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The function given is:

\[
F(s) = \frac{e^{-7s}}{s^2 + 4s - 5}
\]

### Steps to find the inverse Laplace transform:

1. **Handle the exponential term:** The presence of \( e^{-7s} \) indicates a time shift in the inverse Laplace transform. This follows from the shifting theorem:

   \[
   \mathcal{L}^{-1}\left\{ e^{-as} F(s) \right\} = u(t-a) f(t-a)
   \]

   where \( u(t-a) \) is the Heaviside step function. So, after handling the rest of the transform, we will shift the result by 7 units in time.

2. **Factor the denominator:**
   The quadratic \( s^2 + 4s - 5 \) can be factored:
   
   \[
   s^2 + 4s - 5 = (s + 5)(s - 1)
   \]
   
   This allows us to decompose the fraction using partial fractions.

3. **Partial fraction decomposition:**
   We now decompose:

   \[
   \frac{1}{(s+5)(s-1)} = \frac{A}{s+5} + \frac{B}{s-1}
   \]

   To find \( A \) and \( B \), solve:
   
   \[
   1 = A(s-1) + B(s+5)
   \]

   Expanding and solving for \( A \) and \( B \):
   
   \[
   A(s-1) + B(s+5) = A s - A + B s + 5 B = (A+B)s + (-A+5B)
   \]

   Matching the coefficients of like powers of \( s \), we get:

   \[
   A + B = 0 \quad \text{and} \quad -A + 5B = 1
   \]

   Solving these equations:
   
   - From \( A + B = 0 \), we have \( A = -B \).
   - Substitute into \( -A + 5B = 1 \):

     \[
     -(-B) + 5B = 1 \implies B + 5B = 1 \implies 6B = 1 \implies B = \frac{1}{6}
     \]

     Therefore, \( A = -\frac{1}{6} \).

4. **Inverse Laplace of the simpler terms:**
   Now we use the fact that:

   \[
   \mathcal{L}^{-1}\left\{ \frac{1}{s+a} \right\} = e^{-at}
   \]

   Thus, the inverse Laplace transforms of the terms are:

   \[
   \mathcal{L}^{-1}\left\{ \frac{-1/6}{s+5} \right\} = -\frac{1}{6}e^{-5t}, \quad \mathcal{L}^{-1}\left\{ \frac{1/6}{s-1} \right\} = \frac{1}{6}e^{t}
   \]

5. **Combine the results:**
   The inverse Laplace transform before the time shift is:

   \[
   f(t) = -\frac{1}{6}e^{-5t} + \frac{1}{6}e^{t}
   \]

6. **Apply the time shift:**
   Using the shifting theorem, the final solution is:

   \[
   f(t) = u(t-7) \left( -\frac{1}{6}e^{-5(t-7)} + \frac{1}{6}e^{(t-7)} \right)
   \]

---

Would you like more details on any part of the solution? Here are 5 related questions to explore:

1. How does the Laplace shift theorem work for time-domain shifting?
2. Why do we use partial fractions to solve inverse Laplace transforms?
3. How do you factor quadratics when working with Laplace transforms?
4. What is the Heaviside step function, and how does it affect solutions?
5. Can the Laplace transform be applied to systems with non-exponential terms?

**Tip:** Always remember that the time shift in Laplace transforms corresponds to a delay in the time domain through the use of the Heaviside function.

Find the inverse Laplace transform of F(s) = e^(-7s) / (s^2 + 4s - 5).

Find the inverse Laplace transform of F(s) = e^(-7s) / (s^2 + 4s - 5) using time-shifting and partial fraction decomposition. Learn about the shifting theorem and how to apply partial fractions to solve complex Laplace transforms.

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