Unit 7Particular Solutions

Concept 4.127Built for May 2027 AP Exam

Particular Solutions

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

Separation gives a family; an initial condition gives one member. The condition is a point, and using it means substituting both coordinates into the general solution and solving for the constant, not integrating again. After the constant is known, the answer is a specific function, and two questions remain: should it be written explicitly, and on what interval is it valid? An explicit form requires choosing a branch, and the correct branch is the one that passes through the given point. The interval excludes any value where the original equation or the algebra breaks down, which usually means where a denominator vanishes. On a scored part, the branch choice and the domain each carry their own point, and both are easy to forget.

On the AP Exam

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

What you’ll see: The closing part of a differential-equation question, phrased as “find the particular solution … with initial condition,” or as a request for the value of the solution at a later time. This page and Unit 7.3 share their evidence, because the exam never asks for a separated equation without also supplying the condition that pins the constant.
What you’ll do: Substitute the point into the general solution, solve for the constant, write the function explicitly if asked, and state the interval on which it holds.
Memory hook: The point kills every curve but one. One substitution, one constant, one member of the family.
Use it on the exam: For MCQs: Substitute $t = 0$ first. The constant is then usually equal to the initial value, and the options collapse immediately. For FRQs: Write the general solution, then a line beginning “Since $y(0) = \ldots$ ,” then the constant, then the particular solution. Four lines, each one a scored step.

Concept Review

The Family and the Selection

Figure 1 — Every pale curve solves the same differential equation. The condition $y(1) = 3$ selects exactly one of them.

The general solution of $\frac{dy}{dx} = 2x$ is $y = x^{2} + C$ , and each value of $C$ produces a different curve. Substitute $x = 1$ and $y = 3$ :

$3 = 1 + C \qquad\Longrightarrow\qquad C = 2$

So the particular solution is $y = x^{2} + 2$ . The geometry is worth holding onto: the condition is a point, and exactly one curve of the family passes through a given point, because the differential equation assigns exactly one slope there.

Every pale curve solves the same differential equation. The condition $y(1) = 3$ selects exactly one of them.

What the Condition Is Made Of

An initial condition gives a pair of numbers, and both matter. Writing only the $y$ -value leaves the constant undetermined, and writing only the $x$ -value determines nothing at all. The exam occasionally offers an answer that used $C = y_0$ without substituting $x_0$ , which is correct only when the condition is stated at $x = 0$ .

Condition	Substitution	Constant
$y = Ce^{2x} + 1$ , with $y(0) = 5$	$5 = C + 1$	$C = 4$
$y = Ce^{2x} + 1$ , with $y(1) = 5$	$5 = Ce^{2} + 1$	$C = \frac{4}{e^{2}}$
$y = x^{2} + C$ , with $y(1) = 3$	$3 = 1 + C$	$C = 2$
$y^{2} = x^{2} + C$ , with $y(0) = -2$	$4 = C$	$C = 4$ , with the negative branch

The second row is the one that separates careful work from fast work: writing $C = 4$ there is the standard error, and it comes from assuming every condition is stated at the origin.

Choosing the Branch

When the general solution is implicit and quadratic in $y$ , solving for $y$ introduces a square root and therefore two branches. The condition decides. For $y^{2} = x^{2} + 4$ with $y(0) = -2$ , the correct branch is $y = -\sqrt{x^{2}+4}$ , because only that branch gives a negative value at $x = 0$ . A solution that supplies the wrong branch satisfies the differential equation but not the condition, and it usually loses a point even when the algebra is perfect.

Where the Solution Lives

The interval of validity is the largest interval containing $x_0$ on which the solution and its derivative are defined. Two things shorten it: a denominator in the original equation, and a logarithm or square root introduced by the algebra. For $\frac{dy}{dx} = \frac{1}{y}$ with $y(0) = 2$ , the solution $y = \sqrt{2x + 4}$ exists only for $x > -2$ , since the square root requires $2x + 4 \geq 0$ and the original equation requires $y \neq 0$ .

