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

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

Flashcard 3: Find the solution to $\frac{dy}{dt} = -2y$ with initial condition $y(0) = 5$.

Answer: $y(t) = 5e^{-2t}$. Apply initial condition $y(0) = 5$ to general solution.

Flashcard 4: 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 5: 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 6: 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 7: If $\frac{dy}{dt} = ky$, what is the value of $k$ for decay?

Answer: $k < 0$. Negative rate constant produces decreasing exponential solution.

Flashcard 8: 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 9: Which function represents exponential growth: $y = e^{kt}$ or $y = e^{-kt}$?

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

Flashcard 10: 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 11: 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 12: 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 13: 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 14: 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 15: 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 16: 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 17: 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 18: 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 19: What is the carrying capacity in the logistic growth model?

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

Flashcard 20: 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 21: What is the carrying capacity in the logistic growth model?

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

Flashcard 22: Determine the particular solution for $\frac{dy}{dt} = 4y$, $y(0) = 1$.

Answer: $y(t) = e^{4t}$. Initial condition $y(0) = 1$ determines $C = 1$.

Flashcard 23: Determine $y(t)$ for $\frac{dy}{dt} = 5y$, $y(0) = 8$.

Answer: $y(t) = 8e^{5t}$. Growth model with rate 5 and initial condition 8.

Flashcard 24: 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 25: 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 26: 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 27: 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 28: 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 29: 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 30: 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 31: What is the form of a logistic growth differential equation?

Answer: $\frac{dy}{dt} = ky(1 - \frac{y}{L})$. Growth rate decreases as population approaches carrying capacity $L$.

Flashcard 32: 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 33: What is the solution to $\frac{dy}{dt} = 0.1y$ with $y(0) = 10$?

Answer: $y(t) = 10e^{0.1t}$. Slow growth with rate 0.1 and initial value 10.

Flashcard 34: Which function represents exponential growth: $y = e^{kt}$ or $y = e^{-kt}$?

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

Flashcard 35: 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 36: 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 37: State the expression for $y(t)$ if $y$ doubles every 6 years.

Answer: $y(t) = y_0 e^{(t/6)\text{ln}(2)}$. Doubling time of 6 years gives rate $k = \frac{\ln(2)}{6}$.

Flashcard 38: Find the general solution of $\frac{dy}{dt} = 3y$.

Answer: $y(t) = Ce^{3t}$. General solution has arbitrary constant $C$ with rate $k = 3$.

Flashcard 39: What is the particular solution for $\frac{dy}{dt} = -0.3y$, $y(0) = 12$?

Answer: $y(t) = 12e^{-0.3t}$. Exponential decay with rate 0.3 and initial value 12.

Flashcard 40: 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 41: What is the rate constant $k$ if the population doubles in 3 years?

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

Flashcard 42: Solve $\frac{dy}{dt} = 2y$ for $y(0) = 3$. What is $y(t)$?

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

Flashcard 43: 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 44: 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 45: 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 46: 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 47: 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 48: 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 49: 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 50: What is the exponential decay model?

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

Flashcard 51: Which function represents exponential decay: $y = e^{kt}$ or $y = e^{-kt}$?

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

Flashcard 52: 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 53: 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 54: What is the exponential decay model?

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

Flashcard 55: 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 56: 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 57: 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 58: What is the exponential growth model?

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

Flashcard 59: 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 60: 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 61: 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.