When can I drop the absolute value in |y|?

Use the initial condition to determine the sign of y. If the given point has y>0, then |y|=y and you can drop the bars; if y<0, then |y|=-y. The solution stays on one side of y=0 because a continuous solution cannot cross zero here without violating the equation.

Do I need a +C on both sides of the equation?

No. Although both integrals technically produce constants, you can combine them into a single constant on one side (usually the x-side). Writing one C is standard and fully correct, and it saves time. Never omit the constant entirely, though — you need it to satisfy the initial condition.

Should I apply the initial condition before or after solving for y?

You may do either, but substituting right after integrating — before solving for y — is usually cleaner because it avoids exponentiating or manipulating an unknown constant. Just never substitute the initial condition before you integrate, since the constant of integration doesn't exist yet.

How is separation of variables graded on the free-response section?

Points are typically distributed across separating the variables correctly, antidifferentiating both sides, including the constant of integration, using the initial condition, and solving for y. Show the separated equation explicitly so you earn early steps even if a later antiderivative slips.

AP-CALCBC-7.6-7.7

U7.6 Separation of Variables

Q: Do I need a +C on both sides of the equation?

No. Although both integrals technically produce constants, you can combine them into a single constant on one side (usually the x-side). Writing one C is standard and fully correct, and it saves time. Never omit the constant entirely, though — you need it to satisfy the initial condition.

Q: Should I apply the initial condition before or after solving for y?

You may do either, but substituting right after integrating — before solving for y — is usually cleaner because it avoids exponentiating or manipulating an unknown constant. Just never substitute the initial condition before you integrate, since the constant of integration doesn't exist yet.

Q: How is separation of variables graded on the free-response section?

Points are typically distributed across separating the variables correctly, antidifferentiating both sides, including the constant of integration, using the initial condition, and solving for y. Show the separated equation explicitly so you earn early steps even if a later antiderivative slips.

Master separable differential equations for AP Calculus BC: separate variables, integrate both sides, and use initial conditions to find particular solutions.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U7.6 Separation of Variables, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

A separable differential equation is one you can rearrange so that all the

y

stuff lives on one side with

dy

and all the

x

stuff lives on the other with

dx

. Once separated, you integrate both sides — and suddenly a scary-looking equation like

\frac{dy}{dx}=xy

becomes a routine integration problem.

This lesson shows you the exact algebra-then-integrate procedure, how to handle the constant of integration correctly, and how to lock down a particular solution using an initial condition. These skills are tested every year, both on multiple-choice items and as a signature step inside free-response questions, so getting the mechanics automatic pays off directly.

What makes an equation separable

A first-order differential equation is separable if you can write it in the form

\frac{dy}{dx}=f(x)\cdot g(y)

— a product of a function of

x

alone and a function of

y

alone. The test is purely algebraic: can you factor the right-hand side so that

x

-terms and

y

-terms are multiplied, not added?

For example,

\frac{dy}{dx}=xy

is separable because it is

x

times

y

. So is

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

, since it equals

x^2\cdot\frac{1}{y}

. But

\frac{dy}{dx}=x+y

is not separable — a sum cannot be split into a clean product.

Once you confirm separability, divide both sides by

g(y)

and multiply by

dx

to gather like variables:

\frac{1}{g(y)}\,dy = f(x)\,dx

Equation	Separable?	Separated form
$\frac{dy}{dx}=xy$	Yes	$\frac{1}{y}dy=x\,dx$
$\frac{dy}{dx}=\frac{x^2}{y}$	Yes	$y\,dy=x^2\,dx$
$\frac{dy}{dx}=x+y$	No	—

Recognizing separability quickly is often the first graded step on an FRQ.

The separate-and-integrate procedure

Solving proceeds in a fixed sequence. First, separate variables so one side has only

y

and

dy

, the other only

x

and

dx

. Second, integrate both sides. Third, add a single constant of integration

C

— you only need one, conventionally placed on the

x

-side. Fourth, solve for

y

explicitly when possible.

Consider

\frac{dy}{dx}=xy

. Separate:

\frac{1}{y}\,dy = x\,dx

. Integrate both sides:

\ln|y| = \frac{x^2}{2} + C

To solve for

y

, exponentiate:

|y| = e^{x^2/2}\cdot e^{C}

. Since

e^{C}

is just a positive constant, replace

\pm e^{C}

with a new constant

k

y = k\,e^{x^2/2}

This is the general solution — a whole family of curves, one for each value of

k

.

A common misconception is writing

+C

on both sides; that is not wrong but wastes time, since you can combine them into one constant. A more serious error is forgetting the constant entirely, which makes it impossible to satisfy an initial condition. Always integrate first, then apply the initial condition — never before.

General versus particular solutions

The general solution contains the arbitrary constant and represents infinitely many curves. A particular solution pins down that constant using an initial condition of the form

y(x_0)=y_0

.

There are two valid orders for using the initial condition, and choosing the efficient one saves algebra. Method one: solve completely for

y

, then substitute. Method two: substitute the initial condition right after integrating, while the equation still contains

\ln|y|

or an implicit form, to find

C

first. On the AP exam, method two is frequently cleaner because it avoids exponentiating an unknown constant.

Using the earlier example with condition

y(0)=3

: at

\ln|y|=\frac{x^2}{2}+C

, substitute

x=0

y=3

\ln 3 = 0 + C

, so

C=\ln 3

. Then

\ln y = \frac{x^2}{2}+\ln 3

, giving

y = 3e^{x^2/2}

.

