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

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: Identifying separable first-order differential equations, step-by-step separation of variables technique, integrating both sides to find implicit and explicit general solutions, and simplifying constants of integration for final results.

You should already know: How to compute indefinite integrals of basic functions, implicit differentiation rules, basic algebraic factoring and rearrangement.

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 AB 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 primary technique AP Calculus AB uses to solve first-order ordinary differential equations (DEs). A general solution is a full family of functions that satisfies the DE, containing one arbitrary constant (matching the order of the DE) to account for all possible solutions. Per the AP Calculus AB CED, differential equations as a unit make up 6-12% of the total exam weight, and separation of variables for general solutions appears in both multiple-choice (MCQ) and free-response (FRQ) sections. The core idea of this method is that any separable DE can be rearranged to group all terms containing the dependent variable (usually $y$) on one side of the equation, and all terms containing the independent variable (usually $x$ or $t$) on the other. Once separated, we can integrate both sides with respect to their respective variables to get an equation relating $y$ and the independent variable, which can often be solved explicitly for $y$ to get the general solution. This method reverses the process of differentiation, starting from a derivative and working back to the original function.

2. Identifying Separable Differential Equations

Before you can apply separation of variables, you first need to confirm that your differential equation is actually separable. A first-order differential equation is always written in the general form: $$\frac{dy}{dx}=f(x,y)$$ A separable DE is one 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 $y$, or: $$\frac{dy}{dx}=g(x)\cdot h(y)$$ There are no exceptions to this rule for AP Calculus AB: if you cannot write the right-hand side in this form, the DE is not separable, and you cannot solve it with separation of variables (and AP will not ask you to solve non-separable DEs with this technique). Common simplifications that reveal hidden separability include using exponent rules to split $e^{x\pm y}=e^x{e}^{\pm y}$, and factoring out common terms from sums to get a product. For example, $\frac{dy}{dx}=xy+5x$ can be factored to $x(y+5)$, which is separable.

Worked Example

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

1. Start with the definition of separability: $f(x,y)$ must factor to $g(x)h(y)$.
2. Simplify (a) using exponent rules: $2xy+e^{x-y}=2xy+e^x{e}^{-y}$. Wait, this is a sum of two products, not a single product, so it cannot be written as $g(x)h(y)$. Wait—wait a minute, did we make a mistake? No: even though each term is separable, their sum is not. So (a) is not separable.
3. Check (b): $x^2+3y^2$ is a sum, and it cannot be factored into a product of a function of $x$ only and a function of $y$ only. Any attempt to factor will leave a cross term with both $x$ and $y$, 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.

3. Step-by-Step Separation of Variables Technique

Once you confirm a DE is separable, you can follow a consistent process to find the general solution. Starting from $\frac{dy}{dx}=g(x)h(y)$, the first step is to separate the variables: move all $y$ terms (including $dy$) to the left side and all $x$ terms (including $dx$) to the right side. To do this, multiply both sides by $dx$ and divide both sides by $h(y)$ (this is valid as long as $h(y)\ne 0$; the case where $h(y)=0$ gives a constant trivial solution, which can be absorbed into the general solution later). After separation, you get: $$\frac{1}{h(y)}dy=g(x)dx$$ Next, integrate both sides: integrate the left side with respect to $y$ (since it has only $y$ terms and $dy$) and the right side with respect to $x$ (since it has only $x$ terms and $dx$). You only need to add one arbitrary constant of integration, usually on the right side, because the difference of two constants is still a single constant. The result is an implicit general solution; if possible, solve for $y$ to get an explicit general solution, which is usually what the AP exam asks for.

Worked Example

Find the general solution of $\frac{dy}{dx}=4x^2{y}^3$, for $y\ne 0$.

1. Confirm separability: $\frac{dy}{dx}=(4x^2)(y^3)$, so it fits the separable form with $g(x)=4x^2$ and $h(y)=y^3$.
2. Separate variables: Divide both sides by $y^3$ and multiply by $dx$, giving $y^{-3}dy=4x^{2}dx$.
3. Integrate both sides: $$\int y^{-3}dy=\int 4x^{2}dx$$ Left side: $\frac{y^{-2}}{-2}=-\frac{1}{2y^2}$. Right side: $4\cdot \frac{x^3}{3}+C=\frac{4x^3}{3}+C$.
4. Solve for $y$ to get explicit general solution: Multiply both sides by $-2$: $\frac{1}{y^2}=-\frac{8x^3}{3}-2C$. Rename the arbitrary constant $K=-2C$, so $\frac{1}{y^2}=K-\frac{8x^3}{3}$. Take the reciprocal of both sides: $y^2=\frac{1}{K-\frac{8x^3}{3}}=\frac{3}{3K-8x^3}$. The general solution can be written as $y^2=\frac{3}{C^{\prime }-8x^3}$ where $C^{\prime }=3K$ is still an arbitrary constant.

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.

4. Simplifying General Solutions with Logarithms

One of the most common forms of separable DEs results in a logarithmic antiderivative on the $y$ side, which requires special simplification to get an explicit general solution. This occurs almost always in applied problems like exponential growth and decay, where the DE has the form $\frac{dy}{dx}=ky$. When we separate this DE, we get $\frac{1}{y}dy=kdx$, and integrating gives $\ln |y|=kx+C$. To solve for $y$, we exponentiate both sides to eliminate the logarithm: $e^{\ln |y|}=e^{kx+C}$, which simplifies to $|y|=e^C{e}^{kx}$. We can absorb the absolute value and the constant $e^C$ into a new arbitrary constant $K=\pm e^C$, so the general solution becomes $y=K{e}^{kx}$. If we allow $K=0$, this also captures the trivial solution $y=0$ that we get when we divided by $y$ earlier, so it is included in the general solution.

