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 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 2: 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 3: 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 4: 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 5: 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 6: 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 7: 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 8: 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 9: 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 10: 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 11: 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 12: 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 13: 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 14: 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 15: 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 16: 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 17: 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 18: 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 19: 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 20: 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 21: 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 22: 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 23: 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 24: 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 25: 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 26: 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 27: 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 28: 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 29: 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 30: What form does a differential equation take after separation?

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