Exponential Models with Differential Equations - AP Calculus BC

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

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

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

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

3. If $\frac{dy}{dt} = ky$, what is the value of $k$ for decay?

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

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

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

5. What is the doubling time formula for exponential growth?

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

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

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

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

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

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

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

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

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

10. What is the expression for $k$ if the population triples in 5 years?

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

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

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

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

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

13. What is the carrying capacity in the logistic growth model?

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

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

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

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

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

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

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

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

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

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

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

19. What is the solution to the differential equation $\frac{dy}{dt} = ky$?

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

20. What is the form of a logistic growth differential equation?

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

21. What is the solution to $\frac{dy}{dt} = 0.1y$ with $y(0) = 10$?

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

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

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

23. What is the form of a separable differential equation?

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

24. State the expression for $y(t)$ if $y$ doubles every 6 years.

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

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

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

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

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

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

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

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

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

29. What is the rate constant $k$ if the population doubles in 3 years?

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

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

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

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

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

32. What is the half-life formula for exponential decay?

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

33. What is the exponential decay model?

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

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

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

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

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

36. What is the exponential growth model?

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

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

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

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