Initial Conditions and Separation of Variables - AP Calculus AB

1. Find the particular solution of $\frac{dy}{dx} = 4y$ with $y(1) = 3$.

   $y = 3e^{4(x-1)}$. General solution $y = Ce^{4x}$, apply $y(1) = 3$ to find $C$.

2. Find $C$ for the solution $y = Ce^{3x}$ given $y(0) = 5$.

   $C = 5$. At $x = 0$: $5 = Ce^0 = C \cdot 1$, so $C = 5$.

3. What technique is used to solve $\frac{dy}{dx} = \frac{x}{y}$?

   Separation of variables. The equation has form $\frac{dy}{dx} = f(x)g(y)$ where variables separate.

4. What does 'separable' mean in the context of differential equations?

   Variables can be separated on opposite sides. The equation can be written as $f(x)dx = g(y)dy$.

5. Solve $\frac{dy}{dx} = y$ for $y(0) = 2$. What is the particular solution?

   $y = 2e^x$. Separate: $\frac{dy}{y} = dx$, integrate, apply initial condition.

6. Given $\frac{dy}{dx} = 3x$, find the particular solution if $y(0) = 2$.

   $y = \frac{3}{2}x^2 + 2$. Integrate $3x$ and apply $y(0) = 2$ to find $C = 2$.

7. Find the particular solution for $\frac{dy}{dx} = 6x$ with $y(2) = 5$.

   $y = 3x^2 - 7$. Integrate: $y = 3x^2 + C$, apply $y(2) = 5$ to find $C$.

8. What is the general solution for $\frac{dy}{dx} = 0$?

   $y = C$. Zero derivative means the function is constant.

9. Identify the integral needed to solve $\frac{dy}{dx} = x^2y$ after separating variables.

   $\frac{1}{y} dy = x^2 dx$. Rearranging gives $\frac{dy}{y} = x^2 dx$ for integration.

10. What is the purpose of integrating both sides after separation of variables?

    To find the antiderivatives, leading to the general solution. Integration reverses differentiation to recover the function.

11. What is the initial condition in a differential equation problem?

    A value that specifies the solution at a particular point, e.g., $y(x_0) = y_0$. This constraint determines the unique particular solution.

12. What is the general solution for $\frac{dy}{dx} = 2y$?

    $y = Ce^{2x}$. This is exponential growth with rate $2$.

13. What is the general solution of $\frac{dy}{dx} = y$?

    $y = Ce^x$. This is the exponential growth equation with rate $1$.

14. What is the particular solution of $\frac{dy}{dx} = 2x - 1$ if $y(0) = 3$?

    $y = x^2 - x + 3$. Integrate: $y = x^2 - x + C$, apply $y(0) = 3$ gives $C = 3$.

15. Given $\frac{dy}{dx} = -3y$, find the particular solution if $y(0) = 2$.

    $y = 2e^{-3x}$. Separate: $\frac{dy}{y} = -3dx$, integrate, apply initial condition.

16. Given $\frac{dy}{dx} = y\text{sin}(x)$, find the general solution.

    $y = Ce^{-\text{cos}(x)}$. Separate: $\frac{dy}{y} = \sin(x) dx$, then integrate both sides.

17. Find the particular solution of $\frac{dy}{dx} = 2xy$ with $y(0) = 1$.

    $y = e^{x^2}$. Separate: $\frac{dy}{y} = 2x dx$, integrate, apply $y(0) = 1$.

18. What is the role of the integration constant when finding the general solution?

    Represents an arbitrary constant for family of solutions. It parameterizes all possible solutions before applying conditions.

19. Find $C$ for $y = \frac{x^3}{3} + C$ given $y(1) = 2$.

    $C = \frac{5}{3}$. At $x = 1$: $2 = \frac{1}{3} + C$, so $C = \frac{5}{3}$.

20. How do you determine the constant of integration using an initial condition?

    Substitute the initial condition into the general solution. This replaces $C$ with a specific numerical value.

21. Given $\frac{dy}{dx} = y \sin(x)$, find the general solution.

    $y = C e^{-\cos(x)}$. Separate: $\frac{dy}{y} = \sin(x) dx$, then integrate both sides.

22. Solve $\frac{dy}{dx} = 2y$ for $y(0) = 3$. What is the particular solution?

    $y = 3e^{2x}$. Separate: $\frac{dy}{y} = 2dx$, integrate, apply initial condition.

23. State the general form of a separable differential equation.

    An equation of the form $\frac{dy}{dx} = g(x)h(y)$.. The function can be factored into separate $x$ and $y$ terms.

24. What is the general solution of $\frac{dy}{dx} = 4$?

    $y = 4x + C$. Direct integration of constant $4$ gives $4x + C$.

25. What does it mean if a solution is 'particular'?

    It satisfies the differential equation and initial condition. It's the unique solution from the family that meets given conditions.

26. Find the particular solution of $\frac{dy}{dx} = 5x^4$ with $y(0) = 7$.

    $y = x^5 + 7$. Integrate: $y = x^5 + C$, apply $y(0) = 7$ to find $C = 7$.

27. What is the role of the constant of integration in solving separable equations?

    It accounts for the family of solutions. Different values of $C$ give different solution curves.

28. Determine the particular solution for $\frac{dy}{dx} = x$ with $y(0) = 1$.

    $y = \frac{x^2}{2} + 1$. Integrate: $y = \frac{x^2}{2} + C$, apply $y(0) = 1$ to find $C = 1$.

29. What is the general solution of $\frac{dy}{dx} = 1$?

    $y = x + C$. Direct integration of the constant function gives $x + C$.

30. Describe the integration process in separation of variables.

    Integrate both sides after separating variables. This finds antiderivatives of both separated expressions.

31. What is the first step in solving a differential equation using separation of variables?

    Separate the variables to opposite sides of the equation. This isolates each variable for independent integration.

32. Determine the particular solution for $\frac{dy}{dx} = x^3$ with $y(1) = 4$.

    $y = \frac{x^4}{4} + \frac{15}{4}$. Integrate: $y = \frac{x^4}{4} + C$, apply $y(1) = 4$ to find $C$.

33. State the process to find particular solutions using initial conditions.

    Integrate, apply initial conditions, solve for $C$. This systematic approach ensures the correct particular solution.

34. Solve $\frac{dy}{dx} = \frac{x}{y}$ given $y(1) = 2$. What is the particular solution?

    $y^2 = x^2 + 3$. Separate and integrate: $\int y dy = \int x dx$, then apply initial condition.

35. Explain why initial conditions are necessary for finding particular solutions.

    To determine the specific value of the integration constant. Without them, we only get the general solution family.

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