AP Calculus BC Practice Quiz: Finding Particular Solutions Using Initial Conditions and Separation of Variables
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Test your understanding with short quizzes. This quiz has 10 questions to check your progress.
Question 1 of 10
All Questions (10)
A) It verifies that the general solution is correct.
B) It determines the domain of the solution.
C) It selects a single, unique function from the family of solutions.
D) It changes the differential equation into a simpler form.
Correct Answer: C
The general solution contains an arbitrary constant (e.g., +C), representing infinitely many possible solution curves. An initial condition provides a specific point $(x_0, y_0)$ that the solution must pass through, which allows for the calculation of the specific value of the constant, thus identifying one particular solution from the infinite family. [cite: 2791, 2792]
A) y = x^3 - x + 2
B) y = x^3 - x - 2
C) y = 6x - 8
D) y = x^3 - x
Correct Answer: B
The general solution is found by integrating $\frac{dy}{dx}$: $y = \int (3x^2 - 1) dx = x^3 - x + C$. Using the initial condition $(2, 4)$, we substitute to find C: $4 = (2)^3 - 2 + C \Rightarrow 4 = 8 - 2 + C \Rightarrow 4 = 6 + C \Rightarrow C = -2$. Therefore, the particular solution is $y = x^3 - x - 2$. [cite: 2790]
A) y^2 = 2x^2 + 7
B) y = \sqrt{2x^2 + 7}
C) y = -\sqrt{2x^2 + 7}
D) y^2 = 2x^2 - 11
Correct Answer: C
Separate the variables: $y \, dy = 2x \, dx$. Integrate both sides: $\int y \, dy = \int 2x \, dx \Rightarrow \frac{1}{2}y^2 = x^2 + C$. Substitute the initial condition $(1, -3)$: $\frac{1}{2}(-3)^2 = (1)^2 + C \Rightarrow \frac{9}{2} = 1 + C \Rightarrow C = \frac{7}{2}$. The implicit solution is $\frac{1}{2}y^2 = x^2 + \frac{7}{2}$, or $y^2 = 2x^2 + 7$. To find the explicit particular solution, we solve for y: $y = \pm\sqrt{2x^2 + 7}$. Since the initial condition has a negative y-value ($y(1) = -3$), we must choose the negative root. Thus, $y = -\sqrt{2x^2 + 7}$. [cite: 2790]
A) y = \int_{0}^{x} \cos(t^2) dt + 5
B) y = \int_{2}^{x} \cos(t^2) dt + 5
C) y = \int_{5}^{x} \cos(t^2) dt + 2
D) y = -\sin(x^2) + 5
Correct Answer: B
A particular solution to $\frac{dy}{dx}=f(x)$ satisfying $y(x_0)=y_0$ is given by the function $y = \int_{x_0}^{x} f(t) dt + y_0$. In this case, $f(x) = \cos(x^2)$, $x_0 = 2$, and $y_0 = 5$. Substituting these values into the formula gives $y = \int_{2}^{x} \cos(t^2) dt + 5$. [cite: 2793]
A) y = e^{x^2} + 2
B) y = 3e^{x^2}
C) y = e^{x^2+3}
D) \ln|y| = x^2 + 3
Correct Answer: B
Separate the variables: $\frac{1}{y} dy = 2x dx$. Integrate both sides: $\int \frac{1}{y} dy = \int 2x dx \Rightarrow \ln|y| = x^2 + C$. To solve for y, exponentiate both sides: $|y| = e^{x^2+C} = e^C \cdot e^{x^2}$. Let $A = \pm e^C$. So, $y = Ae^{x^2}$. Use the initial condition $(0, 3)$: $3 = Ae^{0^2} \Rightarrow 3 = A(1) \Rightarrow A=3$. The particular solution is $y = 3e^{x^2}$. [cite: 2790]
A) (-\infty, \infty)
B) [0, \infty)
C) (0, \infty)
D) (-\infty, 0) \cup (0, \infty)
Correct Answer: C
First, find the particular solution. $y = \int \frac{1}{2\sqrt{x}} dx = \int \frac{1}{2}x^{-1/2} dx = x^{1/2} + C = \sqrt{x} + C$. Using the point $(4, 5)$: $5 = \sqrt{4} + C \Rightarrow 5 = 2 + C \Rightarrow C = 3$. The solution is $y = \sqrt{x} + 3$. The domain of the function $y = \sqrt{x} + 3$ is $[0, \infty)$. However, the original differential equation $\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$ is undefined at $x=0$. Therefore, the domain of the solution to the differential equation must be restricted to $(0, \infty)$. [cite: 2793]
A) y = -\ln(1-\sin x)
B) y = \ln(\sin x + 1)
C) y = -\ln(\cos x)
D) y = \ln(\cos x)
Correct Answer: A
Separate the variables: $e^{-y} dy = \cos x dx$. Integrate both sides: $\int e^{-y} dy = \int \cos x dx \Rightarrow -e^{-y} = \sin x + C$. Use the initial condition $(0, 0)$: $-e^{-0} = \sin(0) + C \Rightarrow -1 = 0 + C \Rightarrow C = -1$. The implicit solution is $-e^{-y} = \sin x - 1$, or $e^{-y} = 1 - \sin x$. To solve for y, take the natural logarithm of both sides: $-y = \ln(1 - \sin x)$. Therefore, $y = -\ln(1 - \sin x)$. [cite: 2790]
A) C = y_0 - G(x_0)
B) C = y_0 + G(x_0)
C) C = G(y_0) - x_0
D) C = f(x_0) - y_0
Correct Answer: A
The general solution is $y = G(x) + C$. To find the particular solution that passes through $(x_0, y_0)$, we substitute these values into the general solution: $y_0 = G(x_0) + C$. Solving for the constant $C$ gives $C = y_0 - G(x_0)$. This aligns with the fact that the particular solution can be written as $y = G(x) - G(x_0) + y_0$, which is equivalent to the definite integral form $y = \int_{x_0}^{x} f(t) dt + y_0$. [cite: 2791, 2792, 2793]
A) y = \ln(x^3)
B) y = \ln(x^3 - 1)
C) y = e^{x^3-1}
D) y = \ln(x^3+1) - \ln(2)
Correct Answer: A
First, rewrite the equation as $\frac{dy}{dx} = \frac{3x^2}{e^y}$. Separate the variables: $e^y dy = 3x^2 dx$. Integrate both sides: $\int e^y dy = \int 3x^2 dx \Rightarrow e^y = x^3 + C$. Use the initial condition $(1, 0)$: $e^0 = (1)^3 + C \Rightarrow 1 = 1 + C \Rightarrow C = 0$. The implicit solution is $e^y = x^3$. To find the explicit solution, take the natural logarithm of both sides: $y = \ln(x^3)$. [cite: 2790]
A) y = \frac{1}{1-x}
B) y = \frac{-1}{x-1}
C) y = \frac{1}{3-x}
D) y = \frac{-1}{x+1}
Correct Answer: B
Separate the variables: $\frac{1}{y^2} dy = dx$. Integrate both sides: $\int y^{-2} dy = \int dx \Rightarrow -y^{-1} = x + C \Rightarrow \frac{-1}{y} = x + C$. Use the initial condition $(2, -1)$: $\frac{-1}{-1} = 2 + C \Rightarrow 1 = 2 + C \Rightarrow C = -1$. The implicit solution is $\frac{-1}{y} = x - 1$. Solving for y gives $y = \frac{-1}{x-1}$. This solution has a vertical asymptote at $x=1$, which creates a domain restriction. [cite: 2790, 2793]