To find the Laplace transform of $t^2 \cos(3t)$, we can use the Laplace transform properties.

General Formula:

The Laplace transform of $t^n f(t)$ is given by:

$\mathcal{L}\{t^n f(t)\} = (-1)^n \frac{d^n}{ds^n} \left[ \mathcal{L}\{f(t)\} \right]$

where $\mathcal{L}\{f(t)\}$ is the Laplace transform of the function $f(t)$.

Step 1: Laplace Transform of $\cos(3t)$

$\mathcal{L}\{\cos(3t)\} = \frac{s}{s^2 + 9}$

Step 2: Apply the formula for $t^2 \cos(3t)$

Now, since $n = 2$ and $f(t) = \cos(3t)$:

$\mathcal{L}\{t^2 \cos(3t)\} = (-1)^2 \frac{d^2}{ds^2} \left[\frac{s}{s^2 + 9}\right]$

Step 3: Differentiate Twice

Let's differentiate $\frac{s}{s^2 + 9}$ twice with respect to $s$:

1. First derivative: $\frac{d}{ds} \left[\frac{s}{s^2 + 9}\right] = \frac{(s^2 + 9) \cdot 1 - s \cdot 2s}{(s^2 + 9)^2} = \frac{9 - s^2}{(s^2 + 9)^2}$

2. Second derivative: [ \frac{d^2}{ds^2} \left[\frac{s}{s^2 + 9}\right] = \frac{d}{ds} \left[\frac{9 - s^2}{(s^2 + 9)^2}\right] ] Using the quotient rule again: $\frac{d}{ds} \left[\frac{9 - s^2}{(s^2 + 9)^2}\right] = \frac{(s^2 + 9)^2 \cdot (-2s) - (9 - s^2) \cdot 2(s^2 + 9) \cdot 2s}{(s^2 + 9)^4}$ Simplifying this yields: $\frac{-2s(s^2 + 9)^2 - 2s(9 - s^2) \cdot 2(s^2 + 9)}{(s^2 + 9)^4}$ After further simplification: $= \frac{2s(3s^2 - 45)}{(s^2 + 9)^3}$

Final Answer:

$\mathcal{L}\{t^2 \cos(3t)\} = \frac{2s(3s^2 - 45)}{(s^2 + 9)^3}$

Do you want to go over this in more detail or have any questions?

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Here are 5 related questions:

1. How would you find the Laplace transform of $t \sin(2t)$?
2. Can you compute $\mathcal{L}^{-1} \left\{ \frac{s}{s^2 + 4} \right\}$?
3. What is the Laplace transform of $t^3 e^{2t}$?
4. How can you apply the convolution theorem using Laplace transforms?
5. Can you determine the Laplace transform of a periodic function?

Tip: The Laplace transform is particularly useful in solving linear differential equations and analyzing systems.