To find the Laplace Transform $L\{ \cos(3t) e^{2t} \}$, we can apply the basic property of the Laplace Transform for functions of the form $e^{at} f(t)$:

Laplace Transform Properties

1. Basic Property: $L\{e^{at} f(t)\} = F(s - a)$, where $F(s)$ is the Laplace Transform of $f(t)$.
2. Laplace Transform of Cosine: $L\{\cos(kt)\} = \frac{s}{s^2 + k^2}$.

Solution Steps

1. First, recognize that $f(t) = \cos(3t)$.
2. The Laplace Transform of $\cos(3t)$ is: $L\{\cos(3t)\} = \frac{s}{s^2 + 3^2} = \frac{s}{s^2 + 9}.$
3. Using the property $L\{e^{2t} \cos(3t)\} = F(s - 2)$: $L\{e^{2t} \cos(3t)\} = \frac{s - 2}{(s - 2)^2 + 9}.$

Final Answer

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

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

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Here are some related questions to explore this topic further:

1. What is the Laplace Transform of $e^{3t} \sin(4t)$?
2. How do you apply the Laplace Transform for functions involving polynomials and exponentials?
3. Can you find the inverse Laplace Transform of $\frac{s - 3}{(s - 3)^2 + 16}$?
4. How does the property $L\{ e^{at} f(t) \} = F(s - a)$ help simplify Laplace Transforms?
5. What are the Laplace Transform properties for functions with derivatives?

Tip: The Laplace Transform is especially useful for solving differential equations by transforming them into algebraic equations in the $s$-domain.