AP Calculus BC 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 BC.

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: Given $y = Ce^{x^2}$, find $C$ if $y(0) = 1$.

Answer: $C = 1$. Substitute $x = 0$ and $y = 1$ into the general solution.

Flashcard 2: Convert $\frac{dy}{dx} = x^2y^3$ into a separable form.

Answer: $\frac{dy}{y^3} = x^2 dx$. Move $y^3$ to the left side and $x^2$ to the right.

Flashcard 3: What is the integration result for $\frac{dy}{y} = k dx$?

Answer: $\text{ln}|y| = kx + C$. Standard integration result for exponential-type equations.

Flashcard 4: Solve for $y$ given $\frac{dy}{dx} = x$ with initial condition $y(1) = 3$.

Answer: $y = \frac{x^2}{2} + \frac{5}{2}$. Integrate $\int x \, dx = \frac{x^2}{2} + C$ and apply condition.

Flashcard 5: What is the general solution for $\frac{dy}{dx} = -y$?

Answer: $y = Ce^{-x}$. Standard exponential decay with rate $1$.

Flashcard 6: Find the particular solution to $\frac{dy}{dx} = \frac{2y}{x}$ with $y(1) = 3$.

Answer: $y = 3x^2$. Separate to $\frac{dy}{y} = \frac{2dx}{x}$, integrate to get $y = Cx^2$.

Flashcard 7: What is the particular solution for $\frac{dy}{dx} = -3y$ with $y(2) = 1$?

Answer: $y = e^{-3x + 6}$. From $y = Ce^{-3x}$, use $y(2) = 1$ to find $C = e^6$.

Flashcard 8: What is the general solution for $\frac{dy}{dx} = x^2$?

Answer: $y = \frac{x^3}{3} + C$. Direct integration of the polynomial function.

Flashcard 9: What is the general solution of $\frac{dy}{dx} = 1$?

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

Flashcard 10: Determine the general solution for $\frac{dy}{dx} = \frac{1}{x}$.

Answer: $y = \text{ln}|x| + C$. Direct integration of the reciprocal function.

Flashcard 11: What is the role of $C$ in the solution $y = Ce^{kx}$?

Answer: Constant for initial conditions. Determines the specific solution from the family of solutions.

Flashcard 12: What is the solution to $\frac{dy}{dx} = 0$?

Answer: $y = C$. When the derivative is zero, the function is constant.

Flashcard 13: What is the general solution of $\frac{dy}{dx} = 1$?

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

Flashcard 14: Given $\frac{dy}{dx} = 4x$, find $C$ if $y(1) = 6$.

Answer: $C = 4$. From $y = 2x^2 + C$ with $y(1) = 6$, so $C = 4$.

Flashcard 15: What is the solution to $\frac{dy}{dx} = 0$?

Answer: $y = C$. When the derivative is zero, the function is constant.

Flashcard 16: What is the particular solution for $\frac{dy}{dx} = 2y$ with $y(0) = 3$?

Answer: $y = 3e^{2x}$. From $y = Ce^{2x}$ with condition $y(0) = 3$ giving $C = 3$.

Flashcard 17: What is the integration result for $\frac{dy}{y} = k dx$?

Answer: $\ln |y| = kx + C$. Standard integration result for exponential-type equations.

Flashcard 18: What is the particular solution for $\frac{dy}{dx} = -3y$ with $y(2) = 1$?

Answer: $y = e^{-3x + 6}$. From $y = Ce^{-3x}$, use $y(2) = 1$ to find $C = e^6$.

Flashcard 19: How do you integrate $y' = \frac{dy}{dx} = x^2 y$ using separation of variables?

Answer: $\frac{dy}{y} = x^2 dx$. Divide both sides by $y$ and multiply by $dx$.

Flashcard 20: What is the role of the integration constant $C$ in differential equations?

Answer: Represents the family of solutions. Each value of $C$ gives a different particular solution.

Flashcard 21: What is the general solution to $\frac{dy}{dx} = ky$ using separation of variables?

Answer: $y = Ce^{kx}$. Standard form for exponential growth/decay differential equations.

Flashcard 22: Determine $C$ for the particular solution of $\frac{dy}{dx} = 5y$ with $y(0) = 7$.

Answer: $C = 7$. From $y = Ce^{5x}$ with $y(0) = 7$, so $C = 7$.

Flashcard 23: Solve $\frac{dy}{dx} = -2xy$ for $y$ using separation of variables.

