# 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: Apr 23, 2017 at 4:11 PM
5. Oct 3, 2008

Thank you!