AP Calculus BC Flashcards: Finding General Solutions Separation Of Variables

What this deck covers

This deck focuses on Finding General Solutions 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: What is the result of integrating $x dx = y dy$?

Answer: $\frac{x^2}{2} = \frac{y^2}{2} + C$. Both sides integrate to $\frac{x^2}{2}$ and $\frac{y^2}{2}$ respectively.

Flashcard 2: What is the purpose of $C$ in a general solution?

Answer: It accounts for the family of solutions. Multiple solutions exist for different initial conditions.

Flashcard 3: Determine the type of equation: $\frac{dy}{dx} = x y + 1$.

Answer: Not separable. The constant term prevents separation of variables.

Flashcard 4: Identify the error: $\frac{dy}{dx} = x^2y^2$ rewritten as $y^2 dy = x^2 dx$.

Answer: Correct: $\frac{1}{y^2} dy = x^2 dx$. The $y^2$ term should move to the left with reciprocal: $\frac{1}{y^2}$.

Flashcard 5: What must be checked after finding a general solution?

Answer: Verify the solution by differentiating and substituting. Solutions must satisfy the original differential equation.

Flashcard 6: Why is verifying a solution important?

Answer: To ensure it satisfies the original equation. Verification confirms the solution is mathematically correct.

Flashcard 7: What is the next step after separating variables in a differential equation?

Answer: Integrate both sides with respect to their variables. Integration produces antiderivatives on each side of the equation.

Flashcard 8: Identify the step after integration: $\text{ln}|y| = x^2 + C$.

Answer: Solve for $y$: $y = e^{x^2 + C}$. Exponentiate both sides to isolate $y$ from the logarithm.

Flashcard 9: What is the general solution for $\frac{dy}{dx} = 2x$?

Answer: $y = x^2 + C$. Direct integration of $\frac{dy}{dx} = 2x$ gives $y = x^2 + C$.

Flashcard 10: What is the general solution for $\frac{dy}{dx} = 3xy$?

Answer: $y = Ce^{\frac{3x^2}{2}}$. Separating gives $\frac{dy}{y} = 3x dx$, integrating yields $\ln|y| = \frac{3x^2}{2} + C$.

Flashcard 11: Identify the integration: $y dy = x dx$.

Answer: $\frac{y^2}{2} = \frac{x^2}{2} + C$. Both sides integrate using the power rule.

Flashcard 12: Identify the next step: $\text{ln}|y| = \text{ln}|x| + C$.

Answer: Express $y$ in terms of $x$: $y = Cx$. Use properties of logarithms: $\ln|y| - \ln|x| = \ln|\frac{y}{x}| = C$.

Flashcard 13: What is the next step after separating variables in a differential equation?

Answer: Integrate both sides with respect to their variables. Integration produces antiderivatives on each side of the equation.

Flashcard 14: Find the general solution: $\frac{dy}{dx} = y\text{e}^x$.

Answer: $y = Ce^{\text{e}^x}$. Separating gives $\frac{dy}{y} = e^x dx$, integrating yields $\ln|y| = e^x + C$.

Flashcard 15: Find the general solution: $\frac{dy}{dx} = y\text{cos}(x)$.

Answer: $y = Ce^{\text{sin}(x)}$. Separating gives $\frac{dy}{y} = \cos(x) dx$, integrating yields $\ln|y| = \sin(x) + C$.

Flashcard 16: Find the general solution: $\frac{dy}{dx} = yx$.

Answer: $y = Ce^{\frac{x^2}{2}}$. Same as previous: separating $\frac{dy}{y} = x dx$ gives this result.

Flashcard 17: Find the general solution: $\frac{dy}{dx} = 3x^2 y$.

Answer: $y = Ce^{x^3}$. Separating gives $\frac{dy}{y} = 3x^2 dx$, integrating yields $\ln|y| = x^3 + C$.

Flashcard 18: What is the result of integrating $x dx = y dy$?