Answer: $y = Ce^{-x^2}$. Separate to $\frac{dy}{y} = -2x \, dx$ and integrate both sides.

Flashcard 24: What is the particular solution for $\frac{dy}{dx} = 2y$ with $y(0) = 3$?

Answer: $y = 3e^{2x}$. From $y = Ce^{2x}$ with condition $y(0) = 3$ giving $C = 3$.

Flashcard 25: What equation results from integrating $\frac{dy}{y} = \frac{1}{x} dx$?

Answer: $\text{ln}|y| = \text{ln}|x| + C$. Standard result from integrating separated variables.

Flashcard 26: What is the role of the integration constant $C$ in differential equations?

Answer: Represents the family of solutions. Each value of $C$ gives a different particular solution.

Flashcard 27: Find the particular solution of $\frac{dy}{dx} = 2x$ with $y(0) = 4$.

Answer: $y = x^2 + 4$. Integrate to get $y = x^2 + C$, then use initial condition.

Flashcard 28: What is the role of $C$ in the solution $y = Ce^{kx}$?

Answer: Constant for initial conditions. Determines the specific solution from the family of solutions.

Flashcard 29: Find the particular solution for $\frac{dy}{dx} = 3x^2$ with $y(0) = 1$.

Answer: $y = x^3 + 1$. Integrate $3x^2$ and apply the initial condition at $x = 0$.

Flashcard 30: What is the general solution to $\frac{dy}{dx} = -ky$?

Answer: $y = Ce^{-kx}$. Standard exponential decay solution with rate $k$.

Flashcard 31: Find the particular solution of $\frac{dy}{dx} = 2x$ with $y(0) = 4$.

Answer: $y = x^2 + 4$. Integrate to get $y = x^2 + C$, then use initial condition.

Flashcard 32: What is the general solution to $\frac{dy}{dx} = ky$ using separation of variables?

Answer: $y = Ce^{kx}$. Standard form for exponential growth/decay differential equations.

Flashcard 33: What is the general solution for $\frac{dy}{dx} = -y$?

Answer: $y = Ce^{-x}$. Standard exponential decay with rate $1$.

Flashcard 34: State the integration result of $\frac{dy}{y} = \frac{1}{x} dx$.

Answer: $\text{ln}|y| = \text{ln}|x| + C$. Direct integration of the separated variables.

Flashcard 35: Determine the particular solution of $\frac{dy}{dx} = y$ with $y(0) = 5$.

Answer: $y = 5e^x$. From general solution $y = Ce^x$, use condition to find $C = 5$.

Flashcard 36: Given $\frac{dy}{dx} = 2y$, find the particular solution with $y(0) = 1$.

Answer: $y = e^{2x}$. From $y = Ce^{2x}$ with initial condition $y(0) = 1$.

Flashcard 37: Given $\frac{dy}{dx} = 3x^2$, what is the particular solution with $y(0) = 2$?

Answer: $y = x^3 + 2$. Integrate to get $y = x^3 + C$, then use $y(0) = 2$ to find $C = 2$.

Flashcard 38: Solve $\frac{dy}{dx} = \frac{y}{x}$ for $y$ using separation of variables.

Answer: $y = Cx$. Separate to $\frac{dy}{y} = \frac{dx}{x}$ and integrate both sides.

Flashcard 39: Solve $\frac{dy}{dx} = \frac{y}{x}$ for $y$ using separation of variables.

Answer: $y = Cx$. Separate to $\frac{dy}{y} = \frac{dx}{x}$ and integrate both sides.

Flashcard 40: Convert $\frac{dy}{dx} = x^2y^3$ into a separable form.

Answer: $\frac{dy}{y^3} = x^2 dx$. Move $y^3$ to the left side and $x^2$ to the right.

Flashcard 41: Given $\frac{dy}{dx} = 2y$, find the particular solution with $y(0) = 1$.

Answer: $y = e^{2x}$. From $y = Ce^{2x}$ with initial condition $y(0) = 1$.

Flashcard 42: Find the particular solution to $\frac{dy}{dx} = \frac{2y}{x}$ with $y(1) = 3$.

Answer: $y = 3x^2$. Separate to $\frac{dy}{y} = \frac{2dx}{x}$, integrate to get $y = Cx^2$.

Flashcard 43: Solve $\frac{dy}{dx} = x^3y^2$ by separating variables.

Answer: $\frac{dy}{y^2} = x^3 dx$. Move $y^2$ to left side and $x^3$ to right side.

Flashcard 44: Determine the particular solution of $\frac{dy}{dx} = y$ with $y(0) = 5$.

