Question

Which of the options correctly represents the Laplace inverse of \( \dfrac{2}{s^3} \)?

• A. $t^2$
• B. $t^3$
• C. $t^{-2}$
• D. $t^{-3}$

Hint:

Remember: $\mathcal{L}\{t^n\} = \dfrac{n!}{s^{n+1}}$. Match the power of $s$ in the denominator to find the corresponding $n$.

Answer 1

The Correct Option is A

To determine the inverse Laplace transform of $\dfrac{2}{s^3}$, we use the standard Laplace transform identity:
\[ \mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}} \text{for } n \geq 0 \]
We need to match the given Laplace expression $\dfrac{2}{s^3}$ with the right-hand side of the above identity.
Let us assume:
\[ \frac{2}{s^3} = \mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}} \Rightarrow n+1 = 3 \Rightarrow n = 2 \]
So, using $\mathcal{L}\{t^2\} = \dfrac{2!}{s^3} = \dfrac{2}{s^3}$
Therefore, the inverse Laplace transform of $\dfrac{2}{s^3}$ is $t^2$.
Checking the incorrect options:
- $\mathcal{L}\{t^3\} = \dfrac{6}{s^4}$ $\Rightarrow$ doesn’t match.
- $t^{-2}$ and $t^{-3}$ are not valid for Laplace transforms in the classical sense (Laplace transforms are defined for $t^n$ where $n \geq 0$).
Thus, the correct answer is $t^2$.