We are tasked with finding the inverse Laplace transform of the given function:

$F(s) = \frac{e^{-9s}}{s^2 + 2s - 8}$

Step 1: Analyze the Laplace transform

The exponential term $e^{-9s}$ suggests a time-shift property of the Laplace transform. The time-shift property is given by:

$\mathcal{L}^{-1}\{e^{-as}F(s)\} = u(t-a)f(t-a)$ where $u(t-a)$ is the Heaviside step function, and $F(s)$ is the Laplace transform of $f(t)$.

Step 2: Simplify the denominator

We need to simplify the denominator $s^2 + 2s - 8$ by factoring it.

$s^2 + 2s - 8 = (s - 2)(s + 4)$

Step 3: Find the inverse Laplace of the factored form

Now we focus on finding the inverse Laplace transform of $\frac{1}{(s - 2)(s + 4)}$, without the exponential shift for now.

We can use partial fraction decomposition:

$\frac{1}{(s - 2)(s + 4)} = \frac{A}{s - 2} + \frac{B}{s + 4}$

Multiplying both sides by $(s - 2)(s + 4)$ gives:

$1 = A(s + 4) + B(s - 2)$

Expanding this:

$1 = A s + 4A + B s - 2B$

Grouping terms:

$1 = (A + B)s + (4A - 2B)$

By comparing coefficients, we get the system of equations:

1. $A + B = 0$
2. $4A - 2B = 1$

From the first equation, $B = -A$. Substituting into the second equation:

$4A - 2(-A) = 1 \implies 4A + 2A = 1 \implies 6A = 1 \implies A = \frac{1}{6}$

Thus, $B = -\frac{1}{6}$.

So, the partial fraction decomposition is:

$\frac{1}{(s - 2)(s + 4)} = \frac{1/6}{s - 2} - \frac{1/6}{s + 4}$

Step 4: Apply the inverse Laplace transform

We now find the inverse Laplace transform of each term:

$\mathcal{L}^{-1}\left\{\frac{1/6}{s - 2}\right\} = \frac{1}{6}e^{2t}$ and $\mathcal{L}^{-1}\left\{\frac{1/6}{s + 4}\right\} = \frac{1}{6}e^{-4t}$

Thus, the inverse Laplace transform of $\frac{1}{(s - 2)(s + 4)}$ is:

$\frac{1}{6}(e^{2t} - e^{-4t})$

Step 5: Account for the time shift

Using the time-shift property, we shift the result by 9 units of time. The final solution becomes:

$\mathcal{L}^{-1}\left\{\frac{e^{-9s}}{(s - 2)(s + 4)}\right\} = u(t - 9) \cdot \frac{1}{6} \left( e^{2(t - 9)} - e^{-4(t - 9)} \right)$

This is the inverse Laplace transform of the given function.

Would you like more details or have any questions? Here are some related questions you might explore:

1. How does the time-shift property work in Laplace transforms?
2. What are the conditions for using partial fraction decomposition in inverse Laplace transforms?
3. Can we apply the same method if the denominator doesn't factor easily?
4. How would we handle a higher-degree polynomial in the denominator?
5. What is the significance of the Heaviside function in solving inverse Laplace transforms?

Tip: When applying the time-shift property, always remember that the Heaviside function introduces a delay in time but keeps the overall form of the solution unchanged.