To solve the given initial value problem using the Laplace transform:

Given the differential equation: $4y'' + 5y' + 17y = 2\cos(2t), \quad y(0) = 0, \quad y'(0) = 0.$

(a) Taking the Laplace transform:

1. Applying the Laplace transform to both sides of the equation: $\mathcal{L}\{4y''\} + \mathcal{L}\{5y'\} + \mathcal{L}\{17y\} = \mathcal{L}\{2\cos(2t)\}.$

2. Using the properties of the Laplace transform for derivatives: $\mathcal{L}\{y''\} = s^2Y(s) - sy(0) - y'(0), \quad \mathcal{L}\{y'\} = sY(s) - y(0).$ Given $y(0) = 0$ and $y'(0) = 0$, we substitute: $4s^2Y(s) + 5sY(s) + 17Y(s) = \frac{2s}{s^2 + 4}.$

3. Factoring out $Y(s)$: $Y(s) \left(4s^2 + 5s + 17\right) = \frac{2s}{s^2 + 4}.$

4. Solving for $Y(s)$: $Y(s) = \frac{2s}{(4s^2 + 5s + 17)(s^2 + 4)}.$

(b) Expressing $y(t)$ using a convolution integral:

We do not perform partial fraction decomposition. Instead, we write $y(t)$ as a convolution integral using: $y(t) = \mathcal{L}^{-1} \left\{ \frac{2s}{(4s^2 + 5s + 17)(s^2 + 4)} \right\} = \int_0^t f(t - \tau)g(\tau) \, d\tau,$ where $f(t)$ and $g(t)$ are the inverse Laplace transforms of $\frac{1}{4s^2 + 5s + 17}$ and $\frac{2s}{s^2 + 4}$ respectively.

Would you like further details or have any questions on this?

Here are five related questions for further exploration:

1. How do you find the inverse Laplace transform of more complex rational functions without partial fraction decomposition?
2. What are the properties and conditions for using convolution integrals with Laplace transforms?
3. How does the initial condition influence the Laplace transform of derivatives?
4. Can you derive a general form of a convolution integral for similar second-order differential equations?
5. What is the significance of the damping factor in second-order differential equations, and how does it affect the solution?

Tip: Always double-check the initial conditions when applying Laplace transforms, as they are crucial for solving differential equations correctly.