AP Calculus AB and BC

Separable Differential Equations: Step-by-Step Method

Q: Are separable differential equations on the AB exam or only BC?

Both. Topics 7.6 and 7.7 are part of the AB and BC curriculum, and separation of variables is a common free-response task on each exam. Only Euler's method (7.5) and logistic models (7.9) in this unit are BC-only.

Q: Do I need a +C on both sides when I integrate?

No. Integrating both sides technically produces two constants, but their difference is a single arbitrary constant, so you write one +C, usually on the x side. A first-order equation has exactly one constant in its general solution.

Q: Should I solve for y before or after using the initial condition?

Use the initial condition first, right after integrating, while the constant is alone on one side. Solve for C, then isolate y. That order keeps the algebra short and avoids sign mistakes.

Q: Why does the AP rubric care about the domain of the solution?

A particular solution has to be one continuous function through the initial point, so it cannot cross a vertical asymptote or a point where the equation is undefined. Stating the correct interval (FUN-7.E.3) is a scored part of the answer.

To solve a separable differential equation, get every y term with dy on one side and every x term with dx on the other, integrate both sides, add a single +C, then solve for y. If an initial condition is given, substitute it to find C right after integrating, then isolate y and state any domain restriction.

Recognizing a separable equation

A differential equation is separable when you can rearrange it so that every $y$ sits with $dy$ on one side and every $x$ sits with $dx$ on the other. Formally, the right side of the equation has to factor into a function of $x$ alone times a function of $y$ alone:

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

Recognition is the entire skill in Topic 7.6, and it comes down to one question: can I split the right side into an $x$ -only piece and a $y$ -only piece? Multiplication and division separate cleanly, so a product or quotient of $x$ and $y$ terms always works. Addition and subtraction do not, so $\frac{dy}{dx} = x + y$ cannot be separated, and neither can anything where $x$ and $y$ are locked inside the same function, like $\frac{dy}{dx} = \sin(xy)$ . On the AP exam, a first-order equation you are asked to solve algebraically will always separate.

Equation	Separable?	Why
$\frac{dy}{dx} = 6xy$	Yes	$(6x)(y)$ : an $x$ -piece times a $y$ -piece
$\frac{dy}{dx} = \frac{x^2}{y}$	Yes	$(x^2)\left(\frac{1}{y}\right)$
$\frac{dy}{dx} = x + y$	No	a sum will not factor into $x$ times $y$
$\frac{dy}{dx} = \sin(xy)$	No	$x$ and $y$ trapped inside one function

Run this factoring test before you write anything else. It decides whether the rest of the method even applies, and if it fails, the exam wants a slope field or Euler's method instead of algebra.

The method, step by step

Once an equation passes the separable test, the procedure is mechanical and identical every time.

Separate. Multiply and divide until all the $y$ terms (including $dy$ ) are on one side and all the $x$ terms (including $dx$ ) are on the other.
Integrate both sides. Antidifferentiate each side with respect to its own variable.
Add one constant. Put a single $+C$ on the $x$ side. You do not need a constant on both sides, because two arbitrary constants collapse into one.
Apply the initial condition (if the problem gives one) and solve for $C$ now, before you isolate $y$ .
Solve for $y$ . Write $y$ as an explicit function of $x$ , and state any domain restriction.

The move in step 1, treating $dy$ and $dx$ as quantities you can multiply across, looks like a shortcut but it is really bookkeeping for the chain rule. Integrating $p(y) \cdot \frac{dy}{dx}$ with respect to $x$ is the same as integrating $p(y)$ with respect to $y$ , which is exactly why the two sides can be antidifferentiated in their own variables.

One constant, not two

Writing $+C$ on both sides is not wrong, just wasted work: subtracting leaves a single constant $C = C_2 - C_1$ . Put one $+C$ on the $x$ side and move on. The general solution of a first-order equation carries exactly one arbitrary constant.

Initial conditions and particular solutions

A general solution like $y = Ae^{3x^2}$ describes infinitely many curves, one for each value of the constant. An initial condition, a single point the solution must pass through, selects exactly one of them: the particular solution (Topic 7.7). This is the difference between a family of curves and the one curve the problem is actually about.

Apply the initial condition right after you integrate, while $C$ is still alone on one side. Substitute the point, solve for $C$ , then isolate $y$ . Solving for $y$ first and substituting afterward gives the same answer, but it drags the constant through more algebra and is where sign errors creep in.

If you have time, verify the result the way Topic 7.2 asks: differentiate your answer, confirm it satisfies the original equation, and check that it passes through the given point. That catches most algebra slips before they cost you.

Domain restrictions: the point students lose

This is the step that separates a full-credit free-response answer from one that drops a point. A particular solution is a single continuous, differentiable function through the initial point, so its domain is the largest interval that contains that point and on which the solution stays defined. The answer is not finished until you state that interval.

Two things cut the interval short. A vertical asymptote is the first: if the solution blows up at some $x = a$ , the particular solution lives only on the side of $a$ that holds the initial condition. The original equation is the second: if $\frac{dy}{dx}$ divides by $y$ , the solution can never reach a point where $y = 0$ , so those points bound the interval. Report the domain explicitly as an interval and confirm the initial $x$ -value sits inside it (FUN-7.E.3).

Mistakes that cost points

A handful of errors show up again and again on graded work.

