option that only matches once

① Differentiate on its own line. Combining the derivative with the substitution in one line is where chain-rule factors disappear.

② Simplify each side independently before comparing. Comparing a partly simplified left side with a partly simplified right side invites a false mismatch.

③ End with a sentence, not a symbol. “The two sides agree for all $x$ , so the function is a solution” is the sentence that carries the point.

Training & Error Diagnosis

Core Archetype (Simplest Form): Show that $y = x^{2}e^{3x}$ is a solution of $\frac{dy}{dx} = 3y + 2xe^{3x}$ .

Standard Execution:

Differentiate. By the product rule, $y' = 2xe^{3x} + 3x^{2}e^{3x}$ .

Substitute. The right side is $3\left(x^{2}e^{3x}\right) + 2xe^{3x}$ .

Compare. Both equal $3x^{2}e^{3x} + 2xe^{3x}$ , so the candidate is a solution for all real $x$ .

State. The identity holds on $(-\infty, \infty)$ .

Exam Variation (Context / Parameter Shift): The relation $y^{2} = 2x + C$ is claimed to satisfy $\frac{dy}{dx} = \frac{1}{y}$ . Verify the claim and state where it holds.

Solution: Differentiating both sides with respect to $x$ gives $2y\frac{dy}{dx} = 2$ , so $\frac{dy}{dx} = \frac{1}{y}$ . The algebra is valid for every $C$ , but the division requires $y \neq 0$ , and $y = 0$ occurs exactly when $2x + C = 0$ , that is at $x = -\frac{C}{2}$ . So the relation satisfies the equation on the two intervals cut off by that point, and a complete answer names the exclusion rather than claiming the whole real line.

Common Traps & Failure Signals:

Chain rule dropped (Signal: $y' = 3e^{2x}$ for $y = 3e^{2x}$ ): the inner rate of $2$ must appear.
Single-point check (Signal: one value of $x$ substituted, then a conclusion): the claim is about an interval.
Comparison never written (Signal: substitution with no equality line): the reader cannot award the point.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Chain Rule Dropped: You differentiated $3e^{2x}$ as $3e^{2x}$ instead of $6e^{2x}$ , which is the single most common way an otherwise correct verification collapses into a false negative on the exam.
Point Substituted: One convenient value is not an interval.
One Side Only: You never computed the right side, so nothing was compared.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; No released free-response part since 2017 scores verification as a step of its own; the skill appears in the multiple-choice section and as an unstated premise inside the free-response parts that follow it
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; Every free-response set that mentions a differential equation states the equation as given rather than asking the student to confirm a candidate
FRQ Years Appeared: None in Released FRQs ---

Practice Examples

MCQ 1

Test Each Option

Which of these functions satisfies the differential equation $\frac{dy}{dx} = 2xy$ ?

MCQ 2

An Implicit Candidate

The relation $y^{2} = x^{3} + 7$ defines $y$ implicitly. Which differential equation does it satisfy?

MCQ 3

Reading the Parameter Off the Equation

For which value of the constant $k$ is $y = e^{kt}$ a solution of $\frac{dy}{dt} = -5y$ ?

FRQ 4

Verification With and Without an Explicit Formula

(a) Show that $y = 4e^{-3x} + 2$ is a solution of $\frac{dy}{dx} = -3(y - 2)$ for all real $x$ .

(b) Show that every relation of the form $y^{2} = 2x + C$ satisfies $\frac{dy}{dx} = \frac{1}{y}$ , and state the intervals on which the verification is valid.

(c) A student checks $y = 2x + 1$ against $\frac{dy}{dx} = 2x$ at $x = 1$ , obtains $2 = 2$ , and concludes that the function is a solution. Explain what is wrong with that conclusion.

Cheatsheet

Core Reference

	Action	Failure if skipped
1	differentiate the candidate on its own line	a chain factor goes missing
2	substitute the candidate into the non-derivative side	the comparison has nothing to compare
3	simplify each side separately	a false mismatch from partial algebra
4	state equality for all $x$ in an interval	the point is not awarded
5	name any excluded value (implicit case)	half the domain claim is lost

Sentences That Score

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Unit 7Checking if a function satisfies a differential equation

Concept 4.119Built for May 2027 AP Exam

Checking if a function satisfies a differential equation

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

AP Calculus AB tests this skill through the representations and reasoning described below.

