General solutions via separation of variables — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: This chapter covers identifying separable first-order ordinary differential equations, separating variable terms to opposite sides of the equation, integrating both sides, combining constants of integration, and finding implicit and explicit general solutions.

You should already know: Basic integration of common functions with u-substitution. Implicit function definition and derivative rules. Algebraic factoring and manipulation of exponential/logarithmic expressions.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus BC style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is General solutions via separation of variables?

Separation of variables is the foundational solution technique for first-order ordinary differential equations (ODEs) tested on both AP Calculus BC multiple choice (MCQ) and free response questions (FRQ), accounting for approximately 2-4% of total exam score per the official AP Course and Exam Description (CED). A general solution of a first-order ODE is the full family of all functions that satisfy the equation, parameterized by a single arbitrary constant of integration (unlike a particular solution, which fixes this constant to match a given initial condition). A separable first-order ODE is any ODE that can be factored into the form $\frac{dy}{dx}=f(x)g(y)$, meaning the derivative of $y$ with respect to $x$ splits into a product of a function that depends only on $x$ and a function that depends only on $y$. This technique works by moving all $y$-terms (with $dy$) and all $x$-terms (with $dx$) to opposite sides of the equation, then integrating each side independently to get the general solution. It is required for almost all applied differential equation problems on the AP exam.

2. Identifying Separable Differential Equations

Before you can use separation of variables, you must first correctly identify if a given ODE is separable. By definition, a first-order ODE (which relates $x$, $y$, and $\frac{dy}{dx}$) is separable if it can be rearranged to fit the factored form $\frac{dy}{dx}=f(x)g(y)$, where $f(x)$ has no dependence on $y$ and $g(y)$ has no dependence on $x$. Many student errors start here: misclassifying non-separable equations as separable, or vice versa. For example, the simple-looking ODE $\frac{dy}{dx}=x+y$ is not separable, because it cannot be factored into the required product form. Another common form of separable ODEs is written as $M(x)+N(y)\frac{dy}{dx}=0$, which rearranges directly to $N(y)dy=-M(x)dx$, which fits the separable definition. When checking for separability, always isolate $\frac{dy}{dx}$ first, then test if the right-hand side can be split into a product of two single-variable functions.

Worked Example

Problem: 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. Step 1: Confirm $\frac{dy}{dx}$ is already isolated, as given.
2. Step 2: Factor the right-hand side to test separability: $\frac{x^{2}y}{e^y\sqrt{x^2+4}}=\left(\frac{x^2}{\sqrt{x^2+4}}\right)\left(\frac{y}{e^y}\right)$.
3. Step 3: 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.
4. Step 4: Rearrange to separate variables: multiply both sides by $\frac{e^y}{y}dx$ (for $y\ne 0$) to get $\frac{e^y}{y}dy=\frac{x^2}{\sqrt{x^2+4}}dx$, which is the standard separated form.

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 — these are almost never separable.

3. Integrating Both Sides 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 key rule here is that you integrate the side with $dy$ with respect to $y$, and the side with $dx$ with respect to $x$ — the differential tells you the variable of integration. A common point of confusion is why we only keep one constant of integration instead of two. When you integrate both sides, you get: $$\int g(y)dy=\int f(x)dx$$ $$G(y)+C_1=F(x)+C_2$$ where $C_1$ and $C_2$ are arbitrary constants. Rearranging gives $G(y)=F(x)+(C_2-C_1)$. Since the difference of two arbitrary constants is still an arbitrary constant, we can replace $C=C_2-C_1$ and just write $G(y)=F(x)+C$, which simplifies work without losing any generality. This integration step gives you a general solution, with the single arbitrary constant required for a first-order ODE.

Worked Example

Problem: Given the separated ODE $\frac{1}{y}dy=2xdx$, find the general solution after integration.

1. Step 1: Confirm the separated form is already correct, with $y$-terms on the left and $x$-terms on the right.
2. Step 2: Integrate both sides with respect to their respective variables: $\int \frac{1}{y}dy=\int 2xdx$.
3. Step 3: Evaluate each integral: the left integral gives $\ln |y|+C_1$, and the right integral gives $x^2+C_2$.
4. Step 4: Combine constants: rearrange to $\ln |y|=x^2+(C_2-C_1)$, then replace the difference with a single arbitrary constant $C$, giving $\ln |y|=x^2+C$.
5. Step 5: Simplify to explicit form if desired: exponentiate both sides to get $|y|=e^{x^2+C}=e^C{e}^{x^2}$, which can be rewritten as $y=K{e}^{x^2}$ 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 it often leads to unnecessary algebra mistakes when simplifying your final solution.

4. 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 specify, either form is acceptable as long as it is correct, though explicit form is preferred if it can be obtained without excessive complication. An important rule to remember: the number of arbitrary constants in a general solution equals the order of the ODE (the highest derivative present). For a first-order ODE, you must have exactly one arbitrary constant — if you have more or fewer, you made a mistake in your work.