When No Solution Exists

If the point sits where the equation is undefined, no solution passes through it. The equation $\frac{dy}{dx} = \frac{1}{y}$ assigns no slope at $y = 0$ , so a condition like $y(0) = 0$ has no solution at all, even though the algebra will happily produce one. Checking that the point is in the domain of the right-hand side takes one line and prevents an answer that cannot exist.

Common Misconception Check

Myth 1: “The constant equals the initial value.” Truth: It does only when the condition is given at $x = 0$ . Otherwise substitute both coordinates.
Myth 2: “Either branch is acceptable after taking a square root.” Truth: The condition selects the branch. The other one solves the equation but not the problem.
Myth 3: “The solution is valid for all real $x$ unless stated otherwise.” Truth: Its interval is limited by every denominator, logarithm, and square root in the work.

Topic-Specific Study Method

One Substitution, One Constant

Borrowed idea: The general solution is a family, and the initial condition chooses one member. What does that mean in practice? You evaluate the family at the given point and solve for the constant. Nothing else changes. Does the answer need to be a function of $x$ ? If the question asks for $y = f(x)$ , solve for $y$ and pick the branch that passes through the point; otherwise leave the relation implicit.

Exam disguise	What you see on the page	What your brain must translate	Required execution
“Find the particular solution”	general solution plus a point	substitute both coordinates	solve for $C$ , then rewrite
Condition at $x = 0$	$y(0) = y_0$	the constant equals $y_0$	one line is enough
Condition away from the origin	$y(1) = 3$	the constant is not $y_0$	substitute $x_0$ as well
Square root appears	$y^{2} = x^{2} + C$	two branches exist	choose by the sign at $x_0$
“On what interval”	an extra part, easy to skip	the domain of the solution	name the excluded endpoint

① Substitute before simplifying. Plugging the point into an already rearranged form is where the wrong branch gets chosen.

② Write the constant as a fraction if it is not an integer. A decimal hides the structure and makes the later derivative harder to check.

③ State the interval in the same sentence as the solution. Splitting them across two places is how the domain point gets lost.

Training & Error Diagnosis

Core Archetype (Simplest Form): Solve $\frac{dy}{dx} = 2x$ with $y(1) = 3$ .

Standard Execution:

General solution. $y = x^{2} + C$ .

Substitute. $3 = 1 + C$ , so $C = 2$ .

Particular solution. $y = x^{2} + 2$ , defined for all real $x$ .

Check. $y' = 2x$ and $y(1) = 3$ , both as required.

Exam Variation (Context / Parameter Shift): Solve $\frac{dy}{dx} = 2x\left(y - 1\right)$ with $y(0) = 3$ .

Solution: Separating gives $\frac{dy}{y-1} = 2x\,dx$ , so $\ln|y-1| = x^{2} + C$ and $y - 1 = Ae^{x^{2}}$ . The condition $y(0) = 3$ gives $3 - 1 = A$ , so $A = 2$ and $y = 1 + 2e^{x^{2}}$ . Since $y(0) = 3 > 1$ , the solution stays above the equilibrium $y = 1$ , which is consistent with $A > 0$ , and the solution is defined for all real $x$ because the exponential never vanishes.

Common Traps & Failure Signals:

Constant set equal to the value (Signal: $C = 5$ from $y(1) = 5$ ): the $x$ -coordinate was never substituted.
Wrong branch (Signal: a positive square root for a negative initial value): the sign at $x_0$ decides.
No interval (Signal: “for all $x$ ” with a denominator in the equation): the domain was never checked.

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

Coordinate Skipped: You substituted the $y$ -value only.
Branch Guessed: You took the positive square root by habit, and the resulting curve solves the equation while missing the point you were given.
Domain Ignored: You reported a solution on the whole real line when a denominator or a square root had already removed part of it, which is the one error that turns a correct formula into an incorrect answer.

