# Find Inverse Laplace transform of 1/s^2

The inverse Laplace transform of 1/s^2 is equal to t. In this post, we will learn how to find the inverse Laplace transform of 1 divided by s2.

## How to Find Inverse Laplace of 1/s2

To find the inverse Laplace transform of 1 by s square, we will use the formula of the Laplace transform of tn which is given by

L{tn} = $\dfrac{n!}{s^{n+1}}$ where n=0, 1, 2, 3, …

Now we take the inverse Laplace transform on both sides. By doing so, we get that

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

This implies that

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

Put n=1 in the above formula in order to get the inverse Laplace transform of 1/s2. Thus, we obtain that

t1 = 1! L-1(1/s2)

⇒ t = L-1(1/s2).

Therefore, the inverse Laplace transform formula of 1/s^2 is equal to L-1(1/s2) = t, that is, t is the inverse Laplace of 1 by s^2.

Inverse Laplace of 1

Inverse Laplace transform of a constant

Inverse Laplace of 1/s

## FAQs

Q1: What is the inverse Laplace transform of 1/s2?

Answer: The inverse Laplace transform of 1/s2 is t.

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