The problem is asking to find the Laplace transform of the convolution of two functions, $f(t) = \cos(2t)$ and $g(t) = e^{-3t}$.

Step 1: Convolution of the Two Functions

The convolution of two functions $f(t)$ and $g(t)$ is defined as: $(f * g)(t) = \int_0^t f(\tau) g(t - \tau) \, d\tau$

However, we don't need to compute this directly since we are asked to find the Laplace transform of the convolution.

Step 2: Laplace Transform of Convolution

One of the properties of the Laplace transform is: $\mathcal{L}\{f * g\}(s) = \mathcal{L}\{f(t)\}(s) \cdot \mathcal{L}\{g(t)\}(s)$ This means we can take the Laplace transform of each function individually, then multiply the results.

Step 3: Laplace Transform of $f(t) = \cos(2t)$

Using the Laplace transform table for $\cos(at)$: $\mathcal{L}\{\cos(2t)\}(s) = \frac{s}{s^2 + 4}$

Step 4: Laplace Transform of $g(t) = e^{-3t}$

Using the Laplace transform table for $e^{-at}$: $\mathcal{L}\{e^{-3t}\}(s) = \frac{1}{s + 3}$

Step 5: Multiply the Laplace Transforms

Now, we multiply the two Laplace transforms: $\mathcal{L}\{f * g\}(s) = \frac{s}{s^2 + 4} \cdot \frac{1}{s + 3}$ This gives the final result: $\mathcal{L}\{f * g\}(s) = \frac{s}{(s^2 + 4)(s + 3)}$

Thus, the Laplace transform of the convolution of the two functions is: $\boxed{\frac{s}{(s^2 + 4)(s + 3)}}$

Would you like more details or further clarification on any part of this process?

Here are some related questions for further practice:

1. What is the Laplace transform of $\sin(2t)$?
2. How do you compute the convolution $(f * g)(t)$ directly?
3. What is the inverse Laplace transform of $\frac{s}{(s^2 + 4)(s + 1)}$?
4. What is the Laplace transform of $t \cos(2t)$?
5. How does the convolution property simplify solving differential equations?

Tip: The convolution theorem is a powerful tool for simplifying the analysis of linear systems using Laplace transforms.