Unit 7Separation of Variables

Concept 4.78Built for May 2027 AP Exam

Separation of Variables

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

Separation of variables turns one differential equation into two ordinary integrals. You move every $y$ to the left with $dy$ , every $x$ to the right with $dx$ , and integrate each side against its own variable. The result is an equation relating $y$ and $x$ with one constant of integration. Rearrange it into an explicit formula only if the question asks for one, because many correct answers stay implicit. Two details decide the score. The constant is written once, on one side, since putting it on both sides merely renames it; and the integral of $\frac{1}{y-a}$ carries an absolute value that must survive until the constant absorbs it. After that, an initial condition selects the single member of the family.

On the AP Exam

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

What you’ll see: The middle two parts of a differential-equation question. A slope-field sketch usually comes first, then a part that says “use separation of variables to find an expression for $M(t)$ ,” and occasionally a follow-up asking for a limit or a value at a later time.
What you’ll do: Move the variables, integrate both sides, evaluate any constant from the given condition, and then either solve for the function or leave it in a form the question accepts.
Memory hook: $y$ goes with $dy$ , $x$ goes with $dx$ . If a variable is on the wrong side after the split, the equation was not separated.
Use it on the exam: For MCQs: Differentiate the offered answers instead of integrating. Only the correct family differentiates back to the given $\frac{dy}{dx}$ for every $x$ . For FRQs: Write the separated line before integrating. The split is often a scored step, so an answer that jumps straight to the solution can lose a point even when the solution is right.

Concept Review

Why Separating Works

Figure 1 — Four lines, no cleverness. The variables are split, each side is integrated against its own variable, and one constant is carried along.

The method rests on the chain rule read backwards. If $y$ is a function of $x$ , then $\frac{dy}{dx}$ is the derivative of $y$ with respect to $x$ , and the expression $\frac{1}{y-a}\frac{dy}{dx}$ is the derivative of $\ln|y-a|$ . So integrating that expression with respect to $x$ is the same as integrating $\frac{1}{y-a}$ with respect to $y$ . Writing $dy$ and $dx$ on opposite sides is shorthand for that substitution, not a licence to treat derivatives as fractions.

Four lines, no cleverness. The variables are split, each side is integrated against its own variable, and one constant is carried along.

The Mechanics in Four Moves

Move	What you write	What can go wrong
1	isolate $\frac{dy}{dx}$ if it is not already alone	the equation is solved for the wrong quantity
2	divide by every factor containing $y$ , multiply by $dx$	a stray $y$ survives on the right
3	integrate each side against its own variable	the constant is written twice, or not at all
4	solve for $y$ if an explicit form is requested	a sign is lost when the logarithm is undone

Step 2 is where most errors happen, and it is the step that is easiest to check: after the split, the left side must contain no $x$ and the right side no $y$ .

The Logarithmic Case and the Absolute Value

The integral $\int \frac{dy}{y-a} = \ln|y-a|$ needs the absolute value because $y-a$ can be negative, and the logarithm of a negative number is not real. The absolute value is not decoration; it is what allows a single formula to cover solutions that stay below $a$ as well as those above it.

Undoing the logarithm gives $|y-a| = e^{C}e^{kx}$ , and the absolute value turns into a $\pm$ that the constant absorbs:

$y - a = \pm e^{C}e^{kx} = Ae^{kx}, \qquad A \neq 0$

That is why the general solution of $\frac{dy}{dx} = k(y-a)$ is written $y = a + Ae^{kx}$ with $A$ allowed to be any real number. The sign of $A$ records which side of the equilibrium the solution lives on, and it comes from the initial condition rather than from the integration.

One Constant, Not Two

Integrating both sides produces a constant on each side, $F(y) = G(x) + C_1 + C_2$ . The two constants combine into one, because a constant plus a constant is a constant. Writing $C$ twice is not wrong arithmetic, but it is a signal that the writer does not yet see the equation as a single relation, and on a scored part it costs clarity if not points.

When the Answer Stays Implicit

Not every equation can be solved for $y$ in closed form. $\frac{dy}{dx} = \frac{x}{y+1}$ gives $\left(y+1\right)^{2} = x^{2} + C$ , which can be solved, but $\frac{dy}{dx} = \frac{\sin y}{x}$ gives $\ln|x| = \int \frac{dy}{\sin y}$ and no elementary rearrangement exists. The exam accepts an implicit answer whenever the explicit one is not required, and it prefers a correct implicit answer to a wrong explicit one produced by forcing a square root.

