Verifying solutions for differential equations — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: Definition of ordinary differential equations, order of ODEs, verifying explicit general solutions, verifying particular solutions with initial conditions, and verifying implicit solutions to first-order ODEs.

You should already know: Basic differentiation rules including the chain rule, implicit differentiation of relations between $x$ and $y$, notation for constants of integration.

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 Verifying solutions for differential equations?

A differential equation is any equation that relates an unknown function to one or more of its derivatives. Verifying a solution to a differential equation is the foundational process of confirming that a given function (explicit or implicit) satisfies the original equation when its derivatives are substituted back into the equation. Per the 2019 AP Calculus AB Course and Exam Description (CED), this is the opening topic for Unit 7: Differential Equations, and makes up approximately 2-3% of the total exam score. Questions testing this skill appear in both the multiple-choice (MCQ) and free-response (FRQ) sections of the exam. Often, verification is paired with other topics like separation of variables or exponential growth, where you may be asked to confirm a solution you found or work backwards from a given solution to find an unknown constant. Unlike solving a differential equation from scratch, verification only requires proficiency with differentiation, which makes it a straightforward point if you follow the correct process.

2. Verifying Explicit Solutions to First-Order Differential Equations

First-order differential equations are the most common type you will encounter on the AP Calculus AB exam, defined as differential equations that contain only the first derivative of the unknown function $y(x)$, with no higher-order derivatives. The general form of a first-order ordinary differential equation (ODE) is: $$\frac{dy}{dx}=F(x,y)$$ An explicit solution is a function of the form $y=f(x)$ that satisfies the equation. A general solution includes an arbitrary constant of integration, meaning there are infinitely many solutions corresponding to different values of the constant. The verification process follows three consistent steps: 1) Differentiate the given solution $y$ with respect to $x$ to get $\frac{dy}{dx}$. 2) Substitute $y=f(x)$ into the right-hand side function $F(x,y)$ of the differential equation. 3) Confirm that the two resulting expressions are algebraically equivalent. If they are, the function is a solution; if not, it is not.

Worked Example

Problem: Verify that $y=3e^{x^2}$ is a solution to the differential equation $\frac{dy}{dx}=2xy$.

1. First, compute the first derivative of the given explicit solution using the chain rule: $\frac{dy}{dx}=3e^{x^2}\cdot \frac{d}{dx}(x^2)=3e^{x^2}\cdot 2x=6x{e}^{x^2}$.
2. Next, substitute the given $y=3e^{x^2}$ into the right-hand side (RHS) of the differential equation: $2xy=2x(3e^{x^2})=6x{e}^{x^2}$.
3. Compare the derivative from step 1 to the RHS from step 2: $\frac{dy}{dx}=6x{e}^{x^2}=2xy$, so the two sides are identical for all $x$.
4. Conclusion: $y=3e^{x^2}$ is indeed a solution to the given differential equation.

Exam tip: On MCQs that ask which of four functions is a solution, you can save time by eliminating obviously wrong options after computing the derivative instead of checking every option fully.

3. Verifying Particular Solutions with Initial Conditions

A particular solution is a solution to a differential equation that has no arbitrary constants, because the constant is fixed by an initial condition of the form $y(x_0)=y_0$. When asked to verify a particular solution on the AP exam, you are required to complete two separate checks: first, confirm that the function satisfies the differential equation (using the same process as for explicit solutions above), and second, confirm that it satisfies the given initial condition. This is a very common FRQ question, and exam graders award separate points for each check, so skipping one will cost you credit even if your solution is correct. Initial conditions are used to narrow the infinite set of general solutions down to one specific solution that fits the context of the problem.

Worked Example

Problem: Verify that $y=2\sin x+4$ is the particular solution to $\frac{dy}{dx}=2\cos x$ with initial condition $y(0)=4$.

1. First, check that the function satisfies the differential equation: compute $\frac{dy}{dx}=\frac{d}{dx}(2\sin x+4)=2\cos x$, which matches the right-hand side of the ODE exactly.
2. Next, check that the initial condition holds by substituting $x=0$ into the solution: $y(0)=2\sin (0)+4=2(0)+4=4$.
3. This result matches the given initial condition $y(0)=4$.
4. Since both the differential equation and the initial condition are satisfied, $y=2\sin x+4$ is confirmed as the correct particular solution.

