AP Calculus AB Practice Quiz: Verifying Solutions for Differential Equations
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Test your understanding with short quizzes. This quiz has 9 questions to check your progress.
Question 1 of 9
All Questions (9)
A) y = e^(2x)
B) y = 2x
C) y = x^2
D) y = e^x
Correct Answer: A
To verify if a function is a solution, we use its derivative. For y = e^(2x), the derivative is dy/dx = 2e^(2x). Substituting y and dy/dx into the differential equation gives 2e^(2x) = 2(e^(2x)), which is a true statement. Therefore, y = e^(2x) is a solution.
A) y = 12x^2
B) y = 4x^3 + C
C) y = x^4 + C
D) y = x^3 + C
Correct Answer: C
We need to find a function y whose derivative is 4x^3. Let's test the options by taking their derivatives. For y = x^4 + C, the derivative is y' = 4x^3. Substituting this into the differential equation gives 4x^3 = 4x^3, which is true. The constant C represents one of the infinitely many general solutions.
A) y = Ce^(2x)
B) y = Ce^(-2x)
C) y = C sin(2x)
D) y = 2x + C
Correct Answer: B
To verify the general solution y = Ce^(-2x), first find its derivative: y' = -2Ce^(-2x). Now, substitute y and y' into the differential equation: (-2Ce^(-2x)) + 2(Ce^(-2x)) = 0. This simplifies to -2Ce^(-2x) + 2Ce^(-2x) = 0, which results in 0 = 0. This holds true for any constant C, confirming it is a general solution.
A) Finding the integral of the function and comparing it to the equation.
B) Solving the differential equation from scratch.
C) Graphing the function to see if it is continuous.
D) Calculating the function's derivatives and substituting them into the differential equation.
Correct Answer: D
As stated in the content, derivatives are used to verify that a function is a solution to a given differential equation. This involves taking the necessary derivatives of the proposed solution and substituting them, along with the original function, into the differential equation to see if the equality holds.
A) Yes, because its second derivative is the negative of the original function.
B) No, because its first derivative is not zero.
C) Yes, because the function involves trigonometric identities.
D) No, because y'' + y simplifies to 2cos(x) - 2sin(x).
Correct Answer: A
To verify, we need the first and second derivatives. Given y = sin(x) + cos(x). The first derivative is y' = cos(x) - sin(x). The second derivative is y'' = -sin(x) - cos(x). Now substitute y and y'' into the equation: (-sin(x) - cos(x)) + (sin(x) + cos(x)) = 0. This simplifies to 0 = 0, so the function is a solution.
A) A differential equation can only have one unique solution.
B) All solutions to differential equations must be polynomials.
C) There may be infinitely many general solutions to a differential equation.
D) Verification of a solution requires integration.
Correct Answer: C
The arbitrary constant 'C' can be any real number, and for each value, the function y = x^3 + C is a valid solution because its derivative is always 3x^2. This demonstrates that a single differential equation can have an infinite family of solutions, known as general solutions.
A) y = 2x
B) y = Cx^2
C) y = C/x^2
D) y = e^(2x)
Correct Answer: B
Let's test the function y = Cx^2. First, find the derivative: y' = 2Cx. Now, substitute y and y' into the differential equation: x(2Cx) - 2(Cx^2) = 0. This simplifies to 2Cx^2 - 2Cx^2 = 0, which gives 0 = 0. Since this is true for any constant C, y = Cx^2 is a general solution.
A) Yes, because 1 = (x + 1) - x simplifies to 1 = 1.
B) No, because 1 = (x + 1) - x simplifies to x = 0.
C) Yes, because the derivative of x + 1 is x.
D) No, because the function does not contain an exponential term.
Correct Answer: A
To verify if y = x + 1 is a solution, first find its derivative, which is y' = 1. Then, substitute y = x + 1 and y' = 1 into the differential equation: 1 = (x + 1) - x. The right side simplifies to 1, resulting in the true statement 1 = 1. Therefore, y = x + 1 is a solution.
A) The derivative is not needed; only the value of C matters.
B) The derivative, y' = Ce^x - 1, is substituted into the equation to confirm the equality holds.
C) The derivative is set to zero to find critical points, which verifies the solution.
D) The derivative must be equal to the original function y for it to be a valid solution.
Correct Answer: B
The core method of verification is to use the derivative. For y = Ce^x - x - 1, the derivative is y' = Ce^x - 1. Substituting y' and y into the equation gives (Ce^x - 1) - (Ce^x - x - 1) = x. Simplifying the left side, we get Ce^x - 1 - Ce^x + x + 1 = x, which reduces to x = x. This confirms the solution is valid.