Two details the AP exam rewards: keep the domain in mind — the particular solution is only valid on the interval containing

x_0

where the function is continuous — and choose the correct sign or branch based on the initial condition (for instance, if

y(0)=3>0

, drop the absolute value as

y>0

How the exam tests it and common traps

On multiple-choice, you may be handed a general solution and asked which matches a differential equation, or given an initial condition and asked for

y

at a specific point. On free-response, separation of variables is a multi-point workhorse: separating correctly, antidifferentiating correctly, including

+C

, using the initial condition, and solving for

y

are each often worth points.

The most frequent trap is dropping the constant of integration or applying the initial condition before integrating. Another is mishandling absolute values — if the integral gives

\ln|y|

, you must justify removing the bars using the sign implied by the initial condition. A third is antidifferentiation slips, especially forgetting the chain-rule-driven factors like

\frac{1}{2}

when integrating

x

Step	Points-earning action	Trap to avoid
Separate	Correct $dy$ / $dx$ split	Splitting a sum
Integrate	Both antiderivatives right	Missing $+C$
Apply IC	Substitute after integrating	Substituting too early
Solve for $y$	Correct branch/sign	Wrong absolute-value sign

Write legibly and show the separated equation explicitly; readers award the separation step even before the integration is finished.

Key terms

Separable differential equation.: A first-order equation writable as $\frac{dy}{dx}=f(x)\cdot g(y)$ , so variables can be split onto opposite sides.
Separation of variables.: The technique of rearranging a separable equation into $\frac{1}{g(y)}dy=f(x)dx$ and integrating both sides.
General solution.: The family of all solutions to a differential equation, containing an arbitrary constant of integration.
Particular solution.: A single specific solution obtained by using an initial condition to determine the constant.
Initial condition.: A given value $y(x_0)=y_0$ used to solve for the constant of integration.
Constant of integration.: The $+C$ added after antidifferentiating; a single combined constant suffices for both sides.

Worked example

Solve the differential equation

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

with the initial condition

y(0)=4

, and give the particular solution.

First check separability: the right side is

2x\cdot\frac{1}{y}

, a product of an

x

-function and a

y

-function, so it is separable.

Separate the variables by multiplying both sides by

y\,dx

y\,dy = 2x\,dx

Integrate both sides:

\frac{y^2}{2} = x^2 + C

Apply the initial condition immediately. Substitute

x=0

and

y=4

\frac{4^2}{2} = 0 + C \implies 8 = C

\frac{y^2}{2} = x^2 + 8

, which gives

y^2 = 2x^2 + 16

.

Solve for

y

and pick the correct branch. Taking the square root gives

y = \pm\sqrt{2x^2+16}

. Since

y(0)=4

is positive, choose the positive root:

y = \sqrt{2x^2 + 16}

This is the particular solution. Quick check: at

x=0

y=\sqrt{16}=4

, matching the initial condition.

Practice questions

Which of the following is the general solution to

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

$y = x^3 + C$
$y = Ce^{x^3}$
$y = e^{x^3} + C$
$y = Cx^3$

Answer: $y = Ce^{x^3}$

Separate:

\frac{1}{y}dy = 3x^2 dx

. Integrate:

\ln|y| = x^3 + C

. Exponentiate:

|y| = e^{x^3}e^{C}

, so

y = Ce^{x^3}

where

C

absorbs

\pm e^{C}

. The additive-constant options fail because the constant appears as a multiplicative factor after exponentiating.

Solve

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

given

y(0)=1

. Show your steps and give the particular solution.

Answer: $y = \sqrt{\sin x + 1}$

Separate:

2y\,dy = \cos x\,dx

. Integrate:

y^2 = \sin x + C

. Apply

y(0)=1

1 = \sin 0 + C = C

, so

C=1

and

y^2 = \sin x + 1

. Since

y(0)=1>0

, take the positive root:

y=\sqrt{\sin x+1}

. Verify: at

x=0

y=\sqrt{0+1}=1

A quantity satisfies

\frac{dy}{dx}=xy^2

with

y(0)=-1

. Find the particular solution and state where it is valid.

Answer: $y = \frac{-1}{1+\frac{x^2}{2}} = \frac{-2}{2+x^2}$

Separate:

y^{-2}dy = x\,dx

. Integrate:

-y^{-1} = \frac{x^2}{2}+C

, so

-\frac{1}{y}=\frac{x^2}{2}+C

. Apply

y(0)=-1

-\frac{1}{-1}=1 = 0 + C

, so

C=1

. Then

-\frac{1}{y}=\frac{x^2}{2}+1

, giving

y = \frac{-1}{\frac{x^2}{2}+1}=\frac{-2}{2+x^2}

. Since

2+x^2

is never zero, this solution is valid for all real

x

FAQ

When can I drop the absolute value in $\ln|y|$ ?: Use the initial condition to determine the sign of $y$ . If the given point has $y>0$ , then $|y|=y$ and you can drop the bars; if $y<0$ , then $|y|=-y$ . The solution stays on one side of $y=0$ because a continuous solution cannot cross zero here without violating the equation.
Do I need a $+C$ on both sides of the equation?: No. Although both integrals technically produce constants, you can combine them into a single constant on one side (usually the $x$ -side). Writing one $C$ is standard and fully correct, and it saves time. Never omit the constant entirely, though — you need it to satisfy the initial condition.
Should I apply the initial condition before or after solving for y?: You may do either, but substituting right after integrating — before solving for $y$ — is usually cleaner because it avoids exponentiating or manipulating an unknown constant. Just never substitute the initial condition before you integrate, since the constant of integration doesn't exist yet.
How is separation of variables graded on the free-response section?: Points are typically distributed across separating the variables correctly, antidifferentiating both sides, including the constant of integration, using the initial condition, and solving for $y$ . Show the separated equation explicitly so you earn early steps even if a later antiderivative slips.

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