Past 10-Year AP Exam Appearance

Testing Frequency: Nearly Every Year; Eight of the nine released exams since 2017 contain a part that supplies an initial condition and asks for the resulting function, and every one of those parts scores the constant separately from the integration
10-Year Total Count: 8 of 9 Exams · 8 FRQ Parts; These are the same eight parts counted on the separation-of-variables page, read from the other end: the exam never asks for one without the other
FRQ Years Appeared: 2017 (Q4c), 2018 (Q6c), 2019 (Q4c), 2021 (Q6c), 2022 (Q5d), 2023 (Q3d), 2024 (Q3c), 2026 (Q3D) ---

Practice Examples

MCQ 1

One Condition, One Curve

The particular solution of $\frac{dy}{dx} = 2x$ with $y(1) = 3$ is

MCQ 2

A Condition Away From the Origin

The general solution of a differential equation is $y = Ce^{2x} + 1$ . If $y(0) = 5$ , then $C$ equals

MCQ 3

Which Branch Survives

The general solution of $\frac{dy}{dx} = \frac{x}{y}$ is $y^{2} = x^{2} + C$ . If $y(0) = -2$ , the particular solution is

FRQ 4

Constant, Branch, Interval

(a) Solve $\frac{dy}{dx} = 2x\left(y - 1\right)$ with $y(0) = 3$ , giving $y$ explicitly.

(b) Solve $\frac{dy}{dx} = \frac{x}{y}$ with $y(0) = -2$ , giving $y$ explicitly and stating the interval on which the solution is valid.

(c) A student applies the condition $y(0) = 0$ to $\frac{dy}{dx} = \frac{1}{y}$ and obtains $y^{2} = 2x$ , then reports $y = \sqrt{2x}$ . Explain why no solution exists for this condition.

Cheatsheet

Core Reference

	Action	Failure if skipped
1	write the general solution with $C$	the constant has nowhere to live
2	substitute both coordinates of the point	$C$ is set equal to $y_0$
3	solve for $C$ exactly	a decimal hides the structure
4	choose the branch that fits the sign	the equation is satisfied, the point is not
5	state the interval of validity	the domain point is lost

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Unit 7Particular Solutions

Concept 4.127Built for May 2027 AP Exam

Particular Solutions

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: The closing part of a differential-equation question, phrased as “find the particular solution … with initial condition,” or as a request for the value of the solution at a later time. This page and Unit 7.3 share their evidence, because the exam never asks for a separated equation without also supplying the condition that pins the constant.
What you’ll do: Substitute the point into the general solution, solve for the constant, write the function explicitly if asked, and state the interval on which it holds.
Memory hook: The point kills every curve but one. One substitution, one constant, one member of the family.
Use it on the exam: For MCQs: Substitute $t = 0$ first. The constant is then usually equal to the initial value, and the options collapse immediately. For FRQs: Write the general solution, then a line beginning “Since $y(0) = \ldots$ ,” then the constant, then the particular solution. Four lines, each one a scored step.

Concept Review

The Family and the Selection

Figure 1 — Every pale curve solves the same differential equation. The condition $y(1) = 3$ selects exactly one of them.

The general solution of $\frac{dy}{dx} = 2x$ is $y = x^{2} + C$ , and each value of $C$ produces a different curve. Substitute $x = 1$ and $y = 3$ :

$3 = 1 + C \qquad\Longrightarrow\qquad C = 2$

Every pale curve solves the same differential equation. The condition $y(1) = 3$ selects exactly one of them.

What the Condition Is Made Of

Condition	Substitution	Constant
$y = Ce^{2x} + 1$ , with $y(0) = 5$	$5 = C + 1$	$C = 4$
$y = Ce^{2x} + 1$ , with $y(1) = 5$	$5 = Ce^{2} + 1$	$C = \frac{4}{e^{2}}$
$y = x^{2} + C$ , with $y(1) = 3$	$3 = 1 + C$	$C = 2$
$y^{2} = x^{2} + C$ , with $y(0) = -2$	$4 = C$	$C = 4$ , with the negative branch

The second row is the one that separates careful work from fast work: writing $C = 4$ there is the standard error, and it comes from assuming every condition is stated at the origin.

Choosing the Branch

Where the Solution Lives

When No Solution Exists

Common Misconception Check

Myth 1: “The constant equals the initial value.” Truth: It does only when the condition is given at $x = 0$ . Otherwise substitute both coordinates.
Myth 2: “Either branch is acceptable after taking a square root.” Truth: The condition selects the branch. The other one solves the equation but not the problem.
Myth 3: “The solution is valid for all real $x$ unless stated otherwise.” Truth: Its interval is limited by every denominator, logarithm, and square root in the work.