Common Misconception Check

Myth 1: “ $\frac{dy}{dx}$ is a fraction, so I can multiply by $dx$ .” Truth: The manipulation is justified by substitution, not by fraction algebra. It gives the right answer, but the justification matters when the integral is not elementary.
Myth 2: “Every problem needs an explicit $y = f(x)$ .” Truth: An implicit relation is a complete answer unless the question asks for a function.
Myth 3: “The absolute value can be dropped immediately.” Truth: It can be absorbed into the constant only after you write the exponential form.

Topic-Specific Study Method

Split, Then Integrate Each Side

Borrowed idea: Get all the $y$ on one side, all the $x$ on the other. Then integrate both sides. That is the method. It works because $dy$ and $dx$ can be treated as separate objects here. You get an equation, not a function. Solve for $y$ only if the question asks. The constant appears once, on one side, though both integrals produced one.

Exam disguise	What you see on the page	What your brain must translate	Required execution
Rate equation in context	$\frac{dh}{dt} = -\frac{h}{10}$	$h$ with $dh$ , $t$ with $dt$	integrate, then apply $h(0)$
Cooling or limited growth	$\frac{dH}{dt} = k\left(A - H\right)$	the shift by $A$ is part of the variable	integrate $\frac{dH}{H-A}$
Product on the right	$\frac{dy}{dx} = 2x\left(y+3\right)$	divide by $\left(y+3\right)$ first	two logs, one constant
Implicit answer expected	“find an expression for $y$ ”	an implicit form may be enough	stop before forcing a root
Second patient, new constant	same equation, different $y(0)$	the family is unchanged	reapply the condition, do not re-integrate

① Separate before integrating. The unsplit equation cannot be integrated, and attempting it is the most common way a correct idea stalls.

② Keep the absolute value until the constant appears. Dropping it early can silently delete half the solution set.

③ Apply the initial condition to the general solution, not to the separated line. Substituting too early turns a family into a single curve before the constant exists.

Training & Error Diagnosis

Core Archetype (Simplest Form): Find the general solution of $\frac{dy}{dx} = 2xy$ .

Standard Execution:

Separate. $\frac{dy}{y} = 2x\,dx$ , valid for $y \neq 0$ .

Integrate. $\ln|y| = x^{2} + C$ .

Exponentiate. $|y| = e^{C}e^{x^{2}}$ , so $y = Ae^{x^{2}}$ with $A \neq 0$ .

Note the missing member. $y = 0$ is also a solution, so the general solution is $y = Ae^{x^{2}}$ for any real $A$ .

Exam Variation (Context / Parameter Shift): Water drains from a barrel so that $\frac{dh}{dt} = -\frac{h}{10}$ , with $h(0) = 5$ feet. Find $h(t)$ and the time at which the height reaches $1$ foot.

Solution: Separating gives $\frac{dh}{h} = -\frac{1}{10}dt$ , so $\ln|h| = -\frac{t}{10} + C$ and $h = Ae^{-t/10}$ . The condition $h(0) = 5$ fixes $A = 5$ , giving $h(t) = 5e^{-t/10}$ . Setting $h = 1$ gives $e^{-t/10} = \frac{1}{5}$ , so $-\frac{t}{10} = \ln\frac{1}{5}$ and $t = 10\ln 5 \approx 16.09$ seconds. Note that the absolute value is legitimate to drop here because the height is positive for all $t$ , and the answer must be justified on that ground rather than assumed.

Common Traps & Failure Signals:

Unseparated integrand (Signal: an $x$ still on the left after the split): the variables did not actually separate.
Absolute value dropped early (Signal: $\ln(y-a)$ written with no bars and no justification): the sign information is lost.
Condition applied too early (Signal: a number substituted into the separated line): the constant never gets determined properly.

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

Split Skipped: You integrated the unsplit equation.
Constant Doubled: You carried $C$ on both sides and then solved for one of them, which is bookkeeping with no mathematical content and a frequent source of algebra slips.
Explicit Forced: You took a square root to produce $y = f(x)$ when the question accepted the implicit relation, and the branch you chose was wrong for the given initial condition.

