The inverse Laplace transform of 1/(s^3) is definitively t^2/2. This conclusion is derived from the Laplace transform table, which states that the transform of t^n, where n is a positive integer, is n!/s^(n+1). For n=2, this results in (2!/s^(2+1)), leading to the inverse transform being (1/2)t^2. The discussion emphasizes the importance of correctly interpreting the Laplace transform table to avoid confusion between the transform and its inverse.
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tsslaporte said:Find the Laplace of 1/(s^3)
is there some special rule for cube?
The answer is t^2/2
Looking at the Laplace Table t^n looks similar but its not it exactly.
What should I do?
LCKurtz said:You mean find the inverse transform.
So what does the table give you for ##t^n##? Can you modify it?
HallsofIvy said:Are you asking for the "Laplace transform" or the "Inverse Laplace transform"? The standard notation uses "t" for the function and "s" for its Laplace transform.
A table of Laplace transforms, such as the one at http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf, will tell you that the Laplace transform of t^n, for n a positive integer, is n!/s^{n+1}.
So the inverse Laplace transform of 1/s^3= (1/2)(2/s^3)=(1/2)(2!/s^(2+1) is (1/2)t^2.