Finding General Solutions: Separation of Variables - AP Calculus BC

1. What is the result of integrating $x dx = y dy$?

   $\frac{x^2}{2} = \frac{y^2}{2} + C$. Both sides integrate to $\frac{x^2}{2}$ and $\frac{y^2}{2}$ respectively.

2. What is the purpose of $C$ in a general solution?

   It accounts for the family of solutions. Multiple solutions exist for different initial conditions.

3. Why is verifying a solution important?

   To ensure it satisfies the original equation. Verification confirms the solution is mathematically correct.

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

   $y = x^2 + C$. Direct integration of $\frac{dy}{dx} = 2x$ gives $y = x^2 + C$.

5. Identify the integration: $y dy = x dx$.

   $\frac{y^2}{2} = \frac{x^2}{2} + C$. Both sides integrate using the power rule.

6. What is the next step after separating variables in a differential equation?

   Integrate both sides with respect to their variables. Integration produces antiderivatives on each side of the equation.

7. Find the general solution: $\frac{dy}{dx} = y\text{cos}(x)$.

   $y = Ce^{\text{sin}(x)}$. Separating gives $\frac{dy}{y} = \cos(x) dx$, integrating yields $\ln|y| = \sin(x) + C$.

8. Find the general solution: $\frac{dy}{dx} = yx$.

   $y = Ce^{\frac{x^2}{2}}$. Same as previous: separating $\frac{dy}{y} = x dx$ gives this result.

9. What is the integration result for $y dy = 2x dx$?

   $\frac{y^2}{2} = x^2 + C$. Integration gives $\frac{y^2}{2} = 2 \cdot \frac{x^2}{2} + C = x^2 + C$.

10. Find the general solution: $\frac{dy}{dx} = 3x^2 y$.

    $y = Ce^{x^3}$. Separating gives $\frac{dy}{y} = 3x^2 dx$, integrating yields $\ln|y| = x^3 + C$.

11. Find the general solution: $\frac{dy}{dx} = y^2$.

    $y = -\frac{1}{C-x}$. Separating gives $\frac{dy}{y^2} = dx$, integrating yields $-\frac{1}{y} = x + C$.

12. Identify the integral to solve after separating variables: $\frac{dy}{dx} = x \frac{1}{y}$

    $\text{Integrate: } \frac{1}{y} dy = x dx$. Separation gives $\frac{dy}{y} = x dx$, then integrate both sides.

13. Identify the integral to solve after separating variables: $\frac{dy}{dx} = x \frac{1}{y}$.

    $\text{Integrate: } \frac{1}{y} dy = x dx$. Separation gives $\frac{dy}{y} = x dx$, then integrate both sides.

14. Find the general solution: $\frac{dy}{dx} = y^2 \text{sin}(x)$.

    $y = \frac{1}{C - \text{cos}(x)}$. Separating gives $\frac{dy}{y^2} = \sin(x) dx$, integrating yields $-\frac{1}{y} = -\cos(x) + C$.

15. What is the result of integrating both sides: $f(y) dy = g(x) dx$?

    Find $F(y) = G(x) + C$, where $F$ and $G$ are antiderivatives. Integration produces two functions related by a constant.

16. Determine the error: $y dy = x dx$ integrated as $y = x + C$.

    Correct: $\frac{y^2}{2} = \frac{x^2}{2} + C$. Must integrate each term separately: $\int y dy = \frac{y^2}{2}$.

17. What must be checked after finding a general solution?

    Verify the solution by differentiating and substituting. Solutions must satisfy the original differential equation.

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

    $\ln|y| = \frac{x^2}{2} + C$. Antiderivatives are $\ln|y|$ and $\frac{x^2}{2}$ respectively.

19. State the form of a separable differential equation.

    $\frac{dy}{dx} = g(x)h(y)$. This form allows variables to be separated into $\frac{dy}{h(y)} = g(x)dx$.

20. Find the general solution: $\frac{dy}{dx} = xy$.

    $y = Ce^{\frac{x^2}{2}}$. Separating gives $\frac{dy}{y} = x dx$, integrating yields $\ln|y| = \frac{x^2}{2} + C$.

21. What technique is used to verify the solution of a separable differential equation?

    Differentiate the solution and compare with the original equation. Substitution back into the original equation confirms correctness.

22. Determine the type of equation: $\frac{dy}{dx} = x^2 + y^2$.

    Not separable. Cannot separate $x$ and $y$ terms to opposite sides.

23. What form does the solution to a separable differential equation typically take?

    $F(y) = G(x) + C$. Each side integrates to a function plus an arbitrary constant.

24. Find the general solution: $\frac{dy}{dx} = y\text{e}^x$.

    $y = Ce^{\text{e}^x}$. Separating gives $\frac{dy}{y} = e^x dx$, integrating yields $\ln|y| = e^x + C$.

25. Which method helps find a particular solution after finding a general one?

    Use initial conditions. Initial conditions determine the specific value of the constant $C$.

26. What is the role of the constant $C$ in the solution of a differential equation?

    $C$ represents an arbitrary constant, allowing for all solutions. The constant encompasses all possible initial conditions.

27. Find the general solution: $\frac{dy}{dx} = y\cos(x)$.

    $y = Ce^{\sin(x)}$. Separating gives $\frac{dy}{y} = \cos(x) dx$, integrating yields $\ln|y| = \sin(x) + C$.

28. Identify the final form: $\frac{dy}{y} = x dx$.

    $\text{ln}|y| = \frac{x^2}{2} + C$. Integration of separated form $\frac{dy}{y} = x dx$.

29. Identify the error: $\frac{dy}{dx} = x^2y^2$ rewritten as $y^2 dy = x^2 dx$.

    Correct: $\frac{1}{y^2} dy = x^2 dx$. The $y^2$ term should move to the left with reciprocal: $\frac{1}{y^2}$.

30. Determine the type of equation: $\frac{dy}{dx} = x y + 1$.

    Not separable. The constant term prevents separation of variables.

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

    Rewrite the equation so that all $x$ terms are on one side and $y$ terms on the other. This separates variables to opposite sides of the equation.

32. What does it mean if a differential equation is not separable?

    Variables cannot be isolated on separate sides. The equation cannot be written as $g(x)h(y)$ form.

33. Identify the next step: $\text{ln}|y| = \text{ln}|x| + C$.

    Express $y$ in terms of $x$: $y = Cx$. Use properties of logarithms: $\ln|y| - \ln|x| = \ln|\frac{y}{x}| = C$.

34. Identify the step after integration: $\text{ln}|y| = x^2 + C$.

    Solve for $y$: $y = e^{x^2 + C}$. Exponentiate both sides to isolate $y$ from the logarithm.

35. What must be true about $C$ when using an initial condition $y(x_0) = y_0$?

    $C$ is determined by substituting $x_0$ and $y_0$ into the solution. Substitute the initial condition to solve for the constant.

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

    $y = Ce^x$. This is the standard exponential growth/decay solution.

37. How is a particular solution found from a general solution?

    Use initial conditions to solve for $C$. Initial conditions determine the specific constant value.

38. What is the general solution for $\frac{dy}{dx} = 3xy$?

    $y = Ce^{\frac{3x^2}{2}}$. Separating gives $\frac{dy}{y} = 3x dx$, integrating yields $\ln|y| = \frac{3x^2}{2} + C$.

39. Identify the integration: $\frac{dy}{y} = x dx$.

    $\text{ln}|y| = \frac{x^2}{2} + C$. Standard result from separating $\frac{dy}{y} = x dx$.

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