Past 10-Year AP Exam Appearance

Testing Frequency: Nearly Every Year; Eight of the nine released exams since 2017 contain a part whose required work is to separate the variables and integrate each side, and every one of those parts supplies an initial condition in the same sentence
10-Year Total Count: 8 of 9 Exams · 8 FRQ Parts; The five that name “separation of variables” explicitly and the three that only say “find an expression” are counted together, because the scored work is identical
FRQ Years Appeared: 2017 (Q4c), 2018 (Q6c), 2019 (Q4c), 2021 (Q6c), 2022 (Q5d), 2023 (Q3d), 2024 (Q3c), 2026 (Q3D) ---

Practice Examples

MCQ 1

The Plain Family

The general solution of $\frac{dy}{dx} = 2xy$ is

MCQ 2

A Shifted Equilibrium

The general solution of $\frac{dy}{dx} = 2x\left(y + 3\right)$ is

MCQ 3

Where the Sign Comes From

A solution of $\frac{dy}{dt} = ky$ satisfies $y(0) = -4$ . In the form $y = Ce^{kt}$ , the constant $C$ must be

FRQ 4

From a Rate Equation to a Function

(a) Water drains from a barrel so that $\frac{dh}{dt} = -\frac{h}{10}$ , where $h$ is in feet and $t$ in seconds. Given $h(0) = 5$ , find $h(t)$ and the time at which $h = 1$ .

(b) Find the solution of $\frac{dy}{dx} = \frac{x}{y+1}$ that satisfies $y(0) = 1$ , in explicit form.

(c) A student writes $\ln|y - a| = kx + C_1 + C_2$ after integrating both sides separately. Explain why the two constants can be replaced by one.

Cheatsheet

Core Reference

	Action	Failure if skipped
1	isolate $\frac{dy}{dx}$	the split is performed on the wrong expression
2	divide by every $y$ -factor, multiply by $dx$	variables never separate
3	integrate each side against its own variable	the constant is misplaced
4	keep the absolute value until the exponential form	half the family disappears
5	apply the initial condition to the general solution	the constant stays unknown

Sentences That Score

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Unit 7Separation of Variables

Concept 4.78Built for May 2027 AP Exam

Separation of Variables

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 middle two parts of a differential-equation question. A slope-field sketch usually comes first, then a part that says “use separation of variables to find an expression for $M(t)$ ,” and occasionally a follow-up asking for a limit or a value at a later time.
What you’ll do: Move the variables, integrate both sides, evaluate any constant from the given condition, and then either solve for the function or leave it in a form the question accepts.
Memory hook: $y$ goes with $dy$ , $x$ goes with $dx$ . If a variable is on the wrong side after the split, the equation was not separated.
Use it on the exam: For MCQs: Differentiate the offered answers instead of integrating. Only the correct family differentiates back to the given $\frac{dy}{dx}$ for every $x$ . For FRQs: Write the separated line before integrating. The split is often a scored step, so an answer that jumps straight to the solution can lose a point even when the solution is right.

Concept Review

Why Separating Works

Figure 1 — Four lines, no cleverness. The variables are split, each side is integrated against its own variable, and one constant is carried along.

Four lines, no cleverness. The variables are split, each side is integrated against its own variable, and one constant is carried along.

The Mechanics in Four Moves

Move	What you write	What can go wrong
1	isolate $\frac{dy}{dx}$ if it is not already alone	the equation is solved for the wrong quantity
2	divide by every factor containing $y$ , multiply by $dx$	a stray $y$ survives on the right
3	integrate each side against its own variable	the constant is written twice, or not at all
4	solve for $y$ if an explicit form is requested	a sign is lost when the logarithm is undone

Step 2 is where most errors happen, and it is the step that is easiest to check: after the split, the left side must contain no $x$ and the right side no $y$ .

The Logarithmic Case and the Absolute Value

Undoing the logarithm gives $|y-a| = e^{C}e^{kx}$ , and the absolute value turns into a $\pm$ that the constant absorbs:

$y - a = \pm e^{C}e^{kx} = Ae^{kx}, \qquad A \neq 0$

One Constant, Not Two

When the Answer Stays Implicit

Common Misconception Check

Myth 1: “ $\frac{dy}{dx}$ is a fraction, so I can multiply by $dx$ .” Truth: The manipulation is justified by substitution, not by fraction algebra. It gives the right answer, but the justification matters when the integral is not elementary.
Myth 2: “Every problem needs an explicit $y = f(x)$ .” Truth: An implicit relation is a complete answer unless the question asks for a function.
Myth 3: “The absolute value can be dropped immediately.” Truth: It can be absorbed into the constant only after you write the exponential form.

