AP Calculus AB 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 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 happens to the constant of integration when finding a particular solution?

Answer: It is determined using an initial condition. Initial condition substitutes to find specific $C$ value.

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

Answer: Correct: $\frac{1}{y} dy = x^2 dx$. Variables must be properly separated before integrating.

Flashcard 3: What is the role of the constant $C$ in the solution?

Answer: It accounts for the family of solutions. Represents all possible curves in solution family.

Flashcard 4: What form must a differential equation have to use separation of variables?

Answer: The form must be $N(y) \frac{dy}{dx} = M(x)$. Variables must be separable on opposite sides.

Flashcard 5: How do you solve $\frac{dy}{dx} = x e^{y}$ using separation of variables?

Answer: Rewrite as $e^{-y} dy = x dx$ and integrate. Move exponential to left side before separating.

Flashcard 6: What is the general solution for $\frac{dy}{dx} = 0$?

Answer: $y = C$, where $C$ is a constant. Zero derivative means $y$ is constant function.

Flashcard 7: What is the general solution of $\frac{dy}{dx} = ky$?

Answer: $y = Ce^{kx}$, where $C$ is a constant. Standard exponential growth/decay model solution.

Flashcard 8: Separate variables and integrate: $\frac{dy}{dx} = \frac{2x}{y}$.

Answer: $y dy = 2x dx$; integrate to $y^2 = x^2 + C$. Standard separation and integration process.

Flashcard 9: What is the general solution for $\frac{dy}{dx} = 0$?

Answer: $y = C$, where $C$ is a constant. Zero derivative means $y$ is constant function.

Flashcard 10: How do you verify a general solution found by separation of variables?

Answer: Differentiate and substitute back into the original equation. Confirms the solution satisfies the original equation.

Flashcard 11: What is the result of integrating $e^{-y} dy = x dx$?

Answer: $-e^{-y} = \frac{x^2}{2} + C$. Standard result from integrating separated variables.

Flashcard 12: What is the general solution of $\frac{dy}{dx} = \frac{x}{y}$ after separation of variables?

Answer: $y^2 = x^2 + C$. Result from separating $y dy = x dx$ and integrating.

Flashcard 13: How is the constant of integration $C$ determined in a particular solution?

Answer: By using given initial conditions. Substitution yields specific solution from general form.

Flashcard 14: What happens to the constant of integration when finding a particular solution?

Answer: It is determined using an initial condition. Initial condition substitutes to find specific $C$ value.

Flashcard 15: What is the solution of $\frac{dy}{dx} = y^2$ using separation of variables?

Answer: $-\frac{1}{y} = x + C$. Separate: $y^{-2} dy = dx$, then integrate both sides.

Flashcard 16: What is the result of integrating $e^{-y} dy = x dx$?

Answer: $-e^{-y} = \frac{x^2}{2} + C$. Standard result from integrating separated variables.

Flashcard 17: What form does a differential equation take after separation?

Answer: $N(y) dy = M(x) dx$. Standard separated form ready for integration.

Flashcard 18: What must you do after separating variables in a differential equation?

Answer: Integrate both sides with respect to their variables. This yields the general solution with constant $C$.

Flashcard 19: What is the solution of $\frac{dy}{dx} = y^2$ using separation of variables?

Answer: $-\frac{1}{y} = x + C$. Separate: $y^{-2} dy = dx$, then integrate both sides.

Flashcard 20: What is the purpose of the constant $C$ in a general solution?

Answer: It represents the constant of integration. Accounts for all possible solutions in the family.

Flashcard 21: How do you verify a general solution found by separation of variables?

Answer: Differentiate and substitute back into the original equation. Confirms the solution satisfies the original equation.

Flashcard 22: How do you separate variables for $\frac{dy}{dx} = \frac{x}{\text{ln}(y)}$?

Answer: Rewrite as $\text{ln}(y) dy = x dx$. Move $\ln(y)$ to denominator for proper separation.

Flashcard 23: State the form of a differential equation suitable for separation of variables.

Answer: $N(y) dy = M(x) dx$ form. Variables separated on opposite sides of equation.

Flashcard 24: State the form of a differential equation suitable for separation of variables.

Answer: $N(y) dy = M(x) dx$ form. Variables separated on opposite sides of equation.

Flashcard 25: Find the error in separation: $y \frac{dy}{dx} = \frac{1}{x}$ rewritten as $y dy = \frac{1}{x} dx$.

Answer: Correct form: $\frac{1}{y} dy = \frac{1}{x} dx$. Division by $y$ was missed in the separation.

Flashcard 26: What is $y$ if $\frac{dy}{dx} = xy$ after integrating and solving?

Answer: $y = Ce^{\frac{x^2}{2}}$. Exponential solution from integrating $\frac{dy}{y} = x dx$.

