Verifying solutions for differential equations — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Verifying general and particular solutions to first- and second-order ordinary differential equations, including explicit and implicit solutions, substitution of derivatives into the original DE, and checking satisfaction of initial and boundary conditions.

You should already know: How to compute first and second derivatives using the chain rule. How to perform implicit differentiation. How to simplify algebraic expressions with arbitrary constants.

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 Verifying solutions for differential equations?

A differential equation (DE) is any equation that relates an unknown function $y=f(x)$ to its derivatives (e.g., $y^{\prime },y^{\prime \prime }$). Verifying a solution is the process of confirming that a given candidate function (or family of functions) satisfies the DE and any given initial or boundary conditions. According to the AP Calculus BC Course and Exam Description (CED), this topic is part of Unit 7: Differential Equations, which accounts for 6–12% of the total AP exam score. Verifying solutions appears regularly on both multiple-choice (MCQ) and free-response (FRQ) sections of the exam. It is almost always a low-to-medium difficulty point-earner, but it is often paired with other DE topics like separation of variables or logistic growth. Many students mistakenly treat this as a trivial "plug and chug" task with no nuance, but the AP exam regularly tests implicit solutions, second-order DEs, and checking that particular solutions satisfy initial conditions, not just the DE itself. Unlike solving a DE from scratch, verification is a backwards process: you start with a candidate, so you do not have to find the solution yourself, just confirm it works.

2. Verifying Explicit Solutions to Differential Equations

An explicit solution is a candidate function written in the form $y=f(x)$, where $y$ is fully isolated on one side of the equation. This is the most common form of solution you will encounter on the AP exam. To verify an explicit solution, follow a consistent three-step process: 1) Identify the order of the differential equation, and compute all required derivatives of the candidate (for an $n$-th order DE, you need up to the $n$-th derivative of $y$); 2) Substitute the candidate $y$ and all its computed derivatives into the left-hand side (LHS) of the DE; 3) Simplify the LHS and confirm it equals the right-hand side (RHS) of the DE for all $x$ in the domain of the solution. For general solutions (which include one or more arbitrary constants), the equality will hold regardless of the value of the constant, since constants differentiate to zero.

Worked Example

Problem: Verify that $y=3e^{2x}+5x$ is a solution to the second-order differential equation $y^{\prime \prime }-2y^{\prime }=-10$.

1. Compute the first derivative of the candidate: $y^{\prime }=\frac{d}{dx}\left(3e^{2x}+5x\right)=6e^{2x}+5$.
2. Compute the second derivative (required for this second-order DE): $y^{\prime \prime }=\frac{d}{dx}\left(6e^{2x}+5\right)=12e^{2x}$.
3. Substitute $y^{\prime }$ and $y^{\prime \prime }$ into the LHS of the DE: $y^{\prime \prime }-2y^{\prime }=12e^{2x}-2\left(6e^{2x}+5\right)$.
4. Simplify the LHS: $12e^{2x}-12e^{2x}-10=-10$, which matches the RHS of the DE.
5. Conclusion: The equality holds for all real $x$ (the domain of $y$), so $y=3e^{2x}+5x$ is a valid solution.

Exam tip: On multiple-choice questions asking to identify a valid solution, you can save time by eliminating options immediately if the highest-order derivative does not match after substitution, no need to fully simplify every option.

3. Verifying Implicit Solutions to Differential Equations

Many solutions to differential equations cannot be rewritten to isolate $y$ explicitly as a function of $x$, so we work with implicit solutions where $x$ and $y$ appear on both sides of the equation. To verify an implicit solution, you still need to find the derivative required by the DE, but you use implicit differentiation with respect to $x$ to get $y^{\prime }$ (or higher derivatives) in terms of $x$ and $y$, then compare your result to the DE. You do not need to solve for $y$ explicitly to complete verification, which is a common point of confusion for students. Implicit solution verification is most often tested in FRQ sections, paired with implicit differentiation skills from earlier in the course.

Worked Example

Problem: Verify that $\sin y+2xy=x^2$ is an implicit solution to $\frac{dy}{dx}=\frac{2x-2y}{2x+\cos y}$.

