The equation shown in the image involves the inverse Laplace transform. It is:

$\mathcal{L}^{-1} \left[ \frac{s}{(s^2 - 4)^2} \right]$

Let’s solve step by step:

Step 1: Simplify the Denominator

The term $(s^2 - 4)^2$ can be rewritten as $(s - 2)^2(s + 2)^2$, factoring $s^2 - 4$ into $(s - 2)(s + 2)$.

Thus, the given expression becomes:

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

Step 2: Use Partial Fraction Decomposition

We express the fraction as a sum of partial fractions. Set:

$\frac{s}{(s - 2)^2 (s + 2)^2} = \frac{A}{(s - 2)^2} + \frac{B}{(s - 2)} + \frac{C}{(s + 2)^2} + \frac{D}{(s + 2)}$

Multiply through by the denominator to eliminate fractions:

$s = A(s + 2)^2 + B(s + 2)^2(s - 2) + C(s - 2)^2 + D(s - 2)^2(s + 2)$

Expand and group terms to solve for $A$, $B$, $C$, and $D$. After solving, we find:

$A = \frac{1}{8}, \quad B = \frac{1}{4}, \quad C = \frac{1}{8}, \quad D = -\frac{1}{4}.$

Step 3: Take the Inverse Laplace Transform

Using the linearity of the inverse Laplace transform, we compute each term separately:

(i) Term with $A$:

[ \mathcal{L}^{-1} \left[ \frac{A}{(s - 2)^2} \right] = \frac{1}{8} \mathcal{L}^{-1} \left[ \frac{1}{(s - 2)^2} \right] = \frac{1}{8} t e^{2t}. ]

(ii) Term with $B$:

[ \mathcal{L}^{-1} \left[ \frac{B}{s - 2} \right] = \frac{1}{4} \mathcal{L}^{-1} \left[ \frac{1}{s - 2} \right] = \frac{1}{4} e^{2t}. ]

(iii) Term with $C$:

[ \mathcal{L}^{-1} \left[ \frac{C}{(s + 2)^2} \right] = \frac{1}{8} \mathcal{L}^{-1} \left[ \frac{1}{(s + 2)^2} \right] = \frac{1}{8} t e^{-2t}. ]

(iv) Term with $D$:

[ \mathcal{L}^{-1} \left[ \frac{D}{s + 2} \right] = -\frac{1}{4} \mathcal{L}^{-1} \left[ \frac{1}{s + 2} \right] = -\frac{1}{4} e^{-2t}. ]

Step 4: Combine the Results

Summing all terms:

$\mathcal{L}^{-1} \left[ \frac{s}{(s^2 - 4)^2} \right] = \frac{1}{8} t e^{2t} + \frac{1}{4} e^{2t} + \frac{1}{8} t e^{-2t} - \frac{1}{4} e^{-2t}.$

This is the final result.

Let me know if you'd like a more detailed breakdown!

Here are some follow-up questions to deepen understanding:

1. What is the general approach to solving inverse Laplace transforms involving repeated roots?
2. How is partial fraction decomposition applied to higher-order polynomials?
3. What properties of the Laplace transform make it linear?
4. How does the Laplace transform handle delays in time?
5. Can you derive the inverse Laplace of $\frac{1}{(s - a)^n}$ in general?

Tip: Always verify the partial fraction decomposition by re-expanding the terms to confirm it equals the original expression.