Question

Find the inverse Laplace transform of F(s)=(2s+2)/(s^2+2s+5).

Answer 1

Did you even try?

Answer 2

I did, after completing the square, I got (2s+2)/(s+1)^2+4). But I don't know what to do after that.

Answer 3

Well I do not know this tbh, I can send you a link to were some one has gotten the answer for this if you want it?

Answer 4

Yes, please.

Answer 5

http://answers.yahoo.com/question/index?qid=20111001100530AAkTTDl

Answer 6

I hope this is what you are looking for!

Answer 7

It was somewhat helpful.

Answer 8

I think I can help you now
\[F(s) = \frac{2(s-(-1))}{(s-(-1))^2+2^2}\]
\[L^{-1}F(s) = 2 e^{-t}cos2t\]

Answer 9

But how did you get the final answer?

Answer 10

that the final answer, use Laplace transform table.

Answer 11

I got 2s/(s^2+2^2), but is there another way to find the final answer?

Answer 12

it's not that, friend. Your denominator doesn't lead to your answer.
It's s^2 +2s +5, how can it be = x^2 +4?

Answer 13

*s not x

Answer 14

No, I got s=s+1 from 2(s+1)/(s+1)^2+4, 2s/(s^2+4)=2s/(s^2+2^2)

Answer 15

I think you misunderstand the concept. you have f(x) you have to find Laplace inverse of that f(x) . You try to make it looked like some of the formula in Laplace transform table to get f(x)

Answer 16

So what's the formula for this problem?

Answer 17

wait, I search and send you

Answer 18

look at this http://www.google.com/url?sa=i&source=images&cd=&cad=rja&docid=qo4QeNJmCZdgBM&tbnid=BZ6UaufTtSjCDM:&ved=&url=http%3A%2F%2Fmath.fullerton.edu%2Fmathews%2Fc2003%2FLaplaceTransformMod.html&ei=a1ABUvDKOsfErQHx2IFw&psig=AFQjCNFDh-wPLKpLGwkILudMlyPa0BcqSQ&ust=1375904236021405

Answer 19

look at the line 8, it talks about L^(-1) , the right column is what you have, it will lead to the left column answer. That's what I got from your problem. Our responsibility is try to convert the problem into one of the form of the right column, then use this table to find out the answer. Your prof didn't tell you that?

Answer 20

Thanks for the site. Unfortunately, my professor sucks, that's why I need so much help in here.

Answer 21

ok.