Worked Example

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

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

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.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Factoring $\frac{dy}{dx}=x+y$ as $\frac{dy}{dx}=x\left(1+\frac{y}{x}\right)$ and calling it separable. Why: Students confuse any factoring with the requirement for separation, forgetting that all factors must be functions of only one variable. Correct move: 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 move: 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: 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 move: 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 move: Integrating the left side (with $dy$) with respect to $x$, or the right side (with $dx$) with respect to $y$, e.g., turning $\frac{1}{y}dy=2xdx$ into $\int \frac{1}{y}dx=\int 2xdy$. Why: Students confuse the differential, which tells you the variable of integration, with the variable in the integrand. Correct move: 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 move: Simplifying $y=e^{\ln x+C}$ to $y=x+C$ instead of $y=Cx$. Why: Students forget that $e^{a+C}=e^a{e}^C$, and $e^C$ is a multiplicative arbitrary constant, not an additive constant. Correct move: When exponentiating after integration, any arbitrary constant inside the exponent becomes a multiplicative constant after exponentiation.
• Wrong move: Leaving two arbitrary constants in the final general solution of a first-order ODE. Why: Students think each integral needs its own constant, and forget that two constants can be combined. Correct move: Combine all constants into a single arbitrary constant immediately after integration for any first-order ODE.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

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) $\frac{dy}{dx}=2x+5y$

Worked Solution: To be separable, an ODE must be able to be written as $\frac{dy}{dx}=f(x)g(y)$, where $f$ depends only on $x$ and $g$ depends only on $y$. Option A has a cross term $3xy$ that cannot be factored into the required product form, so it is not separable. Option B's $\sin (x+y)$ cannot be split into a product of single-variable functions, so it is not separable. Option C can be factored as $(x^3)\cdot \left(\frac{1}{y^2+1}\right)$, which fits the definition of a separable ODE. Option D's sum $2x+5y$ cannot be factored into the required product form, so it is not separable. The correct answer is C.

Question 2 (Free Response)

Consider the differential equation $\frac{dy}{dx}=\frac{3x^2(y-2)}{x^2+1}$. (a) Show the differential equation is separable and write it in separated form. (b) Find the general solution of the differential equation. (c) Is the constant solution $y=2$ included in the general solution you found in part (b)? Explain.

Worked Solution: (a) Isolate $\frac{dy}{dx}$ and factor the right-hand side: $\frac{3x^2(y-2)}{x^2+1}=\left(\frac{3x^2}{x^2+1}\right)(y-2)$. This is a product of a function of $x$ only and a function of $y$ only, so the ODE is separable. For $y\ne 2$, separating variables gives $\frac{1}{y-2}dy=\frac{3x^2}{x^2+1}dx$.

(b) Integrate both sides and combine constants: $\int \frac{1}{y-2}dy=\int \frac{3x^2}{x^2+1}dx$. The left integral evaluates to $\ln |y-2|$. For the right integral, use u-substitution with $u=x^2+1$, $du=2xdx$, and rewrite $3x^2=3(x^2+1-1)=3u-3$, leading to $\int \frac{3(u-1)}{u}\cdot \frac{du}{2x}\cdot x=\frac{3}{2}\int \left(1-\frac{1}{u}\right)du=\frac{3}{2}(x^2+1-\ln (x^2+1))+C$. This gives the general solution $\ln |y-2|=\frac{3}{2}{x}^2-\frac{3}{2}\ln (x^2+1)+C$, or in explicit form $y=2+K(x^2+1{)}^{-3/2}{e}^{\frac{3}{2}{x}^2}$ where $K$ is an arbitrary non-zero constant.

(c) The constant solution $y=2$ is a valid solution of the original ODE: substituting $y=2$ gives $\frac{dy}{dx}=0$, and the right-hand side is $\frac{3x^2(0)}{x^2+1}=0$, so it satisfies the ODE. It is not included in the general solution from part (b) because we divided by $(y-2)$ during separation, which requires $y\ne 2$, so it only appears if we allow $K=0$, which is not part of the general form derived from separation.

Question 3 (Application / Real-World Style)

The rate of change of the mass $m$ of a melting ice block (in kilograms) with respect to time $t$ (in hours) is proportional to the current mass of the block. This relationship is described by the differential equation $\frac{dm}{dt}=-km$, where $k>0$ is a constant that depends on temperature. Find the general solution for $m(t)$ in terms of $k$ and the arbitrary constant $C$, and interpret the result in context.

Worked Solution: The ODE is separable: $\frac{dm}{dt}=-km=(-k)\cdot m$, which fits the separable form. Separate variables (for $m\ne 0$): $\frac{1}{m}dm=-kdt$. Integrate both sides: $\int \frac{1}{m}dm=\int -kdt⟹\ln |m|=-kt+C$. Exponentiate both sides: $|m|=e^{-kt+C}=e^C{e}^{-kt}$, so $m(t)=C{e}^{-kt}$ where $C=\pm e^C$ is a positive arbitrary constant corresponding to the initial mass of the ice block. This general solution means the mass of the melting ice block decays exponentially over time, approaching zero mass as time increases, which matches the physical behavior of melting ice.

7. Quick Reference Cheatsheet

8. 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 separation of variables and finding general solutions, you will not be able to correctly set up or solve these problems, or master more advanced topics like Euler's method.

• Particular solutions to separable differential equations
• Exponential growth and decay models
• Logistic differential equations and models
• Slope fields and solution curves