The Laplace transform of t^2, i.e the Laplace of t square is equal to 2/s3. In this article, we will learn how to find the Laplace transform of t2.
The Laplace transform t2 (t square) is denoted by L{t2}, and its formula is given as follows:
L{t2} = 2/s3.
This follows from the Laplace formula of tn: L{tn} = n!/sn+1 by putting n=2.
Main Article: Laplace Transform: Definition, Table, Formulas, Properties & Examples
Table of Contents
What is the Laplace Transform of t2?
Answer: The Laplace transform of t2 is equal to 2/s3.
Proof:
The Laplace transform of f(t) by definition is given by
L{f(t)} = $\int_0^\infty$ f(t) e-st dt.
So to find the Laplace transform of t2 by definition, we need to follow below steps.
Step 1: Put f(t) = t2.
Therefore,
L{t2} = $\int_0^\infty$ t2 e-st dt. …(I)
Step 2: In order to compute the above integral, let us use the theory of the Gamma function Γ(x) = $\int_0^\infty$ tx-1 e-t dx. Assume that
z=st
∴ dz=s dt ⇒ dt = dz/s. Also, t=z/s.
t | z |
0 | 0 |
∞ | ∞ |
Step 3: The equation (I) then implies that
L{t2} = $\int_0^\infty \Big(\dfrac{z}{s} \Big)^2 e^{-z} \dfrac{dz}{s}$
= (1/s2+1) $\int_0^\infty z^{2+1-1} e^{-z} dz$
= (1/s3) $\Gamma(2+1)$, by the definition of the Gamma function.
= (1/s3) × 2! as we know that Γ(n+1) = n!
= 2!/s3
= 2/s3.
Therefore, the Laplace transform of t^2 is equal to 2/s3 and this is proved by the definition of Laplace transforms.
Read Also:
Laplace Transform of eat: | The Laplace transform of eat is 1/(s-a). |
Laplace transform of sin(at): | The Laplace transform of sin(at) is a/(s2+a2). |
Laplace transform of cos(at): | The Laplace transform of cos(at) is s/(s2+a2). |
Laplace transform of a constant: | The Laplace transform of a constant c is c/s. |
Inverse Laplace transform of constant: | The inverse Laplace transform of c is cδ(t), where δ(t) is the Dirac delta function. |
FAQs
Answer: The Laplace transform of t2 is L{t2} = 2/s3.
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.