The inverse Laplace transform of 1/s^3 is t2/2. Here we learn how to find the inverse Laplace of 1/s3.
The inverse Laplace transform of 1/s3 is denoted by L-1(1/s3) and its formula is given by
L-1(1/s3) = t2/2.
Table of Contents
Inverse Laplace of 1/s3
| Answer: The inverse Laplace of 1/s^3 is equal to t2/2. |
Explanation:
We know that the inverse Laplace Transform of 1/sn 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/s3 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 t2/2.
Related Topics:
Table of Inverse Laplace Transformations
Inverse Laplace transform of 1
Inverse Laplace transform of 1/s
Inverse Laplace transform of 1/s2
FAQs
Answer: The inverse Laplace transform of 1/s3 is t2/2, that is, L-1(1/s3) = t2/2.
Answer: L-1(1/s3) = t2/2.
