# General solutions via separation of variables

> AP Calculus BC · AP Calculus BC CED Unit 7
> Source: https://www.owlsprep.com/study/ap-calculus-bc-u7-general-solutions-via-separation-of/

This module covers identifying separable first-order ODEs, separating variables, integrating both sides, combining constants, and finding implicit and explicit general solutions for AP Calculus BC.

**Prerequisites:** Basic integration of common functions with u-substitution; Implicit function definition and derivative rules; Algebraic manipulation of exponential/logarithmic expressions

## Learning objectives

- Identify separable first-order ordinary differential equations
- Apply separation of variables to find general solutions of separable ODEs
- Distinguish between implicit and explicit general solutions
- Avoid common exam pitfalls when solving separable differential equations

## Core Definitions

**Separable first-order ordinary differential equation** — A first-order ODE that can be factored into a product of a function that depends only on $x$ and a function that depends only on $y$.

*Notation:* \frac{dy}{dx} = f(x)g(y)

*Example:* $\frac{dy}{dx} = x^2 y$ is separable; $\frac{dy}{dx} = x+y$ is not.

**General solution** — The full family of all functions that satisfy an ODE, parameterized by one arbitrary constant for each order of the equation. For a first-order ODE, there is exactly one arbitrary constant.

*Example:* $y = K e^{x^2}$ is a general solution for $\frac{dy}{dx} = 2xy$.

Separation of variables is the foundational solution technique for first-order ODEs tested on both AP Calculus BC MCQ and FRQ, accounting for approximately 2-4% of the total exam score per the official Course and Exam Description. It is required for almost all applied differential equation problems on the exam.

> **Exam tip:** Always confirm the number of arbitrary constants in your general solution matches the order of the ODE.

## Identifying Separable ODEs

Before you can use separation of variables, you must first correctly identify if a given ODE is separable. By definition, a first-order ODE is separable if it can be rearranged to fit $\frac{dy}{dx} = f(x)g(y)$, where $f(x)$ has no dependence on $y$ and $g(y)$ has no dependence on $x$.

> **warning**
>
> Sums of terms with both $x$ and $y$ are almost never separable. For example, $\frac{dy}{dx} = x + y$ is not separable, as it cannot be factored into the required product form.

**Worked example:** Determine if $\frac{dy}{dx} = \frac{x^2 y}{e^y \sqrt{x^2 + 4}}$ is separable, and rewrite it in standard separated form if it is.

1. Confirm $\frac{dy}{dx}$ is already isolated, as given.
2. Factor the right-hand side to test separability:
3. $$\frac{x^2 y}{e^y \sqrt{x^2 + 4}} = \left(\frac{x^2}{\sqrt{x^2 + 4}}\right) \left(\frac{y}{e^y}\right)$$
4. Verify that no cross terms of $x$ and $y$ appear: the first factor depends only on $x$, the second only on $y$, so the ODE is separable.
5. Rearrange to separate variables: multiply both sides by $\frac{e^y}{y} dx$ (for $y \neq 0$) to get the standard separated form:
6. $$\frac{e^y}{y} dy = \frac{x^2}{\sqrt{x^2 + 4}} dx$$

> **Exam tip:** On MCQ questions asking to identify a separable equation, eliminate any option where the right-hand side is a sum of terms with both $x$ and $y$ immediately.

## Integration and Combining Constants

Once you have separated variables into the form $g(y) dy = f(x) dx$, the next step is to integrate each side. The differential tells you the variable of integration: integrate the side with $dy$ with respect to $y$, and the side with $dx$ with respect to $x$.

A common point of confusion is why we only keep one constant of integration instead of two. When you integrate both sides, you get:

Rearranging gives $G(y) = F(x) + (C_2 - C_1)$. Since the difference of two arbitrary constants is still an arbitrary constant, we replace $C = C_2 - C_1$ and just write $G(y) = F(x) + C$, which simplifies work without losing any generality.

**Worked example:** Given the separated ODE $\frac{1}{y} dy = 2x dx$, find the general solution after integration.

1. Confirm the separated form is correct, with $y$-terms on the left and $x$-terms on the right.
2. Integrate both sides with respect to their respective variables:
3. $$\int \frac{1}{y} dy = \int 2x dx$$
4. Evaluate each integral: the left integral gives $\ln|y| + C_1$, and the right integral gives $x^2 + C_2$.
5. Combine constants into a single arbitrary constant $C$:
6. $$\ln|y| = x^2 + C$$
7. Simplify to explicit form: exponentiate both sides to get $|y| = e^{x^2 + C} = e^C e^{x^2}$, which can be rewritten as:
8. $$y = K e^{x^2}$$
9. where $K = \pm e^C$ is still an arbitrary constant.

> **Exam tip:** Always combine constants immediately after integration; writing two separate constants is never required on the AP exam and often leads to unnecessary algebra mistakes.

## Implicit vs Explicit General Solutions

After integration, you will have an equation relating $x$, $y$, and the arbitrary constant $C$. If $y$ is not isolated on one side of the equation, this is called an implicit general solution. If you can rearrange to write $y$ explicitly as a function of $x$ and $C$, this is called an explicit general solution. The AP exam will usually specify which form it requires, but if it does not, either form is acceptable as long as it is correct, though explicit form is preferred if it can be obtained easily.