Exam tip: If the question asks you to "verify the particular solution", explicitly state that both checks are satisfied to earn full credit; graders do not assume unshown work is correct.

4. Verifying Implicit Solutions

Not all solutions to differential equations can be written explicitly as $y=f(x)$. Instead, they are given implicitly as a relation between $x$ and $y$ of the form $G(x,y)=C$, where $C$ is a constant. To verify an implicit solution, you use implicit differentiation to find $\frac{dy}{dx}$ directly from the given implicit relation, then substitute that $\frac{dy}{dx}$ into the differential equation to confirm it matches. The core logic of verification is identical to explicit solutions; the only difference is the method used to compute $\frac{dy}{dx}$. After differentiating, you will need to rearrange terms to solve for $\frac{dy}{dx}$ before comparing to the original ODE.

Worked Example

Problem: Verify that $y^2+2xy=10$ is an implicit solution to the differential equation $\frac{dy}{dx}=-\frac{y}{y+x}$.

1. Differentiate both sides of the implicit relation with respect to $x$, using the chain rule for $y^2$ and product rule for $2xy$: $$\frac{d}{dx}\left(y^2+2xy\right)=\frac{d}{dx}(10)$$ $$2y\frac{dy}{dx}+2\left(y+x\frac{dy}{dx}\right)=0$$
2. Simplify and solve for $\frac{dy}{dx}$: Divide both sides by 2: $y\frac{dy}{dx}+y+x\frac{dy}{dx}=0$ Group terms with $\frac{dy}{dx}$: $\frac{dy}{dx}(y+x)+y=0$ Solve for $\frac{dy}{dx}$: $\frac{dy}{dx}=-\frac{y}{y+x}$
3. This result matches the differential equation exactly, so the implicit relation is a valid solution.

Exam tip: After rearranging to solve for $\frac{dy}{dx}$, double-check your sign when moving terms across the equals sign; sign errors are the most common mistake in implicit solution verification.

5. Common Pitfalls (and how to avoid them)

• Wrong move: After finding $\frac{dy}{dx}$ for an explicit solution, stops at finding the derivative and does not compare it to the substituted right-hand side of the ODE. Why: Students confuse the goal of verification, assuming finding the derivative alone is enough. Correct move: Always substitute the original $y$ into the RHS of the ODE, then explicitly show that the two sides are equal.
• Wrong move: When verifying a particular solution, only checks that the function solves the ODE and skips checking the initial condition. Why: Students assume the constant is already correct, so the initial condition check is unnecessary. Correct move: Always complete both checks for a particular solution to earn full credit.
• Wrong move: During implicit differentiation, forget to multiply by $\frac{dy}{dx}$ when differentiating terms with $y$, getting $\frac{d}{dx}(y^2)=2y$ instead of $2y\frac{dy}{dx}$. Why: Students are accustomed to differentiating only functions of $x$, so they automatically drop the chain rule term. Correct move: Multiply by $\frac{dy}{dx}$ immediately every time you differentiate a term containing $y$, before moving to the next term.
• Wrong move: When checking the initial condition $y(2)=5$, substitutes $y=2$ for $x$ instead of $x=2$. Why: Students misread standard initial condition notation, mixing up input and output. Correct move: Remember $y(a)=b$ means "when $x=a$, $y=b$", so always substitute $x=a$ into the solution to check for $y=b$.
• Wrong move: When verifying a general solution with arbitrary constant $C$, substitutes a random numerical value for $C$ instead of leaving it symbolic. Why: Students think $C$ must have a numerical value to complete verification. Correct move: $C$ will always cancel out when comparing sides, so leave it as a symbolic constant during the check.
• Wrong move: When substituting $y$ into an ODE RHS with multiple instances of $y$, misses one instance and leaves it as $y$ instead of substituting. Why: Students overlook multiple $y$ terms in complicated ODEs like $\frac{dy}{dx}=xy+y^2$. Correct move: Circle all instances of $y$ in the ODE RHS before substituting to ensure none are missed.