Topic-Specific Study Method

Split, Then Integrate Each Side

Exam disguise	What you see on the page	What your brain must translate	Required execution
Rate equation in context	$\frac{dh}{dt} = -\frac{h}{10}$	$h$ with $dh$ , $t$ with $dt$	integrate, then apply $h(0)$
Cooling or limited growth	$\frac{dH}{dt} = k\left(A - H\right)$	the shift by $A$ is part of the variable	integrate $\frac{dH}{H-A}$
Product on the right	$\frac{dy}{dx} = 2x\left(y+3\right)$	divide by $\left(y+3\right)$ first	two logs, one constant
Implicit answer expected	“find an expression for $y$ ”	an implicit form may be enough	stop before forcing a root
Second patient, new constant	same equation, different $y(0)$	the family is unchanged	reapply the condition, do not re-integrate

① Separate before integrating. The unsplit equation cannot be integrated, and attempting it is the most common way a correct idea stalls.

② Keep the absolute value until the constant appears. Dropping it early can silently delete half the solution set.

③ Apply the initial condition to the general solution, not to the separated line. Substituting too early turns a family into a single curve before the constant exists.

Training & Error Diagnosis

Core Archetype (Simplest Form): Find the general solution of $\frac{dy}{dx} = 2xy$ .

Standard Execution:

Separate. $\frac{dy}{y} = 2x\,dx$ , valid for $y \neq 0$ .

Integrate. $\ln|y| = x^{2} + C$ .

Exponentiate. $|y| = e^{C}e^{x^{2}}$ , so $y = Ae^{x^{2}}$ with $A \neq 0$ .

Note the missing member. $y = 0$ is also a solution, so the general solution is $y = Ae^{x^{2}}$ for any real $A$ .

Solution: Separating gives $\frac{dh}{h} = -\frac{1}{10}dt$ , so $\ln|h| = -\frac{t}{10} + C$ and $h = Ae^{-t/10}$ . The condition $h(0) = 5$ fixes $A = 5$ , giving $h(t) = 5e^{-t/10}$ . Setting $h = 1$ gives $e^{-t/10} = \frac{1}{5}$ , so $-\frac{t}{10} = \ln\frac{1}{5}$ and $t = 10\ln 5 \approx 16.09$ seconds. Note that the absolute value is legitimate to drop here because the height is positive for all $t$ , and the answer must be justified on that ground rather than assumed.

Common Traps & Failure Signals:

Unseparated integrand (Signal: an $x$ still on the left after the split): the variables did not actually separate.
Absolute value dropped early (Signal: $\ln(y-a)$ written with no bars and no justification): the sign information is lost.
Condition applied too early (Signal: a number substituted into the separated line): the constant never gets determined properly.

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

Split Skipped: You integrated the unsplit equation.
Constant Doubled: You carried $C$ on both sides and then solved for one of them, which is bookkeeping with no mathematical content and a frequent source of algebra slips.
Explicit Forced: You took a square root to produce $y = f(x)$ when the question accepted the implicit relation, and the branch you chose was wrong for the given initial condition.

Past 10-Year AP Exam Appearance

Testing Frequency: Nearly Every Year; Eight of the nine released exams since 2017 contain a part whose required work is to separate the variables and integrate each side, and every one of those parts supplies an initial condition in the same sentence
10-Year Total Count: 8 of 9 Exams · 8 FRQ Parts; The five that name “separation of variables” explicitly and the three that only say “find an expression” are counted together, because the scored work is identical
FRQ Years Appeared: 2017 (Q4c), 2018 (Q6c), 2019 (Q4c), 2021 (Q6c), 2022 (Q5d), 2023 (Q3d), 2024 (Q3c), 2026 (Q3D) ---

Practice Examples

MCQ 1

The Plain Family

The general solution of $\frac{dy}{dx} = 2xy$ is

MCQ 2

A Shifted Equilibrium

The general solution of $\frac{dy}{dx} = 2x\left(y + 3\right)$ is

MCQ 3

Where the Sign Comes From

A solution of $\frac{dy}{dt} = ky$ satisfies $y(0) = -4$ . In the form $y = Ce^{kt}$ , the constant $C$ must be

FRQ 4

From a Rate Equation to a Function

(a) Water drains from a barrel so that $\frac{dh}{dt} = -\frac{h}{10}$ , where $h$ is in feet and $t$ in seconds. Given $h(0) = 5$ , find $h(t)$ and the time at which $h = 1$ .