What you’ll see: Rarely as a free-response part of its own. It appears as a multiple-choice item — “which of these functions satisfies the equation” or “show that $y = \ldots$ is a solution” — and inside the first line of a longer free-response part, where writing down the verification earns a point before any solving begins.
What you’ll do: Differentiate the candidate, substitute it into the side of the equation that does not already contain a derivative, simplify each side on its own line, and state that the two agree.
Memory hook: Differentiate, substitute, compare. Three verbs, three lines, no integration anywhere.
Use it on the exam: For MCQs: Do not try to solve the differential equation. Differentiate each offered function and substitute; the one that produces a true statement is the answer, and the check costs about twenty seconds per option. For FRQs: Write the two sides on separate lines before claiming they match. A reader cannot award the substitution point to work in which the two expressions were never visibly compared.

Concept Review

What “satisfies” Actually Means

Figure 1 — Verification is a comparison of two computed expressions. The candidate goes in, two sides come out, and the claim is that they are the same function.

$\text{for all } x \in I: \qquad f'(x) = F\left(x, f(x)\right)$

The phrase that does the work is for all $x$ in $I$ . An equation that holds at one convenient value has not been verified, and the exam writes distractors that exploit exactly that gap.

Verification is a comparison of two computed expressions. The candidate goes in, two sides come out, and the claim is that they are the same function.

The Three-Line Check

Line	What you write	What it proves
1	$y = 3e^{2x} \Rightarrow y' = 6e^{2x}$	the derivative of the candidate, chain rule included
2	right side $= 2y = 2\left(3e^{2x}\right) = 6e^{2x}$	the equation's own expression, with the candidate substituted
3	$6e^{2x} = 6e^{2x}$ for all $x$	the two sides are the same function

Nothing else is required. The verification is finished the moment line 3 is written, and every additional step risks introducing an error into work that was already complete.

Implicit Candidates

When the candidate arrives as a relation rather than a function, differentiate both sides with respect to the independent variable and keep $\frac{dy}{dx}$ as a symbol. For $y^{2} = x^{3} + 7$ :

$2y\frac{dy}{dx} = 3x^{2} \qquad\Longrightarrow\qquad \frac{dy}{dx} = \frac{3x^{2}}{2y}$

Families and Parameters

A one-parameter family can satisfy an equation for every value of the parameter, and showing that is often the whole task. For $y = Ce^{2x}$ :

$y' = 2Ce^{2x} = 2\left(Ce^{2x}\right) = 2y$

Where the Check Fails: One Point Versus an Interval

Common Misconception Check

Myth 1: “If substituting gives a true statement, the work is done.” Truth: It is done only if the statement is true for every $x$ in an interval. One value proves nothing.
Myth 2: “I can check the answer by integrating the equation instead.” Truth: Integration is a different task with a different failure mode. Verification never requires an antiderivative.
Myth 3: “A candidate that fails must be algebraically wrong.” Truth: Most failures are differentiation errors in the candidate, not errors in the equation. Recompute the derivative before concluding anything.

Topic-Specific Study Method

Differentiate First, Then Substitute

Exam disguise	What you see on the page	What your brain must translate	Required execution
Explicit candidate	$y = 3e^{2x}$ , equation given	differentiate, then substitute	three lines, no integration
Implicit candidate	$y^{2} = x^{3} + 7$	differentiate both sides, keep $y'$	solve for $y'$ , note $y \neq 0$
Parameterised family	$y = Ce^{2x}$ for all $C$	keep $C$ symbolic to the end	the parameter must cancel
“Show that” phrasing	a claim, not a question	the answer is the verification itself	write the comparison explicitly
Distractor built on one point	options that agree at a single $x$	test a second value	reject any option that only matches once

① Differentiate on its own line. Combining the derivative with the substitution in one line is where chain-rule factors disappear.