Answer: $\frac{x^2}{2} = \frac{y^2}{2} + C$. Both sides integrate to $\frac{x^2}{2}$ and $\frac{y^2}{2}$ respectively.

Flashcard 19: What is the integration result for $y dy = 2x dx$?

Answer: $\frac{y^2}{2} = x^2 + C$. Integration gives $\frac{y^2}{2} = 2 \cdot \frac{x^2}{2} + C = x^2 + C$.

Flashcard 20: Find the general solution: $\frac{dy}{dx} = 3x^2 y$.

Answer: $y = Ce^{x^3}$. Separating gives $\frac{dy}{y} = 3x^2 dx$, integrating yields $\ln|y| = x^3 + C$.

Flashcard 21: Find the general solution: $\frac{dy}{dx} = y^2$.

Answer: $y = -\frac{1}{C-x}$. Separating gives $\frac{dy}{y^2} = dx$, integrating yields $-\frac{1}{y} = x + C$.

Flashcard 22: What does it mean if a differential equation is not separable?

Answer: Variables cannot be isolated on separate sides. The equation cannot be written as $g(x)h(y)$ form.

Flashcard 23: Identify the integral to solve after separating variables: $\frac{dy}{dx} = x \frac{1}{y}$

Answer: $\text{Integrate: } \frac{1}{y} dy = x dx$. Separation gives $\frac{dy}{y} = x dx$, then integrate both sides.

Flashcard 24: Identify the integral to solve after separating variables: $\frac{dy}{dx} = x \frac{1}{y}$.

Answer: $\text{Integrate: } \frac{1}{y} dy = x dx$. Separation gives $\frac{dy}{y} = x dx$, then integrate both sides.

Flashcard 25: Find the general solution: $\frac{dy}{dx} = y^2 \text{sin}(x)$.

Answer: $y = \frac{1}{C - \text{cos}(x)}$. Separating gives $\frac{dy}{y^2} = \sin(x) dx$, integrating yields $-\frac{1}{y} = -\cos(x) + C$.

Flashcard 26: What is the result of integrating both sides: $f(y) dy = g(x) dx$?

Answer: Find $F(y) = G(x) + C$, where $F$ and $G$ are antiderivatives. Integration produces two functions related by a constant.

Flashcard 27: Determine the error: $y dy = x dx$ integrated as $y = x + C$.

Answer: Correct: $\frac{y^2}{2} = \frac{x^2}{2} + C$. Must integrate each term separately: $\int y dy = \frac{y^2}{2}$.

Flashcard 28: Identify the integration: $y dy = x dx$.

Answer: $\frac{y^2}{2} = \frac{x^2}{2} + C$. Both sides integrate using the power rule.

Flashcard 29: What must be checked after finding a general solution?

Answer: Verify the solution by differentiating and substituting. Solutions must satisfy the original differential equation.

Flashcard 30: State the integration result for $\frac{1}{y} dy = x dx$.

Answer: $\ln|y| = \frac{x^2}{2} + C$. Antiderivatives are $\ln|y|$ and $\frac{x^2}{2}$ respectively.

Flashcard 31: Find the general solution: $\frac{dy}{dx} = yx$.

Answer: $y = Ce^{\frac{x^2}{2}}$. Same as previous: separating $\frac{dy}{y} = x dx$ gives this result.

Flashcard 32: What is the result of integrating both sides: $f(y) dy = g(x) dx$?

Answer: Find $F(y) = G(x) + C$, where $F$ and $G$ are antiderivatives. Integration produces two functions related by a constant.

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

Answer: $y = Ce^x$. This is the standard exponential growth/decay solution.

Flashcard 34: State the form of a separable differential equation.

Answer: $\frac{dy}{dx} = g(x)h(y)$. This form allows variables to be separated into $\frac{dy}{h(y)} = g(x)dx$.

Flashcard 35: What is the general solution for $\frac{dy}{dx} = 2x$?

Answer: $y = x^2 + C$. Direct integration of $\frac{dy}{dx} = 2x$ gives $y = x^2 + C$.

Flashcard 36: Find the general solution: $\frac{dy}{dx} = xy$.