(b) Find the solution of $\frac{dy}{dx} = \frac{x}{y+1}$ that satisfies $y(0) = 1$ , in explicit form.

(c) A student writes $\ln|y - a| = kx + C_1 + C_2$ after integrating both sides separately. Explain why the two constants can be replaced by one.

Cheatsheet

Core Reference

	Action	Failure if skipped
1	isolate $\frac{dy}{dx}$	the split is performed on the wrong expression
2	divide by every $y$ -factor, multiply by $dx$	variables never separate
3	integrate each side against its own variable	the constant is misplaced
4	keep the absolute value until the exponential form	half the family disappears
5	apply the initial condition to the general solution	the constant stays unknown

Sentences That Score

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Unit 7Separation of Variables

Concept 4.78Built for May 2027 AP Exam

Separation of Variables

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 middle two parts of a differential-equation question. A slope-field sketch usually comes first, then a part that says “use separation of variables to find an expression for $M(t)$ ,” and occasionally a follow-up asking for a limit or a value at a later time.
What you’ll do: Move the variables, integrate both sides, evaluate any constant from the given condition, and then either solve for the function or leave it in a form the question accepts.
Memory hook: $y$ goes with $dy$ , $x$ goes with $dx$ . If a variable is on the wrong side after the split, the equation was not separated.
Use it on the exam: For MCQs: Differentiate the offered answers instead of integrating. Only the correct family differentiates back to the given $\frac{dy}{dx}$ for every $x$ . For FRQs: Write the separated line before integrating. The split is often a scored step, so an answer that jumps straight to the solution can lose a point even when the solution is right.

Concept Review

Why Separating Works

Figure 1 — Four lines, no cleverness. The variables are split, each side is integrated against its own variable, and one constant is carried along.

Four lines, no cleverness. The variables are split, each side is integrated against its own variable, and one constant is carried along.

The Mechanics in Four Moves

Move	What you write	What can go wrong
1	isolate $\frac{dy}{dx}$ if it is not already alone	the equation is solved for the wrong quantity
2	divide by every factor containing $y$ , multiply by $dx$	a stray $y$ survives on the right
3	integrate each side against its own variable	the constant is written twice, or not at all
4	solve for $y$ if an explicit form is requested	a sign is lost when the logarithm is undone

Step 2 is where most errors happen, and it is the step that is easiest to check: after the split, the left side must contain no $x$ and the right side no $y$ .

The Logarithmic Case and the Absolute Value

Undoing the logarithm gives $|y-a| = e^{C}e^{kx}$ , and the absolute value turns into a $\pm$ that the constant absorbs:

$y - a = \pm e^{C}e^{kx} = Ae^{kx}, \qquad A \neq 0$

One Constant, Not Two

When the Answer Stays Implicit

Common Misconception Check

Myth 1: “ $\frac{dy}{dx}$ is a fraction, so I can multiply by $dx$ .” Truth: The manipulation is justified by substitution, not by fraction algebra. It gives the right answer, but the justification matters when the integral is not elementary.
Myth 2: “Every problem needs an explicit $y = f(x)$ .” Truth: An implicit relation is a complete answer unless the question asks for a function.
Myth 3: “The absolute value can be dropped immediately.” Truth: It can be absorbed into the constant only after you write the exponential form.

Topic-Specific Study Method

Split, Then Integrate Each Side

Exam disguise	What you see on the page	What your brain must translate	Required execution
Rate equation in context	$\frac{dh}{dt} = -\frac{h}{10}$	$h$ with $dh$ , $t$ with $dt$	integrate, then apply $h(0)$
Cooling or limited growth	$\frac{dH}{dt} = k\left(A - H\right)$	the shift by $A$ is part of the variable	integrate $\frac{dH}{H-A}$
Product on the right	$\frac{dy}{dx} = 2x\left(y+3\right)$	divide by $\left(y+3\right)$ first	two logs, one constant
Implicit answer expected	“find an expression for $y$ ”	an implicit form may be enough	stop before forcing a root
Second patient, new constant	same equation, different $y(0)$	the family is unchanged	reapply the condition, do not re-integrate

① Separate before integrating. The unsplit equation cannot be integrated, and attempting it is the most common way a correct idea stalls.

