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Partial fractions - having 1 in the numerator?

  1. Oct 7, 2008 #1
    This is probably a "basic" question, but I can't seem to remember how to do partial fractions problems where there is only a 1 in the numerator.

    For example (just making this up), let's say I have:

    1/s(s+4)(s+5)

    So what I'd do is 1/s(s+4)(s+5) = A/s + B/(s+4) + C/(s+5) as one would expect, but this can't be right, correct? It seems a lot of coefficients don't "work out."

    For example, after doing cross multiplications and combining terms, you would basically get:

    For s^2 terms: (A+B+C) = 0 [1]
    For s terms: (6A + 5B + 4C) = 0 [2]
    For no-s terms: 5A = 1 [3]

    Then putting [3]--> [1], then getting B=-C - 1/5, and putting that in [2]. You'll end up with A being equal to 1/5, and C being equal to -1/5. Plug this all into [1], so then that means B is equal to zero. What am I missing here? Is this right?

    Matlab says it should be C=-1 and B=1, which means that A is zero then? Huh? I am pretty confused.

    Thanks.
     
    Last edited: Oct 7, 2008
  2. jcsd
  3. Oct 7, 2008 #2

    Mark44

    Staff: Mentor

    You have decomposed the expression correctly, but have made a mistake in your work. Not only that, but the results you reported from Matlab are wrong, too.
    Your equation [1] is correct, equations [2] and [3] are wrong. The coefficient for the s terms is 9A + 5B + 4C, and for the constant term, the coefficient is 20.

    So the three equations are:
    Code (Text):

    A + B + C = 0
    9A + 5B + 4C = 0
    20A = 1
     
    From these, you should get A = 1/20, B = -1/4, and C = 1/5
    Mark
     
  4. Oct 8, 2008 #3
    Thanks Mark.
     
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