1. Differentiate both sides of the implicit candidate with respect to $x$, applying the chain rule to $y$-terms and the product rule to $2xy$: $\frac{d}{dx}(\sin y)+\frac{d}{dx}(2xy)=\frac{d}{dx}(x^2)$.
2. Expand each derivative term: $(\cos y)y^{\prime }+2y+2x{y}^{\prime }=2x$.
3. Isolate $y^{\prime }$ by factoring: $y^{\prime }(2x+\cos y)=2x-2y$, so $y^{\prime }=\frac{2x-2y}{2x+\cos y}$.
4. Compare the computed $y^{\prime }$ to the RHS of the original DE: they are identical. The candidate is therefore a valid implicit solution.

Exam tip: Always write out the $y^{\prime }$ term explicitly after differentiating any $y$-term—forgetting the chain rule $y^{\prime }$ is the most common error on implicit verification problems.

4. Verifying Particular Solutions with Initial Conditions

A general solution to an $n$-th order DE includes $n$ arbitrary constants, meaning it describes an infinite family of solutions. A particular solution fixes the value of these constants to match given initial conditions (for first-order DEs: $y(x_0)=y_0$) or boundary conditions. To verify a particular solution, you must complete two separate checks: first, confirm the candidate satisfies the DE itself, and second, confirm it satisfies the given initial or boundary condition. AP exam questions regularly require both checks, and award separate points for each step, so skipping the initial condition check will cost you points.

Worked Example

Problem: Given that $y=C{e}^{-2x}+x-1$ is the general solution to $y^{\prime }+2y=2x-1$, confirm that $y=3e^{-2x}+x-1$ is the particular solution satisfying $y(0)=2$.

1. First, confirm the candidate satisfies the DE: Compute $y^{\prime }=-6e^{-2x}+1$. Substitute into the LHS of the DE: $y^{\prime }+2y=(-6e^{-2x}+1)+2(3e^{-2x}+x-1)=2x-1$, which equals the RHS. The candidate is a valid solution to the DE.
2. Next, check the initial condition: Evaluate $y$ at $x=0$: $y(0)=3e^0+0-1=3-1=2$, which matches the given initial condition $y(0)=2$.
3. Conclusion: $y=3e^{-2x}+x-1$ is the correct particular solution.

Exam tip: If you solve for a particular solution from scratch on the exam, you can use this verification process to check your work before moving on, which catches arithmetic errors with the constant $C$.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Forgetting to apply the chain rule when differentiating composite candidates like $y=e^{3x^2}$, leading to an incorrect derivative that fails verification. Why: Students rush differentiation and write $\frac{dy}{dx}=3e^{3x^2}$ instead of $6x{e}^{3x^2}$, relying on memory instead of step-by-step work. Correct move: Always write out the chain rule step explicitly for every composite function before substituting into the DE.
• Wrong move: When verifying implicit solutions, trying to simplify the candidate before differentiating, leading to an incorrect derivative. Why: Students confuse implicit verification with other implicit problems and try to solve for $y$ early. Correct move: Always differentiate the implicit candidate first to isolate $y^{\prime }$, then compare to the DE.
• Wrong move: Only verifying that the candidate satisfies the differential equation, and forgetting to check the initial condition when asked for a particular solution. Why: Students assume any solution to the DE is automatically the required particular solution. Correct move: When asked to verify a particular solution, always explicitly show both the DE check and the initial condition check.
• Wrong move: When verifying a DE of the form $y^{\prime }=f(x,y)$, only computing $y^{\prime }$ from the candidate and forgetting to substitute $y$ into $f(x,y)$ to check equality. Why: The DE is already solved for $y^{\prime }$, so students stop after computing the derivative from the candidate. Correct move: After computing $y^{\prime }$ from the candidate, substitute the candidate’s $y$ into $f(x,y)$, simplify, and confirm it matches your computed $y^{\prime }$.
• Wrong move: For a second-order DE, only substituting $y^{\prime }$ and forgetting to include $y^{\prime \prime }$ and $y$ in the substitution. Why: Students stop differentiation after the first derivative and miss the higher-order term required by the DE. Correct move: Before starting verification, write down all derivatives you need based on the order of the DE to avoid missing terms.
• Wrong move: When asked to verify a general solution, solving for the arbitrary constant even though it is not required. Why: Students are used to finding particular solutions, so they automatically solve for $C$ out of habit. Correct move: For general solution verification, confirm the DE holds for any value of the constant (constants always differentiate to zero, so equality will hold regardless of the constant’s value).