② Keep the absolute value until the constant appears. Dropping it early can silently delete half the solution set.

③ Apply the initial condition to the general solution, not to the separated line. Substituting too early turns a family into a single curve before the constant exists.

Training & Error Diagnosis

Core Archetype (Simplest Form): Find the general solution of $\frac{dy}{dx} = 2xy$ .

Standard Execution:

Separate. $\frac{dy}{y} = 2x\,dx$ , valid for $y \neq 0$ .

Integrate. $\ln|y| = x^{2} + C$ .

Exponentiate. $|y| = e^{C}e^{x^{2}}$ , so $y = Ae^{x^{2}}$ with $A \neq 0$ .

Note the missing member. $y = 0$ is also a solution, so the general solution is $y = Ae^{x^{2}}$ for any real $A$ .

Solution: Separating gives $\frac{dh}{h} = -\frac{1}{10}dt$ , so $\ln|h| = -\frac{t}{10} + C$ and $h = Ae^{-t/10}$ . The condition $h(0) = 5$ fixes $A = 5$ , giving $h(t) = 5e^{-t/10}$ . Setting $h = 1$ gives $e^{-t/10} = \frac{1}{5}$ , so $-\frac{t}{10} = \ln\frac{1}{5}$ and $t = 10\ln 5 \approx 16.09$ seconds. Note that the absolute value is legitimate to drop here because the height is positive for all $t$ , and the answer must be justified on that ground rather than assumed.

Common Traps & Failure Signals:

Unseparated integrand (Signal: an $x$ still on the left after the split): the variables did not actually separate.
Absolute value dropped early (Signal: $\ln(y-a)$ written with no bars and no justification): the sign information is lost.
Condition applied too early (Signal: a number substituted into the separated line): the constant never gets determined properly.

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

Split Skipped: You integrated the unsplit equation.
Constant Doubled: You carried $C$ on both sides and then solved for one of them, which is bookkeeping with no mathematical content and a frequent source of algebra slips.
Explicit Forced: You took a square root to produce $y = f(x)$ when the question accepted the implicit relation, and the branch you chose was wrong for the given initial condition.

Past 10-Year AP Exam Appearance

Testing Frequency: Nearly Every Year; Eight of the nine released exams since 2017 contain a part whose required work is to separate the variables and integrate each side, and every one of those parts supplies an initial condition in the same sentence
10-Year Total Count: 8 of 9 Exams · 8 FRQ Parts; The five that name “separation of variables” explicitly and the three that only say “find an expression” are counted together, because the scored work is identical
FRQ Years Appeared: 2017 (Q4c), 2018 (Q6c), 2019 (Q4c), 2021 (Q6c), 2022 (Q5d), 2023 (Q3d), 2024 (Q3c), 2026 (Q3D) ---

Practice Examples

MCQ 1

The Plain Family

The general solution of $\frac{dy}{dx} = 2xy$ is

MCQ 2

A Shifted Equilibrium

The general solution of $\frac{dy}{dx} = 2x\left(y + 3\right)$ is

MCQ 3

Where the Sign Comes From

A solution of $\frac{dy}{dt} = ky$ satisfies $y(0) = -4$ . In the form $y = Ce^{kt}$ , the constant $C$ must be

FRQ 4

From a Rate Equation to a Function

(a) Water drains from a barrel so that $\frac{dh}{dt} = -\frac{h}{10}$ , where $h$ is in feet and $t$ in seconds. Given $h(0) = 5$ , find $h(t)$ and the time at which $h = 1$ .

(b) Find the solution of $\frac{dy}{dx} = \frac{x}{y+1}$ that satisfies $y(0) = 1$ , in explicit form.

(c) A student writes $\ln|y - a| = kx + C_1 + C_2$ after integrating both sides separately. Explain why the two constants can be replaced by one.

Cheatsheet

Core Reference

	Action	Failure if skipped
1	isolate $\frac{dy}{dx}$	the split is performed on the wrong expression
2	divide by every $y$ -factor, multiply by $dx$	variables never separate
3	integrate each side against its own variable	the constant is misplaced
4	keep the absolute value until the exponential form	half the family disappears
5	apply the initial condition to the general solution	the constant stays unknown

Sentences That Score

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