Topic-Specific Study Method

One Substitution, One Constant

Exam disguise	What you see on the page	What your brain must translate	Required execution
“Find the particular solution”	general solution plus a point	substitute both coordinates	solve for $C$ , then rewrite
Condition at $x = 0$	$y(0) = y_0$	the constant equals $y_0$	one line is enough
Condition away from the origin	$y(1) = 3$	the constant is not $y_0$	substitute $x_0$ as well
Square root appears	$y^{2} = x^{2} + C$	two branches exist	choose by the sign at $x_0$
“On what interval”	an extra part, easy to skip	the domain of the solution	name the excluded endpoint

① Substitute before simplifying. Plugging the point into an already rearranged form is where the wrong branch gets chosen.

② Write the constant as a fraction if it is not an integer. A decimal hides the structure and makes the later derivative harder to check.

③ State the interval in the same sentence as the solution. Splitting them across two places is how the domain point gets lost.

Training & Error Diagnosis

Core Archetype (Simplest Form): Solve $\frac{dy}{dx} = 2x$ with $y(1) = 3$ .

Standard Execution:

General solution. $y = x^{2} + C$ .

Substitute. $3 = 1 + C$ , so $C = 2$ .

Particular solution. $y = x^{2} + 2$ , defined for all real $x$ .

Check. $y' = 2x$ and $y(1) = 3$ , both as required.

Exam Variation (Context / Parameter Shift): Solve $\frac{dy}{dx} = 2x\left(y - 1\right)$ with $y(0) = 3$ .

Solution: Separating gives $\frac{dy}{y-1} = 2x\,dx$ , so $\ln|y-1| = x^{2} + C$ and $y - 1 = Ae^{x^{2}}$ . The condition $y(0) = 3$ gives $3 - 1 = A$ , so $A = 2$ and $y = 1 + 2e^{x^{2}}$ . Since $y(0) = 3 > 1$ , the solution stays above the equilibrium $y = 1$ , which is consistent with $A > 0$ , and the solution is defined for all real $x$ because the exponential never vanishes.

Common Traps & Failure Signals:

Constant set equal to the value (Signal: $C = 5$ from $y(1) = 5$ ): the $x$ -coordinate was never substituted.
Wrong branch (Signal: a positive square root for a negative initial value): the sign at $x_0$ decides.
No interval (Signal: “for all $x$ ” with a denominator in the equation): the domain was never checked.

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

Coordinate Skipped: You substituted the $y$ -value only.
Branch Guessed: You took the positive square root by habit, and the resulting curve solves the equation while missing the point you were given.
Domain Ignored: You reported a solution on the whole real line when a denominator or a square root had already removed part of it, which is the one error that turns a correct formula into an incorrect answer.

Past 10-Year AP Exam Appearance

Testing Frequency: Nearly Every Year; Eight of the nine released exams since 2017 contain a part that supplies an initial condition and asks for the resulting function, and every one of those parts scores the constant separately from the integration
10-Year Total Count: 8 of 9 Exams · 8 FRQ Parts; These are the same eight parts counted on the separation-of-variables page, read from the other end: the exam never asks for one without the other
FRQ Years Appeared: 2017 (Q4c), 2018 (Q6c), 2019 (Q4c), 2021 (Q6c), 2022 (Q5d), 2023 (Q3d), 2024 (Q3c), 2026 (Q3D) ---

Practice Examples

MCQ 1

One Condition, One Curve

The particular solution of $\frac{dy}{dx} = 2x$ with $y(1) = 3$ is

MCQ 2

A Condition Away From the Origin

The general solution of a differential equation is $y = Ce^{2x} + 1$ . If $y(0) = 5$ , then $C$ equals

MCQ 3

Which Branch Survives

The general solution of $\frac{dy}{dx} = \frac{x}{y}$ is $y^{2} = x^{2} + C$ . If $y(0) = -2$ , the particular solution is

FRQ 4

Constant, Branch, Interval

(a) Solve $\frac{dy}{dx} = 2x\left(y - 1\right)$ with $y(0) = 3$ , giving $y$ explicitly.

