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 s^{2}.

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## How to Find Inverse Laplace of 1/s^{2}

To find the inverse Laplace transform of 1 by s square, we will use the formula of the Laplace transform of t^{n} which is given by

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

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

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

This implies that

t^{n} = 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/s^{2}. Thus, we obtain that

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

⇒ t = L^{-1}(1/s^{2}).

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

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## FAQs

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

^{2}?Answer: The inverse Laplace transform of 1/s^{2} is t.

This article is written by Dr. T. Mandal, Ph.D in Mathematics. On Mathstoon.com you will find Maths from very basic level to advanced level. Thanks for visiting.