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

Which of the following candidates is a valid solution to the differential equation $y^{\prime \prime }+4y=0$? A) $y=\sin (2x)$ B) $y=\sin (x)$ C) $y=e^{2x}$ D) $y=2x^2$

Worked Solution: To solve, we compute $y^{\prime \prime }$ for each candidate and substitute into the DE to check if the LHS equals 0. For option A: $y=\sin (2x)$, so $y^{\prime }=2\cos (2x)$ and $y^{\prime \prime }=-4\sin (2x)$. Substituting gives $y^{\prime \prime }+4y=-4\sin (2x)+4\sin (2x)=0$, which matches the RHS. Checking the other options confirms they do not satisfy the equality: Option B gives $3\sin (x)\ne 0$, Option C gives $8e^{2x}\ne 0$, and Option D gives $4+8x^2\ne 0$. Correct answer: A

Question 2 (Free Response)

Consider the differential equation $\frac{dy}{dx}=y-x$. (a) Verify that $y=C{e}^x+x+1$ is a general solution to the differential equation. (b) Find the value of $C$ that gives a particular solution satisfying $y(1)=3e+2$, then verify your candidate is the correct particular solution. (c) Show that $y=x+1$ is a solution to the DE, and explain what value of $C$ gives this solution.

Worked Solution: (a) Differentiate the candidate: $\frac{dy}{dx}=C{e}^x+1$. Substitute $y$ into the RHS of the DE: $y-x=(C{e}^x+x+1)-x=C{e}^x+1$. This equals the computed $\frac{dy}{dx}$, so the candidate is a valid general solution that holds for any constant $C$. (b) Substitute $x=1,y=3e+2$ into the general solution: $3e+2=Ce+1+1=Ce+2$, so $C=3$. The candidate is $y=3e^x+x+1$. Verify: $\frac{dy}{dx}=3e^x+1$, and $y-x=3e^x+1=\frac{dy}{dx}$, so it satisfies the DE. Checking the initial condition gives $y(1)=3e+1+1=3e+2$, which matches. This is the correct particular solution. (c) Differentiate $y=x+1$ to get $\frac{dy}{dx}=1$. Substitute into the DE RHS: $y-x=(x+1)-x=1=\frac{dy}{dx}$, so it is a valid solution. This solution corresponds to $C=0$, since substituting $C=0$ into the general solution gives $y=x+1$.

Question 3 (Application / Real-World Style)

In a logistic growth model for a population of wild deer, the rate of growth of the population is given by $\frac{dP}{dt}=0.2P\left(1-\frac{P}{1000}\right)$, where $P(t)$ is the number of deer at time $t$ in years. The candidate solution for this model with initial population $P(0)=200$ is $P(t)=\frac{1000}{1+4e^{-0.2t}}$. Verify that this candidate is the correct solution to the model, including the initial condition.

Worked Solution: First check the initial condition: $P(0)=\frac{1000}{1+4e^0}=\frac{1000}{5}=200$ deer, which matches the given initial population. Next compute the derivative with the quotient rule: $\frac{dP}{dt}=\frac{0(1+4e^{-0.2t})-1000(-0.8e^{-0.2t})}{(1+4e^{-0.2t}{)}^2}=\frac{800e^{-0.2t}}{(1+4e^{-0.2t}{)}^2}$. Now simplify the RHS of the DE: $0.2P\left(1-\frac{P}{1000}\right)=\frac{200}{1+4e^{-0.2t}}\cdot \frac{4e^{-0.2t}}{1+4e^{-0.2t}}=\frac{800e^{-0.2t}}{(1+4e^{-0.2t}{)}^2}=\frac{dP}{dt}$. The candidate satisfies both the DE and the initial condition. In context, this means the candidate function correctly models the deer population growth over time as described by the logistic model.

7. Quick Reference Cheatsheet

8. What's Next

Verifying solutions to differential equations is the foundational prerequisite for all other differential equation topics in Unit 7. Before you learn to solve DEs from scratch using separation of variables, Euler’s method, or logistic growth analysis, you need to understand what a solution actually is and how to confirm it works—this gives you a built-in check to catch errors when you find your own solutions. Without mastering this verification process, you won’t be able to confidently confirm your solutions to separable DEs or logistic growth models, which are high-weight FRQ topics on the AP exam. The skills you build here (substituting derivatives, implicit differentiation, checking initial conditions) also carry over to the study of parametric and polar curves, where you regularly work with derivatives of implicit relations. Next topics to study: Separable differential equations Euler's method Logistic differential equations Slope fields and solution curves