General solutions via separation of variables

AP Calculus AB· AP Calculus AB CED — Differential Equations· 14 min read

1. Identifying Separable First-Order Differential Equations★★☆☆☆⏱ 3 min

📘 Definition

Separable Differential Equation

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

A first-order differential equation where the right-hand side can be factored completely into a product of a function that depends only on $x$ and a function that depends only on $y$ .

Before applying separation of variables, you must confirm your DE is separable. Many DEs appear non-separable at first, but can be rewritten using exponent rules (e.g. $e^{x + y} = e^{x} e^{y}$ ) or factoring to reveal the required product form. AP Calculus AB will only ask you to solve separable DEs with this method.

📐 Worked Example

Determine whether each of the following differential equations is separable: (a) $\frac{d y}{d x} = 2 x y + e^{x - y}$ (b) $\frac{d y}{d x} = x^{2} + 3 y^{2}$

1
Start with the definition: the right-hand side must factor into a product of one function of $x$ only and one function of $y$ only.
2
Simplify part (a) using exponent rules: $2 x y + e^{x - y} = 2 x y + e^{x} e^{- y}$ . This is a sum of two products, not a single product matching the separable form, so (a) is not separable.
3
Check part (b): $x^{2} + 3 y^{2}$ is a sum that cannot be factored into the required product form. Any attempt to factor leaves cross terms with both variables, so (b) is also not separable.
4
Conclusion: Neither equation is separable.

Exam tip:

If your DE is written as a sum of terms, check if you can factor out a common $x$ or $y$ term from the entire sum to get a product. If not, it is not separable for AP Calculus AB purposes.

2. Step-by-Step Separation of Variables Technique★★★☆☆⏱ 4 min

A general solution is the full family of functions that satisfy a first-order DE, containing one arbitrary constant to account for all possible solutions. Once you confirm a DE is separable, follow this consistent process:

Start from the separable form $\frac{d y}{d x} = g (x) h (y)$
Separate variables: move all $y$ terms (including $d y$ ) to the left, and all $x$ terms (including $d x$ ) to the right, resulting in $\frac{1}{h ( y )} d y = g (x) d x$ (valid when $h (y) \neq = 0$ )
Integrate both sides: integrate the left with respect to $y$ , the right with respect to $x$ . Add only one arbitrary constant (the difference of two constants is still one constant)
If possible, solve explicitly for $y$ to get the final general solution; leave it as an implicit solution if an explicit form is not possible

📐 Worked Example

Find the general solution of $\frac{d y}{d x} = 4 x^{2} y^{3}$ , for $y \neq = 0$

1
Confirm separability: $\frac{d y}{d x} = (4 x^{2}) (y^{3})$ , which fits the required separable form with $g (x) = 4 x^{2}$ and $h (y) = y^{3}$ .
2
Separate variables: divide both sides by $y^{3}$ and multiply by $d x$ :
3
$y^{-3} dy = 4x^2 dx$
4
Integrate both sides:
5
$\int y^{-3} dy = \int 4x^2 dx$
6
Evaluate the integrals: left side is $\frac{y ^{- 2}}{- 2} = - \frac{1}{2 y ^{2}}$ . Right side is $4 \cdot \frac{x ^{3}}{3} + C = \frac{4 x ^{3}}{3} + C$ .
7
Solve for $y$ to get an explicit general solution. Multiply both sides by $- 2$ : $\frac{1}{y ^{2}} = - \frac{8 x ^{3}}{3} - 2 C$ . Rename the arbitrary constant $K = - 2 C$ , then rewrite to simplify:
8
$y^2 = \frac{3}{C' - 8x^3}$
9
where $C^{'} = 3 K$ is still an arbitrary constant, which is the final general solution.

Exam tip:

You can rename your arbitrary constant at any point to simplify your final solution—AP exam graders accept any valid name for the constant, as long as it is clear it is arbitrary.

3. Simplifying General Solutions with Logarithms★★★☆☆⏱ 3 min

A very common case of separable DEs (especially for exponential growth/decay and cooling problems) gives a logarithmic antiderivative on the $y$ side, which requires special simplification. For the common DE form $\frac{d y}{d x} = k y$ , separating gives $\frac{1}{y} d y = k d x$ , and integrating gives $ln ∣ y ∣ = k x + C$ .

To get an explicit solution, exponentiate both sides: $∣ y ∣ = e^{k x + C} = e^{C} e^{k x}$ . We then absorb the absolute value and the constant $e^{C}$ into a new arbitrary constant $K = \pm e^{C}$ , giving $y = K e^{k x}$ . Allowing $K = 0$ also captures the trivial solution $y = 0$ from when we divided by $y$ earlier.

