The inverse Laplace transform of \( \frac{2}{s^2} \) can be found by recognizing that the Laplace transform of \( t^n \) is given by:

\[
\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}
\]

In this case, \( \frac{2}{s^2} \) matches the form \( \frac{n!}{s^{n+1}} \) where \( n = 1 \) and \( 1! = 1 \), so we have:

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

Thus, the inverse Laplace transform of \( \frac{2}{s^2} \) is:

\[
2t
\]

Would you like further details on this process or have any other questions?

Here are some related questions:
1. What is the inverse Laplace transform of \( \frac{1}{s^3} \)?
2. How would you find the inverse Laplace transform of \( \frac{5}{s^4} \)?
3. Can you apply the Laplace transform to solve differential equations?
4. What is the Laplace transform of \( t^2 \)?
5. How do we handle inverse Laplace transforms with complex poles?

**Tip**: When using the Laplace transform, recognizing standard forms like \( \frac{n!}{s^{n+1}} \) is key to simplifying inverse transforms.

Explore the process of finding the inverse Laplace transform of 2/s^2, which involves recognizing key forms of the Laplace transform. The problem explains how the formula for t^n helps simplify the inverse transformation. Suitable for students in advanced high school or undergraduate courses.

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