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Inverse Laplace transform - need help!

  1. Oct 3, 2008 #1
    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. jcsd
  3. Oct 3, 2008 #2

    gabbagabbahey

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    You can't decompose [itex]\frac{1}{(s^2 +1)(s-1)}[/itex] like that. [itex](s^2 +1)[/itex] is a second order polynomial, so its numerator must be of the form [itex]Cs+D[/itex] not just [itex]A[/itex].
     
    Last edited: Oct 3, 2008
  4. Oct 3, 2008 #3
    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?
     
  5. Oct 3, 2008 #4

    gabbagabbahey

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    Look under the section "An irreducible quadratic factor in the denominator" here
    .
     
  6. Oct 3, 2008 #5
    Thank you!
     
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