To find the inverse Laplace transform for each function $F(s)$, we will use properties of Laplace transforms such as linearity and tables of elementary transforms.

Solution

(a) $F(s) = \frac{1}{2s + 6}$

Rewrite $F(s)$ to match a standard form: $F(s) = \frac{1}{2(s + 3)} = \frac{1}{2} \cdot \frac{1}{s + 3}$ The inverse Laplace transform of $\frac{1}{s + a}$ is $e^{-at}$. Thus: $\mathcal{L}^{-1}\left(F(s)\right) = \frac{1}{2} e^{-3t}$

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

Rewrite $F(s)$ in terms of known transforms: $F(s) = \frac{4}{(s - 1)(s + 1)}$ We use partial fraction decomposition: $\frac{4}{(s - 1)(s + 1)} = \frac{2}{s - 1} - \frac{2}{s + 1}$ Now we take the inverse Laplace of each term. Using $\mathcal{L}^{-1} \left(\frac{1}{s - a}\right) = e^{at}$: $\mathcal{L}^{-1}\left(F(s)\right) = 2 e^{t} - 2 e^{-t} = 2 \sinh(t)$

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

Rewrite $F(s)$ as two separate terms: $F(s) = \frac{s}{s^2 + 4} - \frac{8}{s^2 + 4}$ For the first term, $\frac{s}{s^2 + a^2}$ has an inverse transform of $\cos(at)$, and for the second term, $\frac{1}{s^2 + a^2}$ has an inverse transform of $\frac{\sin(at)}{a}$: $\mathcal{L}^{-1}\left(F(s)\right) = \cos(2t) - 4 \sin(2t)$

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

Separate terms and find the inverse Laplace transform of each: $\mathcal{L}^{-1}\left(\frac{1}{2s^2}\right) = \frac{t}{2} \quad \text{and} \quad \mathcal{L}^{-1}\left(\frac{1}{s}\right) = 1$ Thus: $\mathcal{L}^{-1}\left(F(s)\right) = \frac{t}{2} - 1$

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Would you like further details on any part? Here are some related questions for practice:

1. How would you apply the shift theorem to find the inverse Laplace of $\frac{1}{s+5}$?
2. What is the inverse Laplace transform of $\frac{3}{s^2 + 9}$?
3. Can you explain why partial fractions are useful in finding inverse Laplace transforms?
4. How does the linearity property help in breaking down complex functions of $s$?
5. What is the effect of scaling on the inverse Laplace transform?

Tip: Always check if the function of $s$ can be simplified or decomposed before finding the inverse transform; this often makes the process much more straightforward.