The Laplace transform of $t \cos(2t)$ can be found using the formula:

$\mathcal{L} \{ t f(t) \} = -\frac{d}{ds} \left( \mathcal{L} \{ f(t) \} \right)$

where $f(t) = \cos(2t)$ and $\mathcal{L} \{ \cos(2t) \} = \frac{s}{s^2 + 4}$.

Now, differentiate $\frac{s}{s^2 + 4}$ with respect to $s$:

= \frac{s^2 + 4 - 2s^2}{(s^2 + 4)^2} = \frac{-s^2 + 4}{(s^2 + 4)^2}$$ Thus, the Laplace transform of $$t \cos(2t)$$ is: $$\mathcal{L} \{ t \cos(2t) \} = -\frac{-s^2 + 4}{(s^2 + 4)^2} = \frac{s^2 - 4}{(s^2 + 4)^2}$$ So, the Laplace transform of $$t \cos(2t)$$ is: $$\mathcal{L} \{ t \cos(2t) \} = \frac{s^2 - 4}{(s^2 + 4)^2}$$ Would you like more details or further clarification? Here are 5 related questions: 1. How do you find the Laplace transform of $$t \sin(3t)$$? 2. What is the Laplace transform of a general function $$t^n \cos(at)$$? 3. How can you apply the convolution theorem to Laplace transforms? 4. What is the inverse Laplace transform of $$\frac{s}{s^2 + a^2}$$? 5. Can you explain the shifting property in Laplace transforms? **Tip:** Differentiation in the Laplace domain is often used to handle expressions with time multiplication, so memorizing this technique is helpful.