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

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

Flashcard 2: 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 3: 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 4: 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 5: 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 6: 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 7: 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 8: 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 9: 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 10: 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 11: 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 12: 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 13: 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 14: 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 15: 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 16: 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 17: 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 18: 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 19: Why is verifying a solution important?

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

Flashcard 20: 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 21: 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 22: 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 23: 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 24: 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 25: 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 26: What is the general solution for $\frac{dy}{dx} = y$?

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

Flashcard 27: 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 28: Identify the integration: $\frac{dy}{y} = x dx$.

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

Flashcard 29: 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 30: 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 $-\frac{1}{y} = -\cos(x) + C$.