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### Problem Analysis:
We are tasked with solving the second-order differential equation using the **Laplace transform**. The equation given is:

\[
5y'' + 2y' + 16y = 3\cos(2t),
\]

with initial conditions:

\[
y(0) = 0, \quad y'(0) = 0.
\]

The Laplace transform method involves:
1. Taking the Laplace transform of both sides of the equation.
2. Solving for \(\mathcal{L}\{y(t)\}\).
3. Writing the solution \(y(t)\) in terms of a convolution integral.

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### (a) Taking the Laplace Transform:
Using Laplace transform properties:
- \(\mathcal{L}\{y''(t)\} = s^2Y(s) - sy(0) - y'(0)\),
- \(\mathcal{L}\{y'(t)\} = sY(s) - y(0)\),
- \(\mathcal{L}\{y(t)\} = Y(s)\),

we apply the Laplace transform to both sides of the differential equation:

\[
5\mathcal{L}\{y''(t)\} + 2\mathcal{L}\{y'(t)\} + 16\mathcal{L}\{y(t)\} = \mathcal{L}\{3\cos(2t)\}.
\]

Substituting the initial conditions \(y(0) = 0\) and \(y'(0) = 0\):

\[
5[s^2Y(s)] + 2[sY(s)] + 16Y(s) = \frac{3s}{s^2 + 4}.
\]

Simplify the equation:

\[
(5s^2 + 2s + 16)Y(s) = \frac{3s}{s^2 + 4}.
\]

Solve for \(Y(s)\):

\[
Y(s) = \frac{3s}{(5s^2 + 2s + 16)(s^2 + 4)}.
\]

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### (b) Convolution Integral:
We will write \(y(t)\) in terms of a convolution integral. The solution involves:

\[
y(t) = \mathcal{L}^{-1}\{Y(s)\}.
\]

Instead of partial fraction decomposition, we'll directly express \(Y(s)\) as a product of transforms. Observe:

\[
Y(s) = \frac{3s}{5s^2 + 2s + 16} \cdot \frac{1}{s^2 + 4}.
\]

The Laplace transform \(\frac{3s}{5s^2 + 2s + 16}\) corresponds to some function \(h_1(t)\), and \(\frac{1}{s^2 + 4}\) corresponds to \(\sin(2t)\). Thus, by the **convolution theorem**:

\[
y(t) = \int_0^t h_1(\tau)\sin(2(t - \tau)) \, d\tau,
\]

where \(h_1(t)\) is the inverse Laplace transform of \(\frac{3s}{5s^2 + 2s + 16}\).

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### Final Answer:
(a) The Laplace transform is:

\[
\mathcal{L}\{y(t)\}(s) = \frac{3s}{(5s^2 + 2s + 16)(s^2 + 4)}.
\]

(b) The solution in terms of a convolution integral is:

\[
y(t) = \int_0^t h_1(\tau)\sin(2(t - \tau)) \, d\tau,
\]

where \(h_1(t) = \mathcal{L}^{-1}\left(\frac{3s}{5s^2 + 2s + 16}\right)\).

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Do you need further details on the inverse Laplace transform or simplifying the convolution integral? Here are related questions to expand on:

1. How is the convolution theorem applied in Laplace transforms?
2. What is the process to compute \(h_1(t)\) explicitly using partial fractions?
3. Can the result \(y(t)\) be numerically solved or simulated?
4. What physical systems are modeled by such second-order differential equations?
5. How does this method compare to other solution techniques like undetermined coefficients?

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**Tip:** The Laplace transform is powerful for initial value problems, especially when non-homogeneous terms (e.g., \(\cos(2t)\)) are present. Always keep the convolution theorem in mind for compact solutions.

Use the Laplace transform to solve the following initial value problem: 5y'' + 2y' + 16y = 3cos(2t), y(0) = 0, y'(0) = 0.

(a) Take the Laplace transform of both sides of the given differential equation to create the corresponding algebraic equation and then solve for \( \mathcal{L} \{ y(t) \} \). Do not perform partial fraction decomposition since we will write the solution in terms of a convolution integral.

(b) Express the solution \( y(t) \) in terms of a convolution integral.

Learn how to use the Laplace transform to solve the second-order differential equation 5y'' + 2y' + 16y = 3cos(2t) with initial conditions y(0) = 0 and y'(0) = 0. This solution covers applying the Laplace transform, avoiding partial fraction decomposition, and using the convolution theorem to express the result.

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