Answer: $y = 5e^x$. From general solution $y = Ce^x$, use condition to find $C = 5$.

Flashcard 45: Given $\frac{dy}{dx} = 4x$, find $C$ if $y(1) = 6$.

Answer: $C = 4$. From $y = 2x^2 + C$ with $y(1) = 6$, so $C = 4$.

Flashcard 46: What equation results from integrating $\frac{dy}{y} = \frac{1}{x} dx$?

Answer: $\ln |y| = \ln |x| + C$. Standard result from integrating separated variables.

Flashcard 47: What is the general solution for $\frac{dy}{dx} = x^2$?

Answer: $y = \frac{x^3}{3} + C$. Direct integration of the polynomial function.

Flashcard 48: Determine the general solution for $\frac{dy}{dx} = \frac{1}{x}$.

Answer: $y = \text{ln}|x| + C$. Direct integration of the reciprocal function.

Flashcard 49: Determine $C$ for the particular solution of $\frac{dy}{dx} = 5y$ with $y(0) = 7$.

Answer: $C = 7$. From $y = Ce^{5x}$ with $y(0) = 7$, so $C = 7$.

Flashcard 50: Solve for $y$ given $\frac{dy}{dx} = x$ with initial condition $y(1) = 3$.

Answer: $y = \frac{x^2}{2} + \frac{5}{2}$. Integrate $\int x \, dx = \frac{x^2}{2} + C$ and apply condition.

Flashcard 51: What is the solution to $\frac{dy}{dx} = 4y$ using separation of variables?

Answer: $y = Ce^{4x}$. Standard exponential solution with growth rate $k = 4$.

Flashcard 52: State the integration result of $\frac{dy}{y} = \frac{1}{x} dx$.

Answer: $\text{ln}|y| = \text{ln}|x| + C$. Direct integration of the separated variables.

Flashcard 53: What is the purpose of finding a particular solution?

Answer: To satisfy given initial conditions. Eliminates the arbitrary constant using given conditions.

Flashcard 54: How do you integrate $y' = \frac{dy}{dx} = x^2 y$ using separation of variables?

Answer: $\frac{dy}{y} = x^2 dx$. Divide both sides by $y$ and multiply by $dx$.

Flashcard 55: What is the general solution to $\frac{dy}{dx} = -ky$?

Answer: $y = Ce^{-kx}$. Standard exponential decay solution with rate $k$.

Flashcard 56: What is the purpose of using initial conditions in solving differential equations?

Answer: To find particular solutions. They eliminate the arbitrary constant $C$ from general solutions.

Flashcard 57: What is the purpose of finding a particular solution?

Answer: To satisfy given initial conditions. Eliminates the arbitrary constant using given conditions.

Flashcard 58: Solve $\frac{dy}{dx} = x^3y^2$ by separating variables.

Answer: $\frac{dy}{y^2} = x^3 dx$. Move $y^2$ to left side and $x^3$ to right side.

Flashcard 59: What is the first step in solving a differential equation using separation of variables?

Answer: Separate the variables. This allows variables to be on different sides for integration.

Flashcard 60: Given $\frac{dy}{dx} = 3x^2$, what is the particular solution with $y(0) = 2$?

Answer: $y = x^3 + 2$. Integrate to get $y = x^3 + C$, then use $y(0) = 2$ to find $C = 2$.

Flashcard 61: Find the particular solution for $\frac{dy}{dx} = 3x^2$ with $y(0) = 1$.

Answer: $y = x^3 + 1$. Integrate $3x^2$ and apply the initial condition at $x = 0$.

Flashcard 62: Given $y = Ce^{x^2}$, find $C$ if $y(0) = 1$.

Answer: $C = 1$. Substitute $x = 0$ and $y = 1$ into the general solution.

Flashcard 63: What is the solution to $\frac{dy}{dx} = 4y$ using separation of variables?

Answer: $y = Ce^{4x}$. Standard exponential solution with growth rate $k = 4$.

Flashcard 64: What is the purpose of using initial conditions in solving differential equations?

Answer: To find particular solutions. They eliminate the arbitrary constant $C$ from general solutions.

Flashcard 65: Solve $\frac{dy}{dx} = -2xy$ for $y$ using separation of variables.

Answer: $y = Ce^{-x^2}$. Separate to $\frac{dy}{y} = -2x \, dx$ and integrate both sides.

Flashcard 66: What is the first step in solving a differential equation using separation of variables?

Answer: Separate the variables. This allows variables to be on different sides for integration.