Answer: $y = Ce^{\frac{x^2}{2}}$. Separating gives $\frac{dy}{y} = x dx$, integrating yields $\ln|y| = \frac{x^2}{2} + C$.

Flashcard 37: Find the general solution: $\frac{dy}{dx} = xy$.

Answer: $y = Ce^{\frac{x^2}{2}}$. Separating gives $\frac{dy}{y} = x dx$, integrating yields $\ln|y| = \frac{x^2}{2} + C$.

Flashcard 38: What technique is used to verify the solution of a separable differential equation?

Answer: Differentiate the solution and compare with the original equation. Substitution back into the original equation confirms correctness.

Flashcard 39: Determine the type of equation: $\frac{dy}{dx} = x^2 + y^2$.

Answer: Not separable. Cannot separate $x$ and $y$ terms to opposite sides.

Flashcard 40: What form does the solution to a separable differential equation typically take?

Answer: $F(y) = G(x) + C$. Each side integrates to a function plus an arbitrary constant.

Flashcard 41: What form does the solution to a separable differential equation typically take?

Answer: $F(y) = G(x) + C$. Each side integrates to a function plus an arbitrary constant.

Flashcard 42: Find the general solution: $\frac{dy}{dx} = y\text{e}^x$.

Answer: $y = Ce^{\text{e}^x}$. Separating gives $\frac{dy}{y} = e^x dx$, integrating yields $\ln|y| = e^x + C$.

Flashcard 43: Which method helps find a particular solution after finding a general one?

Answer: Use initial conditions. Initial conditions determine the specific value of the constant $C$.

Flashcard 44: State the form of a separable differential equation.

Answer: $\frac{dy}{dx} = g(x)h(y)$. This form allows variables to be separated into $\frac{dy}{h(y)} = g(x)dx$.

Flashcard 45: Identify the integration: $\frac{dy}{y} = x dx$.

Answer: $\ln |y| = \frac{x^2}{2} + C$. Standard result from separating $\frac{dy}{y} = x dx$.

Flashcard 46: State the integration result for $\frac{1}{y} dy = x dx$.

Answer: $\ln |y| = \frac{x^2}{2} + C$. Antiderivatives are $\ln|y|$ and $\frac{x^2}{2}$ respectively.

Flashcard 47: Why is verifying a solution important?

Answer: To ensure it satisfies the original equation. Verification confirms the solution is mathematically correct.

Flashcard 48: Find the general solution: $\frac{dy}{dx} = y^2$.

Answer: $y = -\frac{1}{C-x}$. Separating gives $\frac{dy}{y^2} = dx$, integrating yields $-\frac{1}{y} = x + C$.

Flashcard 49: How is a particular solution found from a general solution?

Answer: Use initial conditions to solve for $C$. Initial conditions determine the specific constant value.

Flashcard 50: What technique is used to verify the solution of a separable differential equation?

Answer: Differentiate the solution and compare with the original equation. Substitution back into the original equation confirms correctness.

Flashcard 51: What is the role of the constant $C$ in the solution of a differential equation?

Answer: $C$ represents an arbitrary constant, allowing for all solutions. The constant encompasses all possible initial conditions.

Flashcard 52: Determine the error: $y dy = x dx$ integrated as $y = x + C$.

Answer: Correct: $\frac{y^2}{2} = \frac{x^2}{2} + C$. Must integrate each term separately: $\int y dy = \frac{y^2}{2}$.

Flashcard 53: Identify the final form: $\frac{dy}{y} = x dx$.

Answer: $\ln |y| = \frac{x^2}{2} + C$. Integration of separated form $\frac{dy}{y} = x dx$.

Flashcard 54: Find the general solution: $\frac{dy}{dx} = y\cos(x)$.

Answer: $y = Ce^{\sin(x)}$. Separating gives $\frac{dy}{y} = \cos(x) dx$, integrating yields $\ln|y| = \sin(x) + C$.

Flashcard 55: Identify the final form: $\frac{dy}{y} = x dx$.

