Table of contents

0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

7. Antiderivatives & Indefinite Integrals

Initial Value Problems

Calculus

7. Antiderivatives & Indefinite Integrals

Initial Value Problems

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Multiple Choice

Solve the initial value problem using the method of Laplace transforms: $y^{''} + 4 y = 0$ , $y (0) = 2$ , $y^{'} (0) = 0$ . What is $y (t)$ ?

y (t) = 2 e^{2 t}

y (t) = 2 \sin (2 t)

y (t) = 2 e^{- 2 t}

y (t) = 2 \cos (2 t)

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Verified step by step guidance

Step 1: Apply the Laplace transform to the given differential equation y'' + 4y = 0. Recall that the Laplace transform of y'' is $ \mathcal{L}\{y''\} = s^2Y(s) - sy(0) - y'(0) $, and the Laplace transform of y is $ \mathcal{L}\{y\} = Y(s) $. Substitute the initial conditions y(0) = 2 and y'(0) = 0 into the transformed equation.

Step 2: After applying the Laplace transform, the equation becomes $ s^2Y(s) - 2 + 4Y(s) = 0 $. Rearrange this equation to solve for $ Y(s) $, the Laplace transform of the solution y(t).

Step 3: Simplify the equation to isolate $ Y(s) $. Combine terms to get $ Y(s)(s^2 + 4) = 2 $, and then divide through by $ s^2 + 4 $ to find $ Y(s) = \frac{2}{s^2 + 4} $.

Step 4: Recognize that $ \frac{2}{s^2 + 4} $ is the Laplace transform of $ 2 \cos(2t) $. Use the inverse Laplace transform to convert $ Y(s) $ back into the time domain, yielding $ y(t) = 2 \cos(2t) $.

Step 5: Verify the solution by substituting $ y(t) = 2 \cos(2t) $ back into the original differential equation y'' + 4y = 0 and checking that it satisfies the initial conditions y(0) = 2 and y'(0) = 0.