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

Which of the following functions is a solution to the differential equation $\frac{dy}{dx}=\frac{4y}{x}$ for $x>0$? A) $y=4x$ B) $y=x^4$ C) $y=e^{4x}$ D) $y=4e^x$

Worked Solution: To find the correct solution, we compute $\frac{dy}{dx}$ for each option, then compare it to $\frac{4y}{x}$. For option A: $\frac{dy}{dx}=4$, and $\frac{4y}{x}=\frac{16x}{x}=16$, which are not equal, so A is eliminated. For option B: $\frac{dy}{dx}=4x^3$, and $\frac{4y}{x}=\frac{4x^4}{x}=4x^3$, which are equal. Confirming the other options: for C, $\frac{dy}{dx}=4e^{4x}$ and $\frac{4y}{x}=\frac{4e^{4x}}{x}$, which are not equal. For D, $\frac{dy}{dx}=4e^x$ and $\frac{4y}{x}=\frac{16e^x}{x}$, which are not equal. The correct answer is B.

Question 2 (Free Response)

Consider the differential equation $\frac{dy}{dx}=2y-4x$. (a) Verify that $y=C{e}^{2x}+2x+1$ is a general solution to the differential equation. (b) Find the value of $C$ that gives a particular solution satisfying the initial condition $y(0)=4$. (c) Verify that your particular solution from part (b) is correct.

Worked Solution: (a) First, compute the derivative of the general solution: $\frac{dy}{dx}=2C{e}^{2x}+2$. Next, substitute $y$ into the right-hand side of the ODE: $$2y-4x=2\left(C{e}^{2x}+2x+1\right)-4x=2C{e}^{2x}+4x+2-4x=2C{e}^{2x}+2$$ This matches $\frac{dy}{dx}$ exactly, so the given function is confirmed as a general solution.

(b) Substitute $x=0$, $y=4$ into the general solution to solve for $C$: $$4=C{e}^0+2(0)+1=C+1⟹C=3$$

(c) The particular solution is $y=3e^{2x}+2x+1$. First check the ODE: $\frac{dy}{dx}=6e^{2x}+2$, and $2y-4x=2(3e^{2x}+2x+1)-4x=6e^{2x}+2=\frac{dy}{dx}$. Next check the initial condition: $y(0)=3e^0+0+1=4$, which matches. The particular solution is confirmed.

Question 3 (Application / Real-World Style)

The rate of change of the value of an investment (in thousands of dollars) is modeled by the differential equation $\frac{dV}{dt}=0.05V$, where $t$ is time in years. An investment analyst claims the value of an initial investment of $10,000isgivenby$V(t) = 10 e^{0.05 t}$. Verify that this function satisfies both the differential equation and the initial condition, and confirm whether the analyst’s claim is valid.

Worked Solution: First, compute the derivative of $V(t)$: $\frac{dV}{dt}=10(0.05e^{0.05t})=0.5e^{0.05t}$. Next, substitute $V(t)$ into the right-hand side of the differential equation: $0.05V(t)=0.05(10e^{0.05t})=0.5e^{0.05t}$, which matches $\frac{dV}{dt}$ exactly. Now check the initial condition at $t=0$: $V(0)=10e^0=10$. Since $V$ is measured in thousands of dollars, this corresponds to an initial value of $10,000, which matches the given initial investment. Both conditions are satisfied, so the analyst’s claim is valid.

7. Quick Reference Cheatsheet

8. What's Next

Mastering verification of differential equation solutions is the non-negotiable foundational prerequisite for all subsequent work in Unit 7. Next, you will learn to sketch slope fields for differential equations, use separation of variables to solve first-order ODEs from scratch, and apply these techniques to exponential growth and decay problems. Without the ability to verify solutions, you will not be able to check your work after solving an ODE, leading to unnecessary point loss on FRQs, and you will struggle to confirm that you found the correct particular solution for initial value problems. This topic also reinforces differentiation skills that are used across the entire AP Calculus AB course. Up next in Unit 7:

• Slope fields for differential equations
• Separable differential equations
• Exponential growth and decay models
• Particular solutions with initial conditions