Inverse Laplace Transform: Solving for Coefficients and Completing the Square

[quasar_4]

Homework Statement

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

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

Homework Equations

We get a table of known inverse Laplace transforms.

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!

 

[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.

 

[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?

 

[gabbagabbahey]

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

 

[quasar_4]

Thank you!