(b) Solve $\frac{dy}{dx} = \frac{x}{y}$ with $y(0) = -2$ , giving $y$ explicitly and stating the interval on which the solution is valid.

(c) A student applies the condition $y(0) = 0$ to $\frac{dy}{dx} = \frac{1}{y}$ and obtains $y^{2} = 2x$ , then reports $y = \sqrt{2x}$ . Explain why no solution exists for this condition.

Cheatsheet

Core Reference

	Action	Failure if skipped
1	write the general solution with $C$	the constant has nowhere to live
2	substitute both coordinates of the point	$C$ is set equal to $y_0$
3	solve for $C$ exactly	a decimal hides the structure
4	choose the branch that fits the sign	the equation is satisfied, the point is not
5	state the interval of validity	the domain point is lost

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Unit 7Particular Solutions

Concept 4.127Built for May 2027 AP Exam

Particular Solutions

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: The closing part of a differential-equation question, phrased as “find the particular solution … with initial condition,” or as a request for the value of the solution at a later time. This page and Unit 7.3 share their evidence, because the exam never asks for a separated equation without also supplying the condition that pins the constant.
What you’ll do: Substitute the point into the general solution, solve for the constant, write the function explicitly if asked, and state the interval on which it holds.
Memory hook: The point kills every curve but one. One substitution, one constant, one member of the family.
Use it on the exam: For MCQs: Substitute $t = 0$ first. The constant is then usually equal to the initial value, and the options collapse immediately. For FRQs: Write the general solution, then a line beginning “Since $y(0) = \ldots$ ,” then the constant, then the particular solution. Four lines, each one a scored step.

Concept Review

The Family and the Selection

Figure 1 — Every pale curve solves the same differential equation. The condition $y(1) = 3$ selects exactly one of them.

The general solution of $\frac{dy}{dx} = 2x$ is $y = x^{2} + C$ , and each value of $C$ produces a different curve. Substitute $x = 1$ and $y = 3$ :

$3 = 1 + C \qquad\Longrightarrow\qquad C = 2$

Every pale curve solves the same differential equation. The condition $y(1) = 3$ selects exactly one of them.

What the Condition Is Made Of

Condition	Substitution	Constant
$y = Ce^{2x} + 1$ , with $y(0) = 5$	$5 = C + 1$	$C = 4$
$y = Ce^{2x} + 1$ , with $y(1) = 5$	$5 = Ce^{2} + 1$	$C = \frac{4}{e^{2}}$
$y = x^{2} + C$ , with $y(1) = 3$	$3 = 1 + C$	$C = 2$
$y^{2} = x^{2} + C$ , with $y(0) = -2$	$4 = C$	$C = 4$ , with the negative branch

The second row is the one that separates careful work from fast work: writing $C = 4$ there is the standard error, and it comes from assuming every condition is stated at the origin.

Choosing the Branch

Where the Solution Lives

When No Solution Exists

Common Misconception Check

Myth 1: “The constant equals the initial value.” Truth: It does only when the condition is given at $x = 0$ . Otherwise substitute both coordinates.
Myth 2: “Either branch is acceptable after taking a square root.” Truth: The condition selects the branch. The other one solves the equation but not the problem.
Myth 3: “The solution is valid for all real $x$ unless stated otherwise.” Truth: Its interval is limited by every denominator, logarithm, and square root in the work.

Topic-Specific Study Method

One Substitution, One Constant

Exam disguise	What you see on the page	What your brain must translate	Required execution
“Find the particular solution”	general solution plus a point	substitute both coordinates	solve for $C$ , then rewrite
Condition at $x = 0$	$y(0) = y_0$	the constant equals $y_0$	one line is enough
Condition away from the origin	$y(1) = 3$	the constant is not $y_0$	substitute $x_0$ as well
Square root appears	$y^{2} = x^{2} + C$	two branches exist	choose by the sign at $x_0$
“On what interval”	an extra part, easy to skip	the domain of the solution	name the excluded endpoint

① Substitute before simplifying. Plugging the point into an already rearranged form is where the wrong branch gets chosen.

