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To find the inverse Laplace transform of \( F(s) = \frac{5e^{-6s}}{s^2 + 25} \), let's proceed with the following steps:

1. **Identify the Shift**: The term \( e^{-6s} \) in the Laplace transform corresponds to a time shift. Specifically, \( e^{-cs} \) in the \( s \)-domain represents a shift of \( c \) units to the right in the \( t \)-domain, which in this case is \( c = 6 \).

2. **Simplify the Remaining Part**: The remaining part of \( F(s) \) after factoring out \( e^{-6s} \) is \( \frac{5}{s^2 + 25} \).

3. **Inverse Laplace of \( \frac{5}{s^2 + 25} \)**: The expression \( \frac{5}{s^2 + 25} \) is a standard Laplace form, which matches \( \frac{\omega}{s^2 + \omega^2} \). Here, \( \omega = 5 \), so the inverse Laplace transform of \( \frac{5}{s^2 + 25} \) is:
   \[
   \mathcal{L}^{-1} \left( \frac{5}{s^2 + 25} \right) = 5 \sin(5t)
   \]

4. **Apply the Time Shift**: Since we have \( e^{-6s} \) in \( F(s) \), we apply a time shift to \( 5 \sin(5t) \) by \( t - 6 \) and multiply by the unit step function \( u_{6}(t) \). This results in:
   \[
   f(t) = 5 \sin(5(t - 6)) \cdot u_{6}(t)
   \]

### Final Answer
\[
f(t) = 5 \sin(5(t - 6)) \, u_{6}(t)
\]

Would you like further details or explanations on this process?

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### Related Questions

1. How do you find the inverse Laplace transform of other functions involving time shifts?
2. What are the general rules for inverse Laplace transforms of sinusoidal functions?
3. How does the unit step function \( u_{c}(t) \) affect the time-domain representation?
4. What is the impact of the parameter \( c \) in \( e^{-cs} \) on the time-domain function?
5. How can we use the Laplace transform to solve differential equations involving step functions?

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**Tip:** When dealing with inverse Laplace transforms, remember that exponential terms like \( e^{-cs} \) indicate a time shift in the final answer.

Find the inverse Laplace transform of \( F(s) = \frac{5e^{-6s}}{s^2 + 25} \)

Learn how to find the inverse Laplace transform of F(s) = 5e^(-6s) / (s^2 + 25) using time-shifting and the unit step function. This problem covers core concepts in Laplace transforms, suitable for college-level students.

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