**Worked example:** Find the general solution of $y^2 \frac{dy}{dx} = \cos x$, give both implicit and explicit forms.

1. Separate variables: multiply both sides by $dx$ to get:
2. $$y^2 dy = \cos x dx$$
3. Integrate both sides and combine constants:
4. $$\int y^2 dy = \int \cos x dx \implies \frac{y^3}{3} = \sin x + C$$
5. This is the implicit general solution.
6. To get the explicit solution, multiply by 3 and take the cube root. Let $K = 3C$ (still an arbitrary constant), so the explicit general solution is:
7. $$y = \sqrt[3]{3\sin x + K}$$
8. Verify: the ODE is first-order, and we have exactly one arbitrary constant, which meets the requirement for a general solution.

**Check your understanding**

Test your understanding of separability:

1. Which of the following differential equations is separable?

   - A) $\frac{dy}{dx} = x^2 + 3xy$
   - B) $\frac{dy}{dx} = \sin(x + y)$
   - C) $\frac{dy}{dx} = \frac{x^3}{y^2 + 1}$
   - D) $\frac{dy}{dx} = 2x + 5y$

   *Why:* Only option C can be factored into $\frac{dy}{dx} = (x^3) \cdot \left(\frac{1}{y^2 + 1}\right)$, which fits the definition of a separable ODE. The other options cannot be split into a product of single-variable functions.

> **Exam tip:** If you are asked for a general solution and can easily simplify to explicit form, always do so — it will never be penalized, and it is easier for graders to confirm correctness.

## Common pitfalls

- **Wrong:** Factoring $\frac{dy}{dx} = x + y$ as $\frac{dy}{dx} = x\left(1 + \frac{y}{x}\right)$ and calling it separable.
  - Why it fails: Students confuse any factoring with the requirement for separation, forgetting that all factors must be functions of only one variable.
  - Correct: After factoring, check every factor to confirm it contains only one variable; any factor with both $x$ and $y$ means the ODE is not separable.
- **Wrong:** Forgetting to check for constant solutions after dividing by a function of $y$ during separation, e.g., when dividing by $y$ to solve $\frac{dy}{dx} = xy$, not noting that $y=0$ is a valid solution.
  - Why it fails: Students focus on rearranging terms and forget that division by zero is undefined, so the case where the divisor equals zero is excluded from the general form.
  - Correct: Whenever you divide by $h(y)$ to separate variables, check if $h(y) = 0$ gives a constant function $y=k$ that satisfies the original ODE, and note it as a solution if it does.
- **Wrong:** Integrating the left side (with $dy$) with respect to $x$, or the right side (with $dx$) with respect to $y$, e.g., turning $\frac{1}{y} dy = 2x dx$ into $\int \frac{1}{y} dx = \int 2x dy$.
  - Why it fails: Students confuse the differential, which tells you the variable of integration, with the variable in the integrand.
  - Correct: Always match the variable of integration to the differential: $dy$ means integrate with respect to $y$, and $dx$ means integrate with respect to $x$.
- **Wrong:** Simplifying $y = e^{\ln x + C}$ to $y = x + C$ instead of $y = Cx$.
  - Why it fails: Students forget that $e^{a + C} = e^a e^C$, and $e^C$ is a multiplicative arbitrary constant, not an additive constant.
  - Correct: When exponentiating after integration, any arbitrary constant inside the exponent becomes a multiplicative constant after exponentiation.
- **Wrong:** Leaving two arbitrary constants in the final general solution of a first-order ODE.
  - Why it fails: Students think each integral needs its own constant, and forget that two constants can be combined.
  - Correct: Combine all constants into a single arbitrary constant immediately after integration for any first-order ODE.

## Cheatsheet

| Category | Formula / Rule | Notes |
| --- | --- | --- |
| Definition of separable first-order ODE | $\frac{dy}{dx} = f(x)g(y)$ | $f(x)$ depends only on $x$, $g(y)$ depends only on $y$ |
| Standard separated form | $g(y) dy = f(x) dx$ | Check if $g(y)=0$ gives an extra constant solution after dividing |
| Integration step | $\int g(y) dy = \int f(x) dx + C$ | One arbitrary constant for first-order ODE; combine two constants from separate integrals into one |
| Implicit general solution | $G(y) = F(x) + C$ | Valid form if $y$ cannot be easily isolated; required if explicitly asked for |
| Exponential simplification | $y = e^{F(x) + C} = C' e^{F(x)}$ | $C' = e^C$ is an arbitrary multiplicative constant; never write $y = e^{F(x)} + C$ |
| Number of arbitrary constants | 1 for any first-order ODE | Equal to the order of the ODE; any more or fewer indicates a mistake |
| Constant solution check | If $y=k$ makes $g(y)=0$, substitute back to test | Division by zero excludes this solution from the general form, so it must be checked separately |
| Explicit general solution | $y = F(x, C)$ | $y$ is isolated in terms of $x$ and $C$; preferred if it can be obtained easily |

## What's next

General solutions via separation of variables is the foundational prerequisite for every differential equation topic that follows in AP Calculus BC Unit 7. Immediately after finding general solutions, you will learn to find particular solutions by applying initial conditions to fix the arbitrary constant $C$, which is tested regularly on both MCQ and FRQ sections of the exam. Separation of variables is also the core technique used to solve exponential growth and decay models, logistic growth models, and related application problems that often make up a full FRQ question. Without mastering this topic, you will not be able to correctly set up or solve these problems, or master more advanced topics like Euler's method.

- [Particular solutions with initial conditions](https://www.owlsprep.com/study/ap-calculus-bc-u7-particular-solutions-with-initial-conditions/)
- [Exponential models with differential equations](https://www.owlsprep.com/study/ap-calculus-bc-u7-exponential-models-with-differential-equations/)
- [Logistic models with differential equations (BC only)](https://www.owlsprep.com/study/ap-calculus-bc-u7-logistic-models-with-differential-equations/)

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