Answer: $\text{ln}|y| = \frac{x^2}{2} + C$. Integration of separated form $\frac{dy}{y} = x dx$.

Flashcard 56: Identify the error: $\frac{dy}{dx} = x^2y^2$ rewritten as $y^2 dy = x^2 dx$.

Answer: Correct: $\frac{1}{y^2} dy = x^2 dx$. The $y^2$ term should move to the left with reciprocal: $\frac{1}{y^2}$.

Flashcard 57: Find the general solution: $\frac{dy}{dx} = y^2 \text{sin}(x)$.

Answer: $y = \frac{1}{C - \text{cos}(x)}$. Separating gives $\frac{dy}{y^2} = \sin(x) dx$, integrating yields $-\frac{1}{y} = -\cos(x) + C$.

Flashcard 58: What is the purpose of $C$ in a general solution?

Answer: It accounts for the family of solutions. Multiple solutions exist for different initial conditions.

Flashcard 59: Determine the type of equation: $\frac{dy}{dx} = x y + 1$.

Answer: Not separable. The constant term prevents separation of variables.

Flashcard 60: What is the role of the constant $C$ in the solution of a differential equation?

Answer: $C$ represents an arbitrary constant, allowing for all solutions. The constant encompasses all possible initial conditions.

Flashcard 61: Determine the type of equation: $\frac{dy}{dx} = x^2 + y^2$.

Answer: Not separable. Cannot separate $x$ and $y$ terms to opposite sides.

Flashcard 62: What must be true about $C$ when using an initial condition $y(x_0) = y_0$?

Answer: $C$ is determined by substituting $x_0$ and $y_0$ into the solution. Substitute the initial condition to solve for the constant.

Flashcard 63: What is the first step in solving a differential equation using separation of variables?

Answer: Rewrite the equation so that all $x$ terms are on one side and $y$ terms on the other. This separates variables to opposite sides of the equation.

Flashcard 64: What does it mean if a differential equation is not separable?

Answer: Variables cannot be isolated on separate sides. The equation cannot be written as $g(x)h(y)$ form.

Flashcard 65: Identify the next step: $\text{ln}|y| = \text{ln}|x| + C$.

Answer: Express $y$ in terms of $x$: $y = Cx$. Use properties of logarithms: $\ln|y| - \ln|x| = \ln|\frac{y}{x}| = C$.

Flashcard 66: Identify the step after integration: $\text{ln}|y| = x^2 + C$.

Answer: Solve for $y$: $y = e^{x^2 + C}$. Exponentiate both sides to isolate $y$ from the logarithm.

Flashcard 67: What must be true about $C$ when using an initial condition $y(x_0) = y_0$?

Answer: $C$ is determined by substituting $x_0$ and $y_0$ into the solution. Substitute the initial condition to solve for the constant.

Flashcard 68: Which method helps find a particular solution after finding a general one?

Answer: Use initial conditions. Initial conditions determine the specific value of the constant $C$.

Flashcard 69: What is the general solution for $\frac{dy}{dx} = y$?

Answer: $y = Ce^x$. This is the standard exponential growth/decay solution.

Flashcard 70: How is a particular solution found from a general solution?

Answer: Use initial conditions to solve for $C$. Initial conditions determine the specific constant value.

Flashcard 71: What is the integration result for $y dy = 2x dx$?

Answer: $\frac{y^2}{2} = x^2 + C$. Integration gives $\frac{y^2}{2} = 2 \cdot \frac{x^2}{2} + C = x^2 + C$.

Flashcard 72: What is the general solution for $\frac{dy}{dx} = 3xy$?

Answer: $y = Ce^{\frac{3x^2}{2}}$. Separating gives $\frac{dy}{y} = 3x dx$, integrating yields $\ln|y| = \frac{3x^2}{2} + C$.

Flashcard 73: Identify the integration: $\frac{dy}{y} = x dx$.

Answer: $\text{ln}|y| = \frac{x^2}{2} + C$. Standard result from separating $\frac{dy}{y} = x dx$.