The function provided in the image is:

$F(s) = \frac{1}{4s^2 - 8s}$

To find the inverse Laplace transform, let's first simplify the expression.

Step 1: Factor the denominator

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

Thus, the function becomes:

$F(s) = \frac{1}{4s(s - 2)}$

Step 2: Decompose into partial fractions

We want to express $\frac{1}{4s(s-2)}$ as:

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

Multiply through by the denominator $4s(s-2)$:

$1 = A(s-2) + Bs$

Now, solve for $A$ and $B$:

• When $s = 0$: $1 = A(0-2) + B(0) \Rightarrow A = -\frac{1}{2}$
• When $s = 2$: $1 = A(2-2) + B(2) \Rightarrow B = \frac{1}{2}$

So the partial fraction decomposition is:

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

Step 3: Apply the inverse Laplace transform

Using the inverse Laplace transform rules:

• $\mathcal{L}^{-1} \left\{ \frac{1}{s} \right\} = 1$
• $\mathcal{L}^{-1} \left\{ \frac{1}{s - a} \right\} = e^{at}$

We get:

$\mathcal{L}^{-1} \left\{ \frac{-\frac{1}{2}}{s} \right\} = -\frac{1}{2} \cdot 1 = -\frac{1}{2}$

$\mathcal{L}^{-1} \left\{ \frac{\frac{1}{2}}{s-2} \right\} = \frac{1}{2}e^{2t}$

Step 4: Combine the results

$\mathcal{L}^{-1} \left\{ F(s) \right\} = -\frac{1}{2} + \frac{1}{2}e^{2t}$

Thus, the inverse Laplace transform of $F(s)$ is:

$f(t) = -\frac{1}{2} + \frac{1}{2}e^{2t}$

Would you like more details on any of these steps, or do you have any questions?

Here are 5 questions for further exploration:

1. How do you find the inverse Laplace transform of other types of rational functions?
2. What is the general method for partial fraction decomposition?
3. Can we apply this method for higher-order polynomials in the denominator?
4. How do the properties of the Laplace transform help in solving differential equations?
5. What is the significance of each term in the inverse Laplace transform result?

Tip: Remember that partial fraction decomposition is crucial for breaking down complex rational expressions into simpler ones that are easier to invert using Laplace tables.