② Write the constant as a fraction if it is not an integer. A decimal hides the structure and makes the later derivative harder to check.

③ State the interval in the same sentence as the solution. Splitting them across two places is how the domain point gets lost.

Training & Error Diagnosis

Core Archetype (Simplest Form): Solve $\frac{dy}{dx} = 2x$ with $y(1) = 3$ .

Standard Execution:

General solution. $y = x^{2} + C$ .

Substitute. $3 = 1 + C$ , so $C = 2$ .

Particular solution. $y = x^{2} + 2$ , defined for all real $x$ .

Check. $y' = 2x$ and $y(1) = 3$ , both as required.

Exam Variation (Context / Parameter Shift): Solve $\frac{dy}{dx} = 2x\left(y - 1\right)$ with $y(0) = 3$ .

Solution: Separating gives $\frac{dy}{y-1} = 2x\,dx$ , so $\ln|y-1| = x^{2} + C$ and $y - 1 = Ae^{x^{2}}$ . The condition $y(0) = 3$ gives $3 - 1 = A$ , so $A = 2$ and $y = 1 + 2e^{x^{2}}$ . Since $y(0) = 3 > 1$ , the solution stays above the equilibrium $y = 1$ , which is consistent with $A > 0$ , and the solution is defined for all real $x$ because the exponential never vanishes.

Common Traps & Failure Signals:

Constant set equal to the value (Signal: $C = 5$ from $y(1) = 5$ ): the $x$ -coordinate was never substituted.
Wrong branch (Signal: a positive square root for a negative initial value): the sign at $x_0$ decides.
No interval (Signal: “for all $x$ ” with a denominator in the equation): the domain was never checked.

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

Coordinate Skipped: You substituted the $y$ -value only.
Branch Guessed: You took the positive square root by habit, and the resulting curve solves the equation while missing the point you were given.
Domain Ignored: You reported a solution on the whole real line when a denominator or a square root had already removed part of it, which is the one error that turns a correct formula into an incorrect answer.

Past 10-Year AP Exam Appearance

Testing Frequency: Nearly Every Year; Eight of the nine released exams since 2017 contain a part that supplies an initial condition and asks for the resulting function, and every one of those parts scores the constant separately from the integration
10-Year Total Count: 8 of 9 Exams · 8 FRQ Parts; These are the same eight parts counted on the separation-of-variables page, read from the other end: the exam never asks for one without the other
FRQ Years Appeared: 2017 (Q4c), 2018 (Q6c), 2019 (Q4c), 2021 (Q6c), 2022 (Q5d), 2023 (Q3d), 2024 (Q3c), 2026 (Q3D) ---

Practice Examples

MCQ 1

One Condition, One Curve

The particular solution of $\frac{dy}{dx} = 2x$ with $y(1) = 3$ is

MCQ 2

A Condition Away From the Origin

The general solution of a differential equation is $y = Ce^{2x} + 1$ . If $y(0) = 5$ , then $C$ equals

MCQ 3

Which Branch Survives

The general solution of $\frac{dy}{dx} = \frac{x}{y}$ is $y^{2} = x^{2} + C$ . If $y(0) = -2$ , the particular solution is

FRQ 4

Constant, Branch, Interval

(a) Solve $\frac{dy}{dx} = 2x\left(y - 1\right)$ with $y(0) = 3$ , giving $y$ explicitly.

(b) Solve $\frac{dy}{dx} = \frac{x}{y}$ with $y(0) = -2$ , giving $y$ explicitly and stating the interval on which the solution is valid.

(c) A student applies the condition $y(0) = 0$ to $\frac{dy}{dx} = \frac{1}{y}$ and obtains $y^{2} = 2x$ , then reports $y = \sqrt{2x}$ . Explain why no solution exists for this condition.

Cheatsheet

Core Reference

	Action	Failure if skipped
1	write the general solution with $C$	the constant has nowhere to live
2	substitute both coordinates of the point	$C$ is set equal to $y_0$
3	solve for $C$ exactly	a decimal hides the structure
4	choose the branch that fits the sign	the equation is satisfied, the point is not
5	state the interval of validity	the domain point is lost

Sentences That Score

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