Initial Conditions and Separation of Variables - AP Calculus BC

1. Convert $\frac{dy}{dx} = x^2y^3$ into a separable form.

   $\frac{dy}{y^3} = x^2 dx$. Move $y^3$ to the left side and $x^2$ to the right.

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

   $y = Ce^{-x}$. Standard exponential decay with rate $1$.

3. What is the particular solution for $\frac{dy}{dx} = -3y$ with $y(2) = 1$?

   $y = e^{-3x + 6}$. From $y = Ce^{-3x}$, use $y(2) = 1$ to find $C = e^6$.

4. Determine the general solution for $\frac{dy}{dx} = \frac{1}{x}$.

   $y = \text{ln}|x| + C$. Direct integration of the reciprocal function.

5. What is the role of $C$ in the solution $y = Ce^{kx}$?

   Constant for initial conditions. Determines the specific solution from the family of solutions.

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

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

7. Given $\frac{dy}{dx} = 4x$, find $C$ if $y(1) = 6$.

   $C = 4$. From $y = 2x^2 + C$ with $y(1) = 6$, so $C = 4$.

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

   $y = C$. When the derivative is zero, the function is constant.

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

   $y = 3e^{2x}$. From $y = Ce^{2x}$ with condition $y(0) = 3$ giving $C = 3$.

10. What is the integration result for $\frac{dy}{y} = k dx$?

    $\ln |y| = kx + C$. Standard integration result for exponential-type equations.

11. How do you integrate $y' = \frac{dy}{dx} = x^2 y$ using separation of variables?

    $\frac{dy}{y} = x^2 dx$. Divide both sides by $y$ and multiply by $dx$.

12. What is the role of the integration constant $C$ in differential equations?

    Represents the family of solutions. Each value of $C$ gives a different particular solution.

13. Find the particular solution of $\frac{dy}{dx} = 2x$ with $y(0) = 4$.

    $y = x^2 + 4$. Integrate to get $y = x^2 + C$, then use initial condition.

14. What is the general solution to $\frac{dy}{dx} = -ky$?

    $y = Ce^{-kx}$. Standard exponential decay solution with rate $k$.

15. What is the general solution to $\frac{dy}{dx} = ky$ using separation of variables?

    $y = Ce^{kx}$. Standard form for exponential growth/decay differential equations.

16. Given $\frac{dy}{dx} = 2y$, find the particular solution with $y(0) = 1$.

    $y = e^{2x}$. From $y = Ce^{2x}$ with initial condition $y(0) = 1$.

17. Solve $\frac{dy}{dx} = \frac{y}{x}$ for $y$ using separation of variables.

    $y = Cx$. Separate to $\frac{dy}{y} = \frac{dx}{x}$ and integrate both sides.

18. Find the particular solution to $\frac{dy}{dx} = \frac{2y}{x}$ with $y(1) = 3$.

    $y = 3x^2$. Separate to $\frac{dy}{y} = \frac{2dx}{x}$, integrate to get $y = Cx^2$.

19. Solve $\frac{dy}{dx} = x^3y^2$ by separating variables.

    $\frac{dy}{y^2} = x^3 dx$. Move $y^2$ to left side and $x^3$ to right side.

20. Determine the particular solution of $\frac{dy}{dx} = y$ with $y(0) = 5$.

    $y = 5e^x$. From general solution $y = Ce^x$, use condition to find $C = 5$.

21. What equation results from integrating $\frac{dy}{y} = \frac{1}{x} dx$?

    $\ln |y| = \ln |x| + C$. Standard result from integrating separated variables.

22. What is the general solution for $\frac{dy}{dx} = x^2$?

    $y = \frac{x^3}{3} + C$. Direct integration of the polynomial function.

23. Determine $C$ for the particular solution of $\frac{dy}{dx} = 5y$ with $y(0) = 7$.

    $C = 7$. From $y = Ce^{5x}$ with $y(0) = 7$, so $C = 7$.

24. Solve for $y$ given $\frac{dy}{dx} = x$ with initial condition $y(1) = 3$.

    $y = \frac{x^2}{2} + \frac{5}{2}$. Integrate $\int x \, dx = \frac{x^2}{2} + C$ and apply condition.

25. State the integration result of $\frac{dy}{y} = \frac{1}{x} dx$.

    $\text{ln}|y| = \text{ln}|x| + C$. Direct integration of the separated variables.

26. What is the purpose of using initial conditions in solving differential equations?

    To find particular solutions. They eliminate the arbitrary constant $C$ from general solutions.

27. What is the purpose of finding a particular solution?

    To satisfy given initial conditions. Eliminates the arbitrary constant using given conditions.

28. Given $\frac{dy}{dx} = 3x^2$, what is the particular solution with $y(0) = 2$?

    $y = x^3 + 2$. Integrate to get $y = x^3 + C$, then use $y(0) = 2$ to find $C = 2$.

29. Find the particular solution for $\frac{dy}{dx} = 3x^2$ with $y(0) = 1$.

    $y = x^3 + 1$. Integrate $3x^2$ and apply the initial condition at $x = 0$.

30. Given $y = Ce^{x^2}$, find $C$ if $y(0) = 1$.

    $C = 1$. Substitute $x = 0$ and $y = 1$ into the general solution.

31. What is the solution to $\frac{dy}{dx} = 4y$ using separation of variables?

    $y = Ce^{4x}$. Standard exponential solution with growth rate $k = 4$.

32. Solve $\frac{dy}{dx} = -2xy$ for $y$ using separation of variables.

    $y = Ce^{-x^2}$. Separate to $\frac{dy}{y} = -2x \, dx$ and integrate both sides.

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

    Separate the variables. This allows variables to be on different sides for integration.

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