To find the Laplace transform $L\left\{\frac{6}{(s^2-1)(s+2)}\right\}$, we'll first simplify the expression and then apply partial fraction decomposition.

Step 1: Factor the denominator

The expression is:

$\frac{6}{(s^2-1)(s+2)}$

Factor $s^2 - 1$:

$s^2 - 1 = (s - 1)(s + 1)$

So the expression becomes:

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

Step 2: Partial fraction decomposition

We express the function as a sum of simpler fractions:

$\frac{6}{(s - 1)(s + 1)(s + 2)} = \frac{A}{s - 1} + \frac{B}{s + 1} + \frac{C}{s + 2}$

Multiplying both sides by the denominator $(s - 1)(s + 1)(s + 2)$ gives:

$6 = A(s + 1)(s + 2) + B(s - 1)(s + 2) + C(s - 1)(s + 1)$

Step 3: Solve for A, B, and C

To find $A$, $B$, and $C$, we substitute convenient values of $s$.

1. Let $s = 1$:

$6 = A(1 + 1)(1 + 2) + 0 + 0 \Rightarrow 6 = 6A \Rightarrow A = 1$

2. Let $s = -1$:

$6 = 0 + B(-1 - 1)(-1 + 2) + 0 \Rightarrow 6 = 2B \Rightarrow B = 3$

3. Let $s = -2$:

$6 = 0 + 0 + C(-2 - 1)(-2 + 1) \Rightarrow 6 = 3C \Rightarrow C = -2$

Thus, the decomposition is:

$\frac{6}{(s - 1)(s + 1)(s + 2)} = \frac{1}{s - 1} + \frac{3}{s + 1} - \frac{2}{s + 2}$

Step 4: Apply the inverse Laplace transform

Now, we apply the inverse Laplace transform to each term:

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

$L^{-1}\left\{\frac{3}{s + 1}\right\} = 3e^{-t}$

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

Final Answer:

Combining these, we get the inverse Laplace transform:

$L^{-1}\left\{\frac{6}{(s - 1)(s + 1)(s + 2)}\right\} = e^{t} + 3e^{-t} - 2e^{-2t}$

Would you like any further details or have any questions?

Here are 5 related questions you might find interesting:

1. How do you perform partial fraction decomposition with repeated factors?
2. What is the Laplace transform of $\frac{1}{s^2(s + 1)}$?
3. How does the initial value theorem apply to Laplace transforms?
4. Can you explain the convolution theorem in the context of Laplace transforms?
5. What are some applications of Laplace transforms in engineering?

Tip: When performing partial fraction decomposition, check for repeated roots and handle them appropriately to avoid errors.