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The given differential equation is:

\[
(D^2 - D - 2)y = 20 \sin(2t)
\]
with the initial conditions:
\[
y(0) = 1 \quad \text{and} \quad y'(0) = 2
\]
where \( D \) represents the differential operator \( \frac{d}{dt} \). The task is to solve this using Laplace transforms.

### Step-by-Step Solution:

1. **Rewrite the equation using \( D \):**

\[
(D^2 - D - 2)y = 20 \sin(2t)
\]
can be rewritten as:

\[
\frac{d^2 y}{dt^2} - \frac{dy}{dt} - 2y = 20 \sin(2t)
\]

2. **Take the Laplace Transform of both sides:**

The Laplace transform of \( y(t) \) is denoted as \( Y(s) \). Using the properties of Laplace transforms:
- \( \mathcal{L}\left\{y'(t)\right\} = s Y(s) - y(0) \)
- \( \mathcal{L}\left\{y''(t)\right\} = s^2 Y(s) - s y(0) - y'(0) \)

Apply this to both sides of the equation:
\[
\mathcal{L}\left\{\frac{d^2 y}{dt^2}\right\} - \mathcal{L}\left\{\frac{dy}{dt}\right\} - 2\mathcal{L}\left\{y\right\} = \mathcal{L}\left\{20 \sin(2t)\right\}
\]

Substituting the initial conditions \( y(0) = 1 \) and \( y'(0) = 2 \):
\[
(s^2 Y(s) - s \cdot 1 - 2) - (s Y(s) - 1) - 2Y(s) = \frac{20 \cdot 2}{s^2 + 4}
\]
Simplify the left-hand side:
\[
s^2 Y(s) - s - 2 - s Y(s) + 1 - 2Y(s) = \frac{40}{s^2 + 4}
\]
Combine terms:
\[
(s^2 - s - 2)Y(s) = \frac{40}{s^2 + 4} + s + 1
\]

3. **Solve for \( Y(s) \):**

\[
Y(s) = \frac{\frac{40}{s^2 + 4} + s + 1}{s^2 - s - 2}
\]

4. **Partial Fraction Decomposition:**

To invert the Laplace transform, decompose the expression on the right-hand side into simpler fractions. However, this process will involve breaking down the rational function and using standard inverse Laplace transforms.

Let me calculate the partial fraction decomposition.The Laplace transform of the solution is:

\[
Y(s) = \frac{s + 1 + \frac{40}{s^2 + 4}}{s^2 - s - 2}
\]

Next, to find \( y(t) \), we need to apply the inverse Laplace transform to this expression. However, this process requires breaking the expression down into simpler fractions using partial fraction decomposition.

I will now compute the partial fractions and find the inverse Laplace transform of each part.The partial fraction decomposition of \( Y(s) \) is:

\[
Y(s) = \frac{s - 6}{s^2 + 4} - \frac{8}{3(s + 1)} + \frac{8}{3(s - 2)}
\]

Now, we can take the inverse Laplace transform of each term:
- \( \frac{s - 6}{s^2 + 4} \) corresponds to a combination of cosine and sine terms.
- \( \frac{8}{3(s + 1)} \) and \( \frac{8}{3(s - 2)} \) correspond to exponential terms.

Let me now compute the inverse Laplace transforms for these terms.The solution to the differential equation in the time domain is:

\[
y(t) = \frac{1}{3} \left[ (8e^{2t} - 9\sin(2t) + 3\cos(2t)) e^t - 8 \right] e^{-t}
\]

This expression represents the solution to the given second-order differential equation using Laplace transforms.

Would you like further details on any step, or do you have any additional questions?

### Related Questions:
1. How does the Laplace transform simplify solving differential equations?
2. What is the significance of partial fraction decomposition in the Laplace transform method?
3. How do the initial conditions affect the Laplace transform solution?
4. Can you explain how to handle non-homogeneous differential equations using Laplace transforms?
5. How can we verify the solution using direct substitution into the original differential equation?

### Tip:
Always remember to check if your solution satisfies the initial conditions by substituting \( t = 0 \) and verifying the given values for \( y(0) \) and \( y'(0) \).

Solve (D^2 - D - 2)y = 20 sin(2t), y(0) = 1 and y'(0) = 2 using Laplace transforms.

Step-by-step solution to solve the second-order differential equation (D^2 - D - 2)y = 20sin(2t) using Laplace transforms. Includes initial conditions y(0) = 1 and y'(0) = 2, covering key concepts such as Laplace transforms, partial fraction decomposition, and inverse transforms.

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