Find the Inverse Laplace of 1/(s^3)

[tsslaporte]

Find the Inverse 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]

Find the Laplace of 1/(s^3)

You mean find the inverse transform.

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?

So what does the table give you for $t^n$? Can you modify it?

 

[tsslaporte]

You mean find the inverse transform.
So what does the table give you for $t^n$? Can you modify it?

Yep Inverse sorry, $t^n$ , n = 1,2,3,... is (n!)/(s^n+1)

2/s^2 +1 ?

 

[HallsofIvy]

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 $(1/2)t^2$.

I just saw your response. I think you are misreading the table. It is not "$s^n+ 1$", it is $s^{n+ 1}$.

 

[LCKurtz]

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 $(1/2)t^2$.

Halls, don't you think we should have let him figure out that step?

 

[tsslaporte]

Thanks,

where did the 1/2 come from?

 

[HallsofIvy]

If you cannot see that then you should not be taking this course. Look at your table of Laplace transforms again.