Inverse Laplace transform

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January 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.

Thanks for your help in advance, Chris
January 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.

Thanks for your help in advance, Chris

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
January 18th 2008, 01:15 PM
reakon

I blame the lecturer's handout, it doesn't include that. Thanks for your help
January 11th 2009, 06:43 AM
mhyann

2/(s-1)^2

my answer to that is....2sint.
2/(s-1)^2
=2{1/s^2+1^2)
*rmmber sint formula is b/s^2+b^
so b there is 1..i cant explain it to u.im not good at it.