📐 Worked Example

Find the general solution of $\frac{d y}{d t} = \frac{y ( 2 - t )}{5}$ , for $y > 0$

1
Confirm separability: $\frac{d y}{d t} = y \cdot \frac{2 - t}{5}$ , which fits the separable form.
2
Separate variables: divide by $y$ and multiply by $d t$ :
3
$\frac{1}{y} dy = \frac{2 - t}{5} dt$
4
Integrate both sides:
5
$\int \frac{1}{y} dy = \int \frac{2 - t}{5} dt$
6
Evaluate the integrals: left side is $ln ∣ y ∣$ . Right side simplifies to $\frac{4 t - t ^{2}}{10} + C$ .
7
Since $y > 0$ , we can drop the absolute value. Exponentiate both sides to solve for $y$ :
8
$y = e^{\frac{4 t - t ^{2}}{10} + C} = e^{C} e^{\frac{4 t - t ^{2}}{10}}$ . Rename $K = e^{C}$ , so the final general solution is:
9
$y = K e^{\frac{4t - t^2}{10}}$

Exam tip:

When you have $ln ∣ y ∣ = f (x) + C$ , remember that $e^{f (x) + C} = e^{C} e^{f (x)}$ , not $e^{f (x)} + C$ —this is one of the most common student mistakes on the AP exam.

4. Concept Check★★☆☆☆⏱ 4 min

✓ Quick check

Test your understanding with this AP-style multiple choice question:

Which of the following is the general solution to the differential equation $\frac{d y}{d t} = \frac{t e ^{t}}{2 y}$ for $y > 0$ ?
- A) $y^{2} = 2 e^{t} (t - 1) + C$
- B) $y^{2} = t e^{t} - e^{t} + C$
- C) $y^{2} = 2 t e^{t} + C$
- D) $y^{2} = e^{t} (t - 1) + C$
Reveal answer
D —
To solve, separate variables to get $2 y d y = t e^{t} d t$ . Integrate the left to get $y^{2}$ , integrate the right using integration by parts to get $e^{t} (t - 1) + C$ .

5. Common Pitfalls

Wrong move:

When solving $ln y = x + C$ , exponentiating to get $y = e^{x} + C$ .

Why:

Students misapply exponent rules, forgetting that the exponent distributes as a product, not a sum.

Correct move:

Always simplify $e^{f (x) + C}$ as $e^{C} e^{f (x)}$ , so the constant becomes a multiplicative factor out front, not an additive term.

Wrong move:

Adding a constant of integration to both sides of the integrated equation and keeping two separate constants in the final general solution.

Why:

Students are taught to add a constant to every indefinite integral, so they add one to each side and do not combine them.

Correct move:

Combine the two constants into a single arbitrary constant, since the difference of two constants is still one constant.

Wrong move:

Integrating the left side (with $y$ terms) with respect to $x$ after separation.

Why:

Students mix up differentials after separation, forgetting which variable goes with which integral.

Correct move:

Always keep $d y$ with $y$ terms, so the side with $y$ terms is integrated with respect to $y$ , and the side with $x$ terms is integrated with respect to $x$ .

Wrong move:

Trying to separate $\frac{d y}{d x} = g (x) + h (y)$ by writing $\frac{1}{h ( y )} d y = g (x) d x$ .

Why:

Students confuse addition and multiplication, and incorrectly attempt to separate non-separable DEs.

Correct move:

Always confirm the right-hand side is a product of a function of $x$ and a function of $y$ before attempting separation; if it is a sum, it is not separable for AP Calculus AB.

Wrong move:

Forgetting the absolute value in $\int \frac{1}{y} d y = ln ∣ y ∣$ , leading to only positive solutions.

Why:

Students memorize the antiderivative as $ln y$ and skip the absolute value.

Correct move:

Always write the absolute value, then absorb the sign into the arbitrary constant when exponentiating, so all solutions are captured.

6. Quick Reference Cheatsheet

Category	Formula	Notes
Separable DE Definition	$\frac{d y}{d x} = g (x) h (y)$	RHS must be product of $x$ -only and $y$ -only functions
Separation Step	$\frac{1}{h ( y )} d y = g (x) d x$	Valid for $h (y) \neq = 0$ ; constant solution exists if $h (c) = 0$
Integration Step	$\int \frac{1}{h ( y )} d y = \int g (x) d x + C$	Only one arbitrary constant needed for first-order DE
Antiderivative of $1/ y$	$\int \frac{1}{y} d y = ln ∣ y ∣ + C$	Always include absolute value; sign absorbed into constant
Exponentiation for Log Solutions	$ln ∣ y ∣ = f (x) + C ⟹ y = K e^{f (x)}$	$K = \pm e^{C}$ is arbitrary; $K = 0$ captures trivial solution
Implicit General Solution	$F (x, y) = C$	Acceptable if $y$ cannot be solved explicitly
Explicit General Solution	$y = f (x, C)$	Preferred by AP exam when solvable

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Identify separable differential equation
2022 · FRQ
Find general solution of separable DE

What's Next

Mastering separation of variables to find general solutions is the essential prerequisite for the next core topics in Unit 7 of AP Calculus AB. Without correctly finding the general solution first, you cannot solve for the constant of integration to get a particular solution that matches a given initial condition, which is required for all real-world differential equation problems on the exam. This topic also underpins all applied differential equation problems on the AP exam, including exponential growth and decay, Newton's law of cooling, and rate problems that require integrating to find a function from its rate of change. Next you will apply the skills you learned here to solve initial value problems and contextual application questions that commonly appear in both sections of the exam.