② Simplify each side independently before comparing. Comparing a partly simplified left side with a partly simplified right side invites a false mismatch.

③ End with a sentence, not a symbol. “The two sides agree for all $x$ , so the function is a solution” is the sentence that carries the point.

Training & Error Diagnosis

Core Archetype (Simplest Form): Show that $y = x^{2}e^{3x}$ is a solution of $\frac{dy}{dx} = 3y + 2xe^{3x}$ .

Standard Execution:

Differentiate. By the product rule, $y' = 2xe^{3x} + 3x^{2}e^{3x}$ .

Substitute. The right side is $3\left(x^{2}e^{3x}\right) + 2xe^{3x}$ .

Compare. Both equal $3x^{2}e^{3x} + 2xe^{3x}$ , so the candidate is a solution for all real $x$ .

State. The identity holds on $(-\infty, \infty)$ .

Exam Variation (Context / Parameter Shift): The relation $y^{2} = 2x + C$ is claimed to satisfy $\frac{dy}{dx} = \frac{1}{y}$ . Verify the claim and state where it holds.

Solution: Differentiating both sides with respect to $x$ gives $2y\frac{dy}{dx} = 2$ , so $\frac{dy}{dx} = \frac{1}{y}$ . The algebra is valid for every $C$ , but the division requires $y \neq 0$ , and $y = 0$ occurs exactly when $2x + C = 0$ , that is at $x = -\frac{C}{2}$ . So the relation satisfies the equation on the two intervals cut off by that point, and a complete answer names the exclusion rather than claiming the whole real line.

Common Traps & Failure Signals:

Chain rule dropped (Signal: $y' = 3e^{2x}$ for $y = 3e^{2x}$ ): the inner rate of $2$ must appear.
Single-point check (Signal: one value of $x$ substituted, then a conclusion): the claim is about an interval.
Comparison never written (Signal: substitution with no equality line): the reader cannot award the point.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Chain Rule Dropped: You differentiated $3e^{2x}$ as $3e^{2x}$ instead of $6e^{2x}$ , which is the single most common way an otherwise correct verification collapses into a false negative on the exam.
Point Substituted: One convenient value is not an interval.
One Side Only: You never computed the right side, so nothing was compared.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; No released free-response part since 2017 scores verification as a step of its own; the skill appears in the multiple-choice section and as an unstated premise inside the free-response parts that follow it
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; Every free-response set that mentions a differential equation states the equation as given rather than asking the student to confirm a candidate
FRQ Years Appeared: None in Released FRQs ---

Practice Examples

MCQ 1

Test Each Option

Which of these functions satisfies the differential equation $\frac{dy}{dx} = 2xy$ ?

MCQ 2

An Implicit Candidate

The relation $y^{2} = x^{3} + 7$ defines $y$ implicitly. Which differential equation does it satisfy?

MCQ 3

Reading the Parameter Off the Equation

For which value of the constant $k$ is $y = e^{kt}$ a solution of $\frac{dy}{dt} = -5y$ ?

FRQ 4

Verification With and Without an Explicit Formula

(a) Show that $y = 4e^{-3x} + 2$ is a solution of $\frac{dy}{dx} = -3(y - 2)$ for all real $x$ .

(b) Show that every relation of the form $y^{2} = 2x + C$ satisfies $\frac{dy}{dx} = \frac{1}{y}$ , and state the intervals on which the verification is valid.

(c) A student checks $y = 2x + 1$ against $\frac{dy}{dx} = 2x$ at $x = 1$ , obtains $2 = 2$ , and concludes that the function is a solution. Explain what is wrong with that conclusion.

Cheatsheet

Core Reference

	Action	Failure if skipped
1	differentiate the candidate on its own line	a chain factor goes missing
2	substitute the candidate into the non-derivative side	the comparison has nothing to compare
3	simplify each side separately	a false mismatch from partial algebra
4	state equality for all $x$ in an interval	the point is not awarded
5	name any excluded value (implicit case)	half the domain claim is lost

Sentences That Score

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