Let's solve the problem step by step to find the inverse Laplace transform of

$\bar{y}(s) = \frac{-8}{s(s^2 - 4)}.$

Step 1: Decompose into Partial Fractions

The denominator of the given function is $s(s^2 - 4) = s(s - 2)(s + 2)$. Let's decompose the fraction into partial fractions: $\frac{-8}{s(s^2 - 4)} = \frac{-8}{s(s - 2)(s + 2)} = \frac{A}{s} + \frac{B}{s - 2} + \frac{C}{s + 2}.$

Multiply through by $s(s - 2)(s + 2)$ to eliminate the denominators: $-8 = A(s - 2)(s + 2) + B(s)(s + 2) + C(s)(s - 2).$

Expand each term: $-8 = A(s^2 - 4) + B(s^2 + 2s) + C(s^2 - 2s).$

Combine terms: $-8 = (A + B + C)s^2 + (2B - 2C)s + (-4A).$

Step 2: Solve for $A$, $B$, and $C$

By comparing coefficients on both sides of the equation, we get:

1. Coefficient of $s^2$: $A + B + C = 0$,
2. Coefficient of $s$: $2B - 2C = 0 \implies B = C$,
3. Constant term: $-4A = -8 \implies A = 2$.

Substitute $A = 2$ and $B = C$ into the first equation: $2 + B + B = 0 \implies 2 + 2B = 0 \implies B = -1.$

Thus, $A = 2$, $B = -1$, and $C = -1$.

Step 3: Rewrite the Fraction

Substitute $A$, $B$, and $C$ back into the partial fraction decomposition: $\frac{-8}{s(s^2 - 4)} = \frac{2}{s} - \frac{1}{s - 2} - \frac{1}{s + 2}.$

Step 4: Apply the Inverse Laplace Transform

Using the standard inverse Laplace transform formulas:

1. $\mathcal{L}^{-1}\left(\frac{1}{s}\right) = 1$,
2. $\mathcal{L}^{-1}\left(\frac{1}{s - a}\right) = e^{at}$,
3. $\mathcal{L}^{-1}\left(\frac{1}{s + a}\right) = e^{-at}$,

we find: $\mathcal{L}^{-1}\left(\frac{2}{s}\right) = 2,$ $\mathcal{L}^{-1}\left(\frac{-1}{s - 2}\right) = -e^{2t},$ $\mathcal{L}^{-1}\left(\frac{-1}{s + 2}\right) = -e^{-2t}.$

Thus, the solution is: $y(t) = 2 - e^{2t} - e^{-2t}.$

Final Answer:

The correct option is (c): $y(t) = 2 - e^{2t} - e^{-2t}.$

Let me know if you'd like further clarification or more examples! 😊

Related Questions:

1. How do partial fractions help in solving inverse Laplace transforms?
2. What are the key properties of the Laplace transform used in solving differential equations?
3. Why is the inverse Laplace transform useful in solving initial value problems?
4. Can Laplace transforms handle discontinuous functions like step or impulse inputs?
5. How do you handle repeated roots when performing partial fraction decomposition?

Tip:

Always double-check coefficients carefully when solving partial fractions, as even small mistakes can lead to incorrect solutions.