The inverse Laplace transform of 1/s^3 is t^{2}/2. Here we learn how to find the inverse Laplace of 1/s^{3}.

The inverse Laplace transform of 1/s^{3} is denoted by L^{-1}(1/s^{3}) and its formula is given by

L^{-1}(1/s^{3}) = t^{2}/2.

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## Inverse Laplace of 1/s^{3}

Answer: The inverse Laplace of 1/s^3 is equal to t^{2}/2. |

**Explanation:**

We know that the inverse Laplace Transform of 1/s^{n} is given by the formula:

L^{-1}$\left(\dfrac{1}{s^{n+1}} \right)$ = $\dfrac{t^n}{n!}$

In order to get the inverse Laplace of 1/s^{3} we need to put **n=2** in the above formula, so that we obtain

L^{-1}$\left(\dfrac{1}{s^{2+1}} \right)$ = $\dfrac{t^2}{2!}$

This implies that

L^{-1}$\left(\dfrac{1}{s^3} \right)$ = $\dfrac{t^2}{2}$.

So the inverse Laplace transform of 1 by s^3 is equal to t^{2}/2.

Related Topics:

Table of Inverse Laplace Transformations

Inverse Laplace transform of 1

Inverse Laplace transform of 1/s

Inverse Laplace transform of 1/s^{2}

## FAQs

**Q1: What is the inverse Laplace of 1/s**

^{3}?Answer: The inverse Laplace transform of 1/s^{3} is t^{2}/2, that is, L^{-1}(1/s^{3}) = t^{2}/2.

**Q2: Find L**

^{-1}(1/s^{3}).Answer: L^{-1}(1/s^{3}) = t^{2}/2.