Flashcard 27: What do you obtain after integrating both sides of a separated equation?

Answer: An implicit solution, often in terms of $y$ and $x$. Integration produces this general form before solving for $y$.

Flashcard 28: How do you separate variables for $\frac{dy}{dx} = \frac{x}{\text{ln}(y)}$?

Answer: Rewrite as $\text{ln}(y) dy = x dx$. Move $\ln(y)$ to denominator for proper separation.

Flashcard 29: State the integral of $y \frac{dy}{dx} = x$ after separation.

Answer: $\frac{1}{2}y^2 = \frac{1}{2}x^2 + C$. Direct integration after variable separation.

Flashcard 30: In separation of variables, what is done after integrating?

Answer: Solve for $y$ if possible to express $y$ explicitly. Convert implicit form to explicit if possible.

Flashcard 31: What does the equation $dy = M(x) dx$ imply after separation?

Answer: Integrate both sides to find $y = \text{integral of } M(x) dx + C$. Direct integration when variables are separated.

Flashcard 32: What is the solution to $\frac{dy}{dx} = x^2 y^3$ after separation?

Answer: $\frac{-1}{2y^2} = \frac{x^3}{3} + C$. Separate: $y^{-3} dy = x^2 dx$, then integrate.

Flashcard 33: What is the purpose of the constant $C$ in a general solution?

Answer: It represents the constant of integration. Accounts for all possible solutions in the family.

Flashcard 34: How is the constant of integration $C$ determined in a particular solution?

Answer: By using given initial conditions. Substitution yields specific solution from general form.

Flashcard 35: What form must a differential equation have to use separation of variables?

Answer: The form must be $N(y) \frac{dy}{dx} = M(x)$. Variables must be separable on opposite sides.

Flashcard 36: Identify the error: $\frac{dy}{dx} = x^2y$ rewritten as $y \frac{dy}{dx} = x^2$.

Answer: Correct: $\frac{1}{y} dy = x^2 dx$. Variables must be properly separated before integrating.

Flashcard 37: State the integral of $y \frac{dy}{dx} = x$ after separation.

Answer: $\frac{1}{2}y^2 = \frac{1}{2}x^2 + C$. Direct integration after variable separation.

Flashcard 38: What is the next step after obtaining $y \frac{dy}{dx} = x$?

Answer: Separate to $y dy = x dx$ and integrate both sides. Variables are now separated for integration.

Flashcard 39: What is the general solution of $\frac{dy}{dx} = \frac{1}{y}$?

Answer: $y^2 = 2x + C$. Integrate $\int y dy = \int dx$ to get this result.

Flashcard 40: Separate variables and integrate: $\frac{dy}{dx} = \frac{2x}{y}$.

Answer: $y dy = 2x dx$; integrate to $y^2 = x^2 + C$. Standard separation and integration process.

Flashcard 41: What must you do after separating variables in a differential equation?

Answer: Integrate both sides with respect to their variables. This yields the general solution with constant $C$.

Flashcard 42: What form does a differential equation take after separation?

Answer: $N(y) dy = M(x) dx$. Standard separated form ready for integration.

Flashcard 43: What is the next step after obtaining $y \frac{dy}{dx} = x$?

Answer: Separate to $y dy = x dx$ and integrate both sides. Variables are now separated for integration.

Flashcard 44: What is the solution to $\frac{dy}{dx} = y^2 x$?

Answer: $-\frac{1}{y} = \frac{x^2}{2} + C$. Separate: $y^{-2} dy = x dx$, then integrate both sides.

Flashcard 45: What is a key requirement for using separation of variables?

Answer: The equation must be factored into $N(y) dy = M(x) dx$. Variables must be separable into this product form.

Flashcard 46: In separation of variables, what is done after integrating?

Answer: Solve for $y$ if possible to express $y$ explicitly. Convert implicit form to explicit if possible.

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

Answer: $-\frac{1}{y} = \frac{x^2}{2} + C$. Separate: $y^{-2} dy = x dx$, then integrate both sides.

Flashcard 48: Separate and integrate: $\frac{dy}{dx} = xy$.

Answer: Rewrite as $\frac{1}{y} dy = x dx$ and integrate. Standard separation technique for this product form.

Flashcard 49: What do you obtain after integrating both sides of a separated equation?

Answer: An implicit solution, often in terms of $y$ and $x$. Integration produces this general form before solving for $y$.

Flashcard 50: What is the general solution of $\frac{dy}{dx} = \frac{1}{x}$?

Answer: $y = \text{ln}|x| + C$. Direct integration of $\frac{1}{x}$ with respect to $x$.

Flashcard 51: What is the general solution of $\frac{dy}{dx} = \frac{1}{y}$?

Answer: $y^2 = 2x + C$. Integrate $\int y dy = \int dx$ to get this result.

