AP Calculus AB Flashcards: Initial Conditions And Separation Of Variables

What this deck covers

This deck focuses on Initial Conditions And Separation Of Variables, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus AB.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: What is the general solution of $\frac{dy}{dx} = 1$?

Answer: $y = x + C$. Direct integration of the constant function gives $x + C$.

Flashcard 2: What is the role of the constant of integration in solving separable equations?

Answer: It accounts for the family of solutions. Different values of $C$ give different solution curves.

Flashcard 3: Find the particular solution for $\frac{dy}{dx} = 6x$ with $y(2) = 5$.

Answer: $y = 3x^2 - 7$. Integrate: $y = 3x^2 + C$, apply $y(2) = 5$ to find $C$.

Flashcard 4: Determine the particular solution for $\frac{dy}{dx} = x$ with $y(0) = 1$.

Answer: $y = \frac{x^2}{2} + 1$. Integrate: $y = \frac{x^2}{2} + C$, apply $y(0) = 1$ to find $C = 1$.

Flashcard 5: Given $\frac{dy}{dx} = y \sin(x)$, find the general solution.

Answer: $y = C e^{-\cos(x)}$. Separate: $\frac{dy}{y} = \sin(x) dx$, then integrate both sides.

Flashcard 6: Solve $\frac{dy}{dx} = \frac{x}{y}$ given $y(1) = 2$. What is the particular solution?

Answer: $y^2 = x^2 + 3$. Separate and integrate: $\int y dy = \int x dx$, then apply initial condition.

Flashcard 7: Solve $\frac{dy}{dx} = 2y$ for $y(0) = 3$. What is the particular solution?

Answer: $y = 3e^{2x}$. Separate: $\frac{dy}{y} = 2dx$, integrate, apply initial condition.

Flashcard 8: Identify the integral needed to solve $\frac{dy}{dx} = x^2y$ after separating variables.

Answer: $\frac{1}{y} dy = x^2 dx$. Rearranging gives $\frac{dy}{y} = x^2 dx$ for integration.

Flashcard 9: What is the particular solution of $\frac{dy}{dx} = 2x - 1$ if $y(0) = 3$?

Answer: $y = x^2 - x + 3$. Integrate: $y = x^2 - x + C$, apply $y(0) = 3$ gives $C = 3$.

Flashcard 10: What is the first step in solving a differential equation using separation of variables?

Answer: Separate the variables to opposite sides of the equation. This isolates each variable for independent integration.

Flashcard 11: State the general form of a separable differential equation.

Answer: An equation of the form $\frac{dy}{dx} = g(x)h(y)$.. The function can be factored into separate $x$ and $y$ terms.

Flashcard 12: Given $\frac{dy}{dx} = 3x$, find the particular solution if $y(0) = 2$.

Answer: $y = \frac{3}{2}x^2 + 2$. Integrate $3x$ and apply $y(0) = 2$ to find $C = 2$.

Flashcard 13: What is the purpose of integrating both sides after separation of variables?

Answer: To find the antiderivatives, leading to the general solution. Integration reverses differentiation to recover the function.

Flashcard 14: Find the particular solution of $\frac{dy}{dx} = 2xy$ with $y(0) = 1$.

Answer: $y = e^{x^2}$. Separate: $\frac{dy}{y} = 2x dx$, integrate, apply $y(0) = 1$.

Flashcard 15: How do you determine the constant of integration using an initial condition?

Answer: Substitute the initial condition into the general solution. This replaces $C$ with a specific numerical value.

Flashcard 16: Find $C$ for the solution $y = Ce^{3x}$ given $y(0) = 5$.

Answer: $C = 5$. At $x = 0$: $5 = Ce^0 = C \cdot 1$, so $C = 5$.

Flashcard 17: What does 'separable' mean in the context of differential equations?

Answer: Variables can be separated on opposite sides. The equation can be written as $f(x)dx = g(y)dy$.

Flashcard 18: Determine the particular solution for $\frac{dy}{dx} = x^3$ with $y(1) = 4$.

Answer: $y = \frac{x^4}{4} + \frac{15}{4}$. Integrate: $y = \frac{x^4}{4} + C$, apply $y(1) = 4$ to find $C$.

Flashcard 19: What does it mean if a solution is 'particular'?

Answer: It satisfies the differential equation and initial condition. It's the unique solution from the family that meets given conditions.

Flashcard 20: Solve $\frac{dy}{dx} = y$ for $y(0) = 2$. What is the particular solution?

Answer: $y = 2e^x$. Separate: $\frac{dy}{y} = dx$, integrate, apply initial condition.

Flashcard 21: What is the initial condition in a differential equation problem?

Answer: A value that specifies the solution at a particular point, e.g., $y(x_0) = y_0$. This constraint determines the unique particular solution.

Flashcard 22: What is the general solution for $\frac{dy}{dx} = 0$?

Answer: $y = C$. Zero derivative means the function is constant.

Flashcard 23: Find $C$ for $y = \frac{x^3}{3} + C$ given $y(1) = 2$.

Answer: $C = \frac{5}{3}$. At $x = 1$: $2 = \frac{1}{3} + C$, so $C = \frac{5}{3}$.

Flashcard 24: Find the particular solution of $\frac{dy}{dx} = 4y$ with $y(1) = 3$.

Answer: $y = 3e^{4(x-1)}$. General solution $y = Ce^{4x}$, apply $y(1) = 3$ to find $C$.

Flashcard 25: What is the general solution of $\frac{dy}{dx} = y$?

Answer: $y = Ce^x$. This is the exponential growth equation with rate $1$.

Flashcard 26: Explain why initial conditions are necessary for finding particular solutions.

Answer: To determine the specific value of the integration constant. Without them, we only get the general solution family.

Flashcard 27: Describe the integration process in separation of variables.

Answer: Integrate both sides after separating variables. This finds antiderivatives of both separated expressions.

Flashcard 28: What is the role of the integration constant when finding the general solution?

Answer: Represents an arbitrary constant for family of solutions. It parameterizes all possible solutions before applying conditions.

Flashcard 29: What is the general solution of $\frac{dy}{dx} = 4$?

Answer: $y = 4x + C$. Direct integration of constant $4$ gives $4x + C$.

Flashcard 30: What technique is used to solve $\frac{dy}{dx} = \frac{x}{y}$?

Answer: Separation of variables. The equation has form $\frac{dy}{dx} = f(x)g(y)$ where variables separate.