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laplace\:f\left(t\right)=e^{-4t}\:cos\:\sqrt{3}\:t\:-\:t^2\:e^{-2t}

The Laplace transform of the function \( f(t) = e^{-4t} \cos(\sqrt{3} t) - t^2 e^{-2t} \) is:

\[
\mathcal{L}\{f(t)\} = \frac{(s + 2)^3(s + 4) - 2(s + 4)^2 - 6}{(s + 2)^3 \left( (s + 4)^2 + 3 \right)}
\]

Would you like a detailed explanation of how this result was obtained, or do you have any further questions?

Here are five related questions to explore further:
1. What is the Laplace transform of \( t^n e^{at} \)?
2. How does the convolution theorem relate to Laplace transforms?
3. How can you use Laplace transforms to solve differential equations?
4. What is the inverse Laplace transform of a rational function?
5. How does the shifting property apply in Laplace transforms?

**Tip:** Understanding the Laplace transform's shifting theorem is useful when dealing with terms like \( e^{at} \).

Discover the Laplace transform of the function f(t) = e^{-4t} cos(√3 t) - t^2 e^{-2t) with a step-by-step solution. This problem covers advanced calculus topics such as the Laplace transform, exponential and trigonometric functions, and the shifting theorem.

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