Worked Example

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

1. Confirm separability: $\frac{dy}{dt}=y\cdot \frac{2-t}{5}$, which fits the separable form.
2. Separate variables: Divide both sides by $y$ and multiply by $dt$, giving $\frac{1}{y}dy=\frac{2-t}{5}dt$.
3. Integrate both sides: $$\int \frac{1}{y}dy=\int \frac{2-t}{5}dt$$ Left side: $\ln |y|$. Right side: $\frac{1}{5}\left(2t-\frac{t^2}{2}\right)+C=\frac{4t-t^2}{10}+C$.
4. Exponentiate both sides to solve for $y$: Since $y>0$, we can drop the absolute value, so $y=e^{\frac{4t-t^2}{10}+C}=e^C{e}^{\frac{4t-t^2}{10}}$. Rename $K=e^C$, so the general solution is $y=K{e}^{\frac{4t-t^2}{10}}$, where $K>0$ is an arbitrary constant.

Exam tip: When you have $\ln |y|=f(x)+C$, always 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 exam.

5. Common Pitfalls (and how to avoid them)

• 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 $dy$ 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 $\frac{dy}{dx}=g(x)+h(y)$ by writing $\frac{1}{h(y)}dy=g(x)dx$. 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 $\int \frac{1}{y}dy=\ln |y|$, 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. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

Which of the following is the general solution to the differential equation $\frac{dy}{dt}=\frac{t{e}^t}{2y}$ for $y>0$? A) $y^2=2e^t(t-1)+C$ B) $y^2=t{e}^t-e^t+C$ C) $y^2=2t{e}^t+C$ D) $y^2=e^t(t-1)+C$

Worked Solution: First, confirm the DE is separable: it can be written as $\frac{dy}{dt}=(t{e}^t)\cdot (\frac{1}{2y})$, so it fits the separable form. Separate variables to get $2ydy=t{e}^{t}dt$. Integrate the left side: $\int 2ydy=y^2+C_1$. Integrate the right side using integration by parts, with $u=t$, $dv=e^{t}dt$: $\int t{e}^{t}dt=t{e}^t-\int e^{t}dt=t{e}^t-e^t+C_2=e^t(t-1)+C_2$. Combine constants: $y^2=e^t(t-1)+(C_2-C_1)=e^t(t-1)+C$. The correct answer is D.

Question 2 (Free Response)

Consider the differential equation $\frac{dy}{dx}=\frac{y(2-x)}{5}$. (a) Is the differential equation separable? Justify your answer. (b) Use separation of variables to find the general solution to the differential equation. (c) Explain why the general solution contains one arbitrary constant.

Worked Solution: (a) Yes, the differential equation is separable. The right-hand side can be factored into $\frac{dy}{dx}=\left(\frac{2-x}{5}\right)\cdot y$, which matches the separable form $\frac{dy}{dx}=g(x)h(y)$, where $g(x)$ depends only on $x$ and $h(y)$ depends only on $y$.

(b) Separate variables: $\frac{1}{y}dy=\frac{2-x}{5}dx$ for $y\ne 0$. Integrate both sides: $\ln |y|=\frac{1}{5}\left(2x-\frac{x^2}{2}\right)+C=\frac{4x-x^2}{10}+C$. Exponentiate both sides: $|y|=e^{\frac{4x-x^2}{10}+C}=e^C{e}^{\frac{4x-x^2}{10}}$. Absorb the absolute value and constant into $K=\pm e^C$, so the general solution is $y=K{e}^{\frac{4x-x^2}{10}}$ for any arbitrary constant $K$ (including $K=0$, which gives the trivial solution $y=0$).

(c) The differential equation is first-order, meaning it only contains the first derivative of $y$ with no higher-order derivatives. An $n$-th order differential equation has a general solution with $n$ arbitrary constants, so a first-order DE has exactly one arbitrary constant that represents the entire family of solutions.

Question 3 (Application / Real-World Style)

A coffee shop owner is modeling the cooling of a freshly brewed cup of coffee. The rate of change of the temperature difference between the coffee and the room is proportional to the current temperature difference, where temperature difference $D$ is measured in degrees Fahrenheit, and time $t$ is measured in minutes. This follows the differential equation $\frac{dD}{dt}=-0.08D$. Find the general solution for $D(t)$, and explain what the arbitrary constant represents in context.

Worked Solution: The DE is separable for $D>0$ (the coffee is warmer than room temperature here). Separate variables to get $\frac{1}{D}dD=-0.08dt$. Integrate both sides: $\ln D=-0.08t+C$ (drop absolute value since $D>0$). Exponentiate: $D=e^{-0.08t+C}=e^C{e}^{-0.08t}=D_0{e}^{-0.08t}$, where $D_0=e^C$ is a positive constant. In context, the arbitrary constant $D_0$ is the initial temperature difference between the coffee and the room at time $t=0$, when the coffee is first brewed. The general solution is $D(t)=D_0{e}^{-0.08t}$ degrees Fahrenheit.

7. Quick Reference Cheatsheet

8. What's Next

Mastering separation of variables to find general solutions is the essential prerequisite for the next core topics in Unit 7: finding particular solutions given an initial condition, which is required for all real-world differential equation problems on the AP exam. Without correctly finding the general solution first, you cannot solve for the constant of integration to get a particular solution that matches your initial condition. 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 MCQ and FRQ sections of the exam.

Finding particular solutions to differential equations Exponential growth and decay models Slope fields and solution curves