Separation of variables is a method for solving certain differential equations by moving all the $y$ terms to one side and all the $x$ terms to the other, then integrating both sides. This gives you a general solution: a family of functions that differ only by a constant of integration $C$ . For AP Calculus, include the constant of integration before using any initial condition.

Why This Matters for the AP Calculus Exam

Differential equations make up a noticeable share of the AP Calculus exam (about 6 to 12 percent on AB and 6 to 9 percent on BC), and separation of variables is the core algebraic method for actually solving them. You can see these problems in both multiple-choice and free-response sections. This topic is the bridge between recognizing a differential equation and writing the function that satisfies it, so it sets up later work with initial conditions, exponential models, and (on BC) logistic models.

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Key Takeaways

A separable differential equation can be written as $\frac{dy}{dx} = g(x)h(y)$ , where one factor depends only on $x$ and the other only on $y$ .
Solve by separating variables, integrating both sides, and adding a single constant of integration.
Always include $+C$ . Leaving it out limits how many points you can earn.
The general solution is a family of functions, not a single curve, because $C$ can take infinitely many values.
Not every differential equation separates. If $x$ and $y$ terms cannot be split apart, this method does not apply.
Check a solution by differentiating it and substituting back into the original equation.

Solving Differential Equations

A solution to a differential equation is a continuous function that satisfies the equation. There can be infinitely many solutions, and together they form a family of curves. If you want to review how to confirm a function is a solution, see Verifying Solutions to Differential Equations.

Going the other direction, from an equation to a solution, is harder. Some differential equations like $2xy' + y = 3x^2$ are difficult to solve and are beyond this course. But one important type, separable differential equations, is well within reach.

Separable Differential Equations

Separable differential equations are first-order equations whose variables can be split so that each side involves only one variable. They are usually written as

$\frac{dy}{dx} = g(x)h(y)$

where $g(x)$ depends only on $x$ and $h(y)$ depends only on $y$ .

The strategy is to integrate both sides to eliminate the derivative:

Separate variables: Rearrange so all $y$ terms (including $dy$ ) are on one side and all $x$ terms (including $dx$ ) are on the other.
Integrate both sides: Antidifferentiate each side with respect to its own variable. Add the constant of integration.
Solve for $y$ : If possible, write $y$ explicitly. Sometimes the answer stays in implicit form.

Finding the constant $C$ from an initial condition is the next step, covered in Finding Particular Solutions Using Initial Conditions.

Worked Practice Problems

Practice Problem #1

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

Separate the variables:

y \, dy = 2x \, dx

Integrate both sides:

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

This gives, with $C$ as the constant of integration:

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

Solve for $y$ explicitly:

y^2 = 2x^2 + 2C

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

So the general solution is

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

The constant $C$ gets pinned down later if an initial condition is given.

Practice Problem #2

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

Move $xy$ to the right side:

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

This one cannot be separated. There is no way to split the right side into a product of a function of $x$ times a function of $y$ , so separation of variables does not apply. Recognizing this quickly saves you time on the exam.

Practice Problem #3

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

Separate the variables:

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

Integrate both sides:

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

This gives:

\ln|y| = \frac{1}{3}x^3 + C

Solve for $y$ :

|y| = e^{\frac{1}{3}x^3 + C} = e^{C}\cdot e^{\frac{1}{3}x^3}

Since $e^C$ is just another positive constant, rename it $C$ :

|y| = Ce^{\frac{1}{3}x^3}

Accounting for the absolute value gives two cases, which combine into:

y = \pm \,Ce^{\frac{1}{3}x^3}

This is the general solution. Notice it is an exponential function, recognizable because the rate of change depends on the dependent variable $y$ , the same behavior that makes $e^x$ its own derivative.

How to Use This on the AP Calculus Exam

Problem Solving

Confirm the equation is separable before starting. If you cannot write it as $g(x)h(y)$ , switch strategies.
Keep your separation clean: every $y$ and $dy$ on one side, every $x$ and $dx$ on the other.
Antidifferentiate carefully. A fraction like $\frac{1}{y}$ integrates to $\ln|y|$ , but not every fraction gives a logarithm.
Write $+C$ the moment you integrate. It is easy to forget and hard to recover later.

Free Response

Show the separation step and the integration step clearly. Skipping work makes it hard for readers to follow your method.
When solving for $y$ , keep track of $\pm$ and absolute values so your general solution is complete.
If the problem later gives an initial condition, you will use it to solve for $C$ (see Topic 7.7).

Common Trap

Assuming every differential equation with a fraction has a logarithmic solution. Some do, many do not. Match the antiderivative to the actual function.

Common Misconceptions

Every differential equation is separable. Many are not. If you cannot split the variables into a product form, this method will not work, and that is a valid thing to state.
The constant of integration is optional. Dropping $C$ changes a whole family of solutions into a single curve and costs points. Always include it.
All fractions integrate to a logarithm. Only $\frac{1}{y}$ -type forms give $\ln|y|$ . Use the power rule or other antiderivatives when the function calls for it.
A general solution is one function. It is an entire family of functions parameterized by $C$ . A single function comes only after applying an initial condition.
You can combine two constants into something complicated. When you have a constant on each side after integrating, merge them into one $C$ . Tracking two separate constants is unnecessary.

Vocabulary

The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.

Term	Definition
antidifferentiation	The process of finding a function whose derivative is a given function; the reverse operation of differentiation, also known as integration.
differential equation	An equation that relates a function to its derivatives, describing how a quantity changes in relation to one or more variables.
general solution	The complete family of solutions to a differential equation, containing arbitrary constants that represent all possible particular solutions.
separation of variables	A method for solving differential equations by rearranging the equation so that all terms involving one variable are on one side and all terms involving the other variable are on the other side.

Frequently Asked Questions

What is separation of variables in AP Calculus?

Separation of variables is a method for solving certain differential equations by rewriting them so all y terms are on one side and all x terms are on the other, then integrating both sides.

How do you know if a differential equation is separable?

A differential equation is separable if it can be rewritten in the form dy/dx = g(x)h(y), with one factor depending only on x and the other depending only on y.

What are the steps for separation of variables?

Separate the variables, integrate both sides, include a constant of integration, and solve for y if the algebra allows it. Some answers may stay in implicit form.

Why do you need + C in a general solution?

The + C represents a whole family of possible solutions. Without it, you only have one solution curve instead of the general solution to the differential equation.

What is the difference between a general solution and a particular solution?

A general solution includes an arbitrary constant C. A particular solution uses an initial condition to find the specific value of C for one solution curve.

How is AP Calculus 7.6 tested?

AP Calculus 7.6 is tested through differential equation problems where you identify separability, separate variables correctly, integrate both sides, and keep clear notation for the constant of integration.

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