AP Calculus BC Flashcards: Exponential Models With Differential Equations

What this deck covers

This deck focuses on Exponential Models With Differential Equations, 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 exponential decay model?

Answer: $y = y_0 e^{-kt}$. Initial value $y_0$ times exponential with negative rate $k$.

Flashcard 2: Identify the integrating factor for the differential equation $\frac{dy}{dt} + Py = Q$.

Answer: $e^{\textstyle \int P \, dt}$. Multiplying factor to convert equation to exact form.

Flashcard 3: What is the solution to the differential equation $\frac{dy}{dt} = ky$?

Answer: $y = Ce^{kt}$. Exponential function with arbitrary constant $C$ and growth rate $k$.

Flashcard 4: What is the expression for $y(t)$ if $\frac{dy}{dt} = ky$ and $y(0) = y_0$?

Answer: $y(t) = y_0 e^{kt}$. Exponential growth formula with given initial condition.

Flashcard 5: Solve $\frac{dy}{dt} = -4y$ given $y(0) = 9$.

Answer: $y(t) = 9e^{-4t}$. Exponential decay with rate $k = 4$ and initial value 9.

Flashcard 6: Which function represents exponential decay: $y = e^{kt}$ or $y = e^{-kt}$?

Answer: $y = e^{-kt}$. Negative exponent causes exponential decrease over time.

Flashcard 7: What is the equilibrium solution for $\frac{dy}{dt} = ky(1 - \frac{y}{L})$?

Answer: $y = 0$ and $y = L$. Points where $\frac{dy}{dt} = 0$, so population remains constant.

Flashcard 8: Solve $\frac{dy}{dt} = 3y$ if $y(0) = 6$. What is $y(t)$?

Answer: $y(t) = 6e^{3t}$. Growth rate 3 with initial condition $y(0) = 6$.

Flashcard 9: What is the solution to $\frac{dy}{dt} = 0.5y$ with $y(0) = 15$?

Answer: $y(t) = 15e^{0.5t}$. Growth model with rate $k = 0.5$ and initial value 15.

Flashcard 10: Find $y(t)$ for $\frac{dy}{dt} = ky$, $y(0) = y_0$.

Answer: $y(t) = y_0 e^{kt}$. Standard exponential solution with initial value specified.

Flashcard 11: Determine $y(t)$ if $\frac{dy}{dt} = -3y$ and $y(0) = 7$.

Answer: $y(t) = 7e^{-3t}$. Decay model with initial condition $y(0) = 7$.

Flashcard 12: Which function represents exponential growth: $y = e^{kt}$ or $y = e^{-kt}$?

Answer: $y = e^{kt}$. Positive exponent causes exponential increase over time.

Flashcard 13: What is the rate constant $k$ if the population doubles in 3 years?

Answer: $k = \frac{\text{ln}(2)}{3}$. Doubling time formula $t_d = \frac{\ln(2)}{k}$ solved for $k$.

Flashcard 14: State the formula for the derivative of $e^{kt}$ with respect to $t$.

Answer: $\frac{d}{dt} e^{kt} = ke^{kt}$. Chain rule applied to exponential function with coefficient $k$.

Flashcard 15: What is the carrying capacity in the logistic growth model?

Answer: $L$. Maximum sustainable population in logistic growth model.

Flashcard 16: What is the solution to $\frac{dy}{dt} = -6y$ for $y(0) = 4$?

Answer: $y(t) = 4e^{-6t}$. Decay with rate 6 and initial condition 4.

Flashcard 17: Find $y(t)$ if $\frac{dy}{dt} = 5y$ and $y(0) = 10$.

Answer: $y(t) = 10e^{5t}$. Initial condition $y(0) = 10$ with growth rate $k = 5$.

Flashcard 18: What is the solution to the separable equation $\frac{dy}{dx} = x^2 y$?

Answer: $y = Ce^{x^3/3}$. Separation gives $\frac{dy}{y} = x^2 dx$, integrate both sides.

Flashcard 19: What is the solution to $\frac{dy}{dt} = -0.2y$ with $y(0) = 20$?

Answer: $y(t) = 20e^{-0.2t}$. Slow decay with rate 0.2 and initial value 20.

Flashcard 20: What is the doubling time formula for exponential growth?

Answer: $t_d = \frac{\text{ln}(2)}{k}$. Time for quantity to double its original value.

Flashcard 21: What is the half-life formula for exponential decay?

Answer: $t_{1/2} = \frac{\text{ln}(2)}{k}$. Time for quantity to reduce to half its original value.

Flashcard 22: What is the form of a separable differential equation?

Answer: $\frac{dy}{dx} = g(y)h(x)$. Variables can be separated: $\frac{dy}{g(y)} = h(x)dx$.

Flashcard 23: Find $y(t)$ if $\frac{dy}{dt} = 7y$ and $y(0) = 2$.

Answer: $y(t) = 2e^{7t}$. Exponential growth with rate 7 and initial condition 2.

Flashcard 24: Find the particular solution for $\frac{dy}{dt} = 9y$, $y(0) = 1$.

Answer: $y(t) = e^{9t}$. Fast growth rate 9 with unit initial condition.

Flashcard 25: Determine $y(t)$ if $\frac{dy}{dt} = -3y$ and $y(0) = 7$.

Answer: $y(t) = 7e^{-3t}$. Decay model with initial condition $y(0) = 7$.

Flashcard 26: What is the expression for $k$ if the population triples in 5 years?

Answer: $k = \frac{\text{ln}(3)}{5}$. Tripling time formula $t = \frac{\ln(3)}{k}$ solved for $k$.

Flashcard 27: Which function represents exponential growth: $y = e^{kt}$ or $y = e^{-kt}$?

Answer: $y = e^{kt}$. Positive exponent causes exponential increase over time.

Flashcard 28: What is the solution to $\frac{dy}{dt} = 0.5y$ with $y(0) = 15$?

Answer: $y(t) = 15e^{0.5t}$. Growth model with rate $k = 0.5$ and initial value 15.

Flashcard 29: If $\frac{dy}{dt} = ky$ and $y(0) = y_0$, express $y(t)$ in terms of $y_0$.

Answer: $y(t) = y_0 e^{kt}$. Standard exponential growth solution with initial value $y_0$.

Flashcard 30: Solve $\frac{dy}{dt} = 2y$ for $y(0) = 3$. What is $y(t)$?

Answer: $y(t) = 3e^{2t}$. Initial condition $y(0) = 3$ gives $C = 3$.