Inverse Laplace transform

• Jan 18th 2008, 12:41 PM
reakon
Inverse Laplace transform
Hi there,

I am going through a worksheet on Integral equations and a problem has arisen. I am up to having:

L[y](s)=2/(s-1)^2 and I need to find y(t). I know that the inverse Laplace transform rules state that 1/s^2 goes to t and that 1/(s^2+a^2) goes to
sin(at) but nothing of the form above.

Does anyone know If I am able to take 1/(s-1) as 1/u and get the inverse Laplace as u=s-1? I can't think of any other way unless there is a Laplace transform rule that I am not aware of.

• Jan 18th 2008, 01:09 PM
CaptainBlack
Quote:

Originally Posted by reakon
Hi there,

I am going through a worksheet on Integral equations and a problem has arisen. I am up to having:

L[y](s)=2/(s-1)^2 and I need to find y(t). I know that the inverse Laplace transform rules state that 1/s^2 goes to t and that 1/(s^2+a^2) goes to
sin(at) but nothing of the form above.

Does anyone know If I am able to take 1/(s-1) as 1/u and get the inverse Laplace as u=s-1? I can't think of any other way unless there is a Laplace transform rule that I am not aware of.

Get a better table of transforms:

$\mathcal{L}\left[ \frac{t^{n-1}e^{at}}{(n-1)!}\right]=\frac{1}{(s-a)^n},\ n=1,2, ..$

RonL
• Jan 18th 2008, 01:15 PM
reakon
I blame the lecturer's handout, it doesn't include that. Thanks for your help
• Jan 11th 2009, 06:43 AM
mhyann
2/(s-1)^2