Adding the constant too late. Put $+C$ in at the integration step and carry it through, not after you exponentiate.
Losing the absolute value. $\int \frac{1}{y}\,dy = \ln|y|$ , not $\ln y$ . The absolute value is what lets the final constant be positive or negative.
Dropping the constant solution. Dividing by $y$ assumes $y \neq 0$ , but $y = 0$ (or whatever makes $h(y) = 0$ ) is often a solution too. It usually reappears when you let the constant equal zero.
Choosing the wrong sign. When solving for $y$ produces a $\pm$ , the initial condition decides which sign to keep.
Skipping the domain. If the problem gives an initial condition, the interval of validity is part of the answer, not an afterthought.

Where this fits in Unit 7

Separation of variables is the only algebraic solving method the course expects you to run by hand, which makes recognizing it the real decision. If the equation separates, use this method. If it does not, the exam turns to qualitative tools: a slope field to picture the family of solutions (Topics 7.3 and 7.4), or Euler's method to approximate a value numerically (Topic 7.5, BC only).

Two famous models are separable equations in disguise. Exponential growth and decay, $\frac{dy}{dt} = ky$ , separates to give $y = y_0 e^{kt}$ (Topic 7.8). The BC logistic equation $\frac{dy}{dt} = ky(a - y)$ separates too, though it needs partial fractions to integrate (Topic 7.9). Both are just this same method applied to a specific right side.

Worked examples

Worked example

Finding a general solution

Find the general solution of $\frac{dy}{dx} = 6xy$ .

Check that it is separable. The right side is $(6x)(y)$ , an $x$ -only factor times a $y$ -only factor, so separation of variables applies (Topic 7.6).
Separate the variables. Divide both sides by $y$ and multiply by $dx$ to get $\frac{1}{y}\,dy = 6x\,dx$ . Dividing by $y$ assumes $y \neq 0$ ; keep that case in mind.
Integrate both sides. $\int \frac{1}{y}\,dy = \int 6x\,dx$ gives $\ln|y| = 3x^2 + C$ , with a single constant on the right.
Solve for $y$ . Exponentiate both sides: $|y| = e^{3x^2 + C} = e^C \cdot e^{3x^2}$ . Drop the absolute value by letting the constant absorb the sign, $A = \pm e^C$ , so $y = Ae^{3x^2}$ .
Recover the lost solution. The excluded case $y = 0$ is itself a solution of the original equation, and it matches $y = Ae^{3x^2}$ when $A = 0$ . Letting $A$ be any real number captures every solution.

$y = Ae^{3x^2}$ , where $A$ is any constant.

Worked example

Particular solution with a domain restriction

Solve $\frac{dy}{dx} = y^2$ with the initial condition $y(0) = 1$ , and state the domain of the solution.

Confirm separability. $y^2 = (1)(y^2)$ is a function of $x$ (the constant $1$ ) times a function of $y$ , so the equation separates.
Separate. $\frac{1}{y^2}\,dy = dx$ , which is $y^{-2}\,dy = dx$ .
Integrate both sides. $\int y^{-2}\,dy = \int 1\,dx$ gives $-\frac{1}{y} = x + C$ .
Apply the initial condition now, while $C$ is alone. Substitute $x = 0$ and $y = 1$ : $-\frac{1}{1} = 0 + C$ , so $C = -1$ .
Solve for $y$ . The equation becomes $-\frac{1}{y} = x - 1$ , so $\frac{1}{y} = 1 - x$ and $y = \frac{1}{1 - x}$ .
State the domain. The solution has a vertical asymptote at $x = 1$ . The initial point $x = 0$ lies to the left of it, so the particular solution is valid only on $x < 1$ .

$y = \frac{1}{1 - x}$ for $x < 1$ .

Worked example

Choosing the correct sign from the initial condition

Solve $\frac{dy}{dx} = -\frac{x}{y}$ with $y(0) = 2$ .

Separate. Multiply both sides by $y$ and by $dx$ : $y\,dy = -x\,dx$ . The equation divides by $y$ , so $y = 0$ is excluded from the start.
Integrate both sides. $\int y\,dy = \int -x\,dx$ gives $\frac{y^2}{2} = -\frac{x^2}{2} + C$ .
Clear fractions. Multiply through by $2$ and rename the constant: $y^2 = -x^2 + C_1$ , so $x^2 + y^2 = C_1$ .
Apply the initial condition. Substitute $x = 0$ and $y = 2$ : $0 + 4 = C_1$ , so $x^2 + y^2 = 4$ .
Solve for $y$ and pick the sign. Taking the square root gives $y = \pm\sqrt{4 - x^2}$ . Because $y(0) = 2$ is positive, keep the positive root: $y = \sqrt{4 - x^2}$ .
State the domain. The expression needs $4 - x^2 > 0$ , and the original equation forbids $y = 0$ , which occurs at $x = \pm 2$ . Both conditions give $-2 < x < 2$ .

$y = \sqrt{4 - x^2}$ for $-2 < x < 2$ .

Frequently asked questions

Are separable differential equations on the AB exam or only BC?

Both. Topics 7.6 and 7.7 are part of the AB and BC curriculum, and separation of variables is a common free-response task on each exam. Only Euler's method (7.5) and logistic models (7.9) in this unit are BC-only.

Do I need a $+C$ on both sides when I integrate?

No. Integrating both sides technically produces two constants, but their difference is a single arbitrary constant, so you write one $+C$ , usually on the $x$ side. A first-order equation has exactly one constant in its general solution.

Should I solve for y before or after using the initial condition?

Use the initial condition first, right after integrating, while the constant is alone on one side. Solve for $C$ , then isolate $y$ . That order keeps the algebra short and avoids sign mistakes.

Why does the AP rubric care about the domain of the solution?

A particular solution has to be one continuous function through the initial point, so it cannot cross a vertical asymptote or a point where the equation is undefined. Stating the correct interval (FUN-7.E.3) is a scored part of the answer.