Flashcard 52: What is the first step in solving a differential equation using separation of variables?

Answer: Rewrite the equation as $N(y) \frac{dy}{dx} = M(x)$. This isolates variables on separate sides for integration.

Flashcard 53: What is $y$ if $\frac{dy}{dx} = xy$ after integrating and solving?

Answer: $y = Ce^{\frac{x^2}{2}}$. Exponential solution from integrating $\frac{dy}{y} = x dx$.

Flashcard 54: What is the first step in solving a differential equation using separation of variables?

Answer: Rewrite the equation as $N(y) \frac{dy}{dx} = M(x)$. This isolates variables on separate sides for integration.

Flashcard 55: Find the general solution for $\frac{dy}{dx} = \frac{x}{y}$.

Answer: $y^2 = x^2 + C$. Separate variables: $y dy = x dx$, then integrate.

Flashcard 56: After separation and integration, what form does the solution take?

Answer: An implicit equation involving $y$, $x$, and $C$. Variables not yet isolated to explicit $y$ form.

Flashcard 57: How do you solve $\frac{dy}{dx} = x e^{y}$ using separation of variables?

Answer: Rewrite as $e^{-y} dy = x dx$ and integrate. Move exponential to left side before separating.

Flashcard 58: Identify the next step: $\frac{dy}{dx} = \frac{3x}{2y}$ rewritten as $2y dy = 3x dx$.

Answer: Integrate both sides: $\frac{y^2}{2} = \frac{3x^2}{2} + C$. Standard integration after proper separation.

Flashcard 59: What does it mean if a solution is implicit?

Answer: It is not solved explicitly for $y$ in terms of $x$. $y$ cannot be isolated algebraically from the equation.

Flashcard 60: Identify the next step: $\frac{dy}{dx} = \frac{3x}{2y}$ rewritten as $2y dy = 3x dx$.

Answer: Integrate both sides: $\frac{y^2}{2} = \frac{3x^2}{2} + C$. Standard integration after proper separation.

Flashcard 61: What is the role of the constant $C$ in the solution?

Answer: It accounts for the family of solutions. Represents all possible curves in solution family.

Flashcard 62: Identify the integration step for $y \frac{dy}{dx} = x^2$.

Answer: Integrate: $\frac{y^2}{2} = \frac{x^3}{3} + C$. Direct integration after separating variables properly.

Flashcard 63: What is the solution to $\frac{dy}{dx} = x^2 y^3$ after separation?

Answer: $\frac{-1}{2y^2} = \frac{x^3}{3} + C$. Separate: $y^{-3} dy = x^2 dx$, then integrate.

Flashcard 64: What is the general solution of $\frac{dy}{dx} = \frac{1}{x}$?

Answer: $y = \text{ln}|x| + C$. Direct integration of $\frac{1}{x}$ with respect to $x$.

Flashcard 65: Separate and integrate: $\frac{dy}{dx} = xy$.

Answer: Rewrite as $\frac{1}{y} dy = x dx$ and integrate. Standard separation technique for this product form.

Flashcard 66: Find the general solution for $\frac{dy}{dx} = \frac{x}{y}$.

Answer: $y^2 = x^2 + C$. Separate variables: $y dy = x dx$, then integrate.

Flashcard 67: What does the equation $dy = M(x) dx$ imply after separation?

Answer: Integrate both sides to find $y = \text{integral of } M(x) dx + C$. Direct integration when variables are separated.

Flashcard 68: What is a key requirement for using separation of variables?

Answer: The equation must be factored into $N(y) dy = M(x) dx$. Variables must be separable into this product form.

Flashcard 69: What does it mean if a solution is implicit?

Answer: It is not solved explicitly for $y$ in terms of $x$. $y$ cannot be isolated algebraically from the equation.

Flashcard 70: After separation and integration, what form does the solution take?

Answer: An implicit equation involving $y$, $x$, and $C$. Variables not yet isolated to explicit $y$ form.

Flashcard 71: What is the general solution of $\frac{dy}{dx} = \frac{x}{y}$ after separation of variables?

Answer: $y^2 = x^2 + C$. Result from separating $y dy = x dx$ and integrating.

Flashcard 72: What is the general solution of $\frac{dy}{dx} = ky$?

Answer: $y = Ce^{kx}$, where $C$ is a constant. Standard exponential growth/decay model solution.

Flashcard 73: Find the error in separation: $y \frac{dy}{dx} = \frac{1}{x}$ rewritten as $y dy = \frac{1}{x} dx$.

Answer: Correct form: $\frac{1}{y} dy = \frac{1}{x} dx$. Division by $y$ was missed in the separation.

Flashcard 74: Identify the integration step for $y \frac{dy}{dx} = x^2$.

Answer: Integrate: $\frac{y^2}{2} = \frac{x^3}{3} + C$. Direct integration after separating variables properly.