AP Calculus BC 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^(5x)
B) y = 5e^x
C) y = x^5
D) y = 5x
Correct Answer: A
To verify if a function is a solution, we use its derivative. Let y = e^(5x). The derivative is y' = 5e^(5x). Substituting y and y' into the differential equation gives 5e^(5x) = 5(e^(5x)), which is a true statement. Therefore, y = e^(5x) is a solution.
A) Integration
B) Finding a limit
C) Differentiation
D) Factoring
Correct Answer: C
According to the provided content, derivatives are used to verify that a function is a solution to a given differential equation. This process involves taking the derivative(s) of the proposed function and substituting them into the equation.
A) By showing that the derivative of Ce^(kx) is k(Ce^(kx)).
B) By setting C=1 and k=1 and showing it works for a specific case.
C) By integrating the differential equation.
D) By showing that the solution is true only for C=0.
Correct Answer: A
To verify the general solution, we find its derivative. The derivative of y = Ce^(kx) with respect to x is y' = C * k * e^(kx), or y' = k(Ce^(kx)). Since y = Ce^(kx), we can substitute y back into the derivative equation to get y' = ky. This confirms the function is a solution for any constant C.
A) The verification process is flawed and allows for multiple answers.
B) The solutions often include an arbitrary constant, C, where each value of C creates a different valid solution.
C) All differential equations are linear.
D) Derivatives can be calculated in infinitely many ways.
Correct Answer: B
A differential equation often has a general solution that contains an arbitrary constant (often denoted by C). Because C can be any real number, there is an infinite family of functions that satisfy the equation, leading to infinitely many solutions.
A) y' + y = 0
B) y'' - y = 0
C) y'' + y = 0
D) y' - 4sin(x) = 0
Correct Answer: C
To find the correct differential equation, we must find the derivatives of the given function, y = 4cos(x). The first derivative is y' = -4sin(x). The second derivative is y'' = -4cos(x). Now we test the options. For option C: y'' + y = (-4cos(x)) + (4cos(x)) = 0. This is a true statement, so y = 4cos(x) is a solution to y'' + y = 0.
A) Yes, because x(6x^2) - 3(2x^3) = 0.
B) No, because x(2x^3) - 3(6x^2) ≠ 0.
C) Yes, because the equation holds true for x=0.
D) No, because a solution must contain an arbitrary constant C.
Correct Answer: A
First, find the derivative of the proposed solution y = 2x^3. The derivative is y' = 6x^2. Next, substitute y and y' into the differential equation: x(y') - 3(y) becomes x(6x^2) - 3(2x^3). Simplifying this expression gives 6x^3 - 6x^3 = 0. Since 0 = 0 is true, the function is a solution.
A) y = e^(3x)
B) y = e^(-3x)
C) y = 2e^(3x) + 4e^(-3x)
D) y = e^(9x)
Correct Answer: D
We verify each option by finding the second derivative. For D, if y = e^(9x), then y' = 9e^(9x) and y'' = 81e^(9x). Substituting into the equation gives y'' - 9y = 81e^(9x) - 9(e^(9x)) = 72e^(9x). Since 72e^(9x) ≠ 0, this function is not a solution. The other options are solutions.
A) Integrate y = C/x.
B) Solve for the constant C.
C) Find the derivative of y = C/x.
D) Substitute x=0 into the equation.
Correct Answer: C
The process of verifying a solution to a differential equation begins by finding the necessary derivatives of the proposed solution. For y = C/x = Cx^(-1), the first derivative is y' = -Cx^(-2) = -C/x^2. This derivative can then be substituted into the differential equation to check for equality.
A) No, because the derivative of C is not zero.
B) Yes, because the derivative of y = x^2 + C is y' = 2x, and substituting this into the equation gives 2x = 2x.
C) No, because the constant C makes the solution invalid.
D) Yes, but only if C = 0.
Correct Answer: B
The function y = x^2 + C represents a family of infinitely many solutions. To verify, we find the derivative: y' = d/dx(x^2 + C) = 2x + 0 = 2x. We then substitute this result into the differential equation y' = 2x, which gives 2x = 2x. This is a true statement for any value of the constant C, confirming it is a general solution.