Homework Help: Inverse Laplace transform - need help!

1. Oct 3, 2008

quasar_4

1. The problem statement, all variables and given/known data

Determine the inverse Laplace transform of 1/((s^2 +1)*(s-1)).

The answer is 1/2*(e^x - cos(x) - sin(x)).

2. Relevant equations

We get a table of known inverse Laplace transforms.

3. The attempt at a solution

I tried to break this up using partial fractions, i.e., A/(s^2 +1 ) + B/(s-1). Then I solved for the coefficients A and B. Now for B I got B=1/2, which gives us 1/2*(s-1) which is just 1/2*e^x. So there's part of the answer already.

I am stuck on the other part. For A (maybe I am not solving for A correctly) I got A=1/(i-1). But then I have this fraction 1/((i-1)*(s^2 +1)). The 1/(s^2+1) is just sin(x). So I don't see why my answer wouldn't just be

1/2*e^x + (1/(i-1))*sin(x). I even tried expanding out the product (i-1)(s^2+1) and looking for a way to complete the square, but I must have missed it, whatever it was.

Where does the other half come from? And the other cos(x)? I know the answer above is right (both from the professor and Maple) but I don't see how they got it. Any help would be wonderful!

2. Oct 3, 2008

gabbagabbahey

You can't decompose $\frac{1}{(s^2 +1)(s-1)}$ like that. $(s^2 +1)$ is a second order polynomial, so its numerator must be of the form $Cs+D$ not just $A$.

Last edited: Oct 3, 2008
3. Oct 3, 2008

quasar_4

Oh, ok! So it is a partial fraction issue.

Does this generalize so that a fraction of the form

1/((s^n+1)*(s-1))

has to have a polynomial of degree n-1 on the numerator? Is there a good website or text that you know of that gives all the rules for partial fractions?

4. Oct 3, 2008

gabbagabbahey

Look under the section "An irreducible quadratic factor in the denominator" http://en.wikipedia.org/wiki/Partial_fraction" [Broken]
.

Last edited by a moderator: May 3, 2017
5. Oct 3, 2008

quasar_4

Thank you!

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