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Homework Help: Dumb partial fractions question

  1. Nov 22, 2006 #1
    dumb partial fractions question....

    suppose i get x+1=A(x-2)+B(x-2)

    how do you then find A and B?
     
  2. jcsd
  3. Nov 22, 2006 #2

    Hootenanny

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    Which values of x could you chose so that the equation simplified to leave you with a single unknown?
     
  4. Nov 22, 2006 #3

    arildno

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    By demanding you've got an IDENTITY there, and remembering that the functions f(x)=1 and g(x)=x are linearly INDEPENDENT functions.
    This will give you two equations for your two unknows A and B.
     
  5. Nov 22, 2006 #4
    But say I use 2 as my x value, it removes both A and B in this equation....
     
  6. Nov 22, 2006 #5

    arildno

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    Oops, I didn't see that closely!
    You've made a mistake somewhere.
     
  7. Nov 22, 2006 #6
    i haven't made a mistake i have to find the inverse laplace transform of s+1/(s^2 - 4s + 4) and im using partial fractions to do this.........but im stuck now......:cry:
     
  8. Nov 22, 2006 #7

    OlderDan

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    The denominator is a perfect square. You need to use a different form for the sum of fractions. Have you seen this before?

    Is that s+1/(s^2 - 4s + 4) or (s+1)/(s^2 - 4s + 4)??

    I assume the second based on your original equation.
     
    Last edited: Nov 22, 2006
  9. Nov 22, 2006 #8
    ok thanks - does this mean i find the inverse laplace transform of 1/(s-2)^2 and add this to the inverse LT of s/(s-2)^2 ??????????

    I can do the first that would be te^2t wouldnt it?

    Not sure about the second part???? :cry:
     
  10. Nov 22, 2006 #9

    OlderDan

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    If you can use tables, go here

    http://www.vibrationdata.com/Laplace.htm

    2.10 confirms your first term, and 2.15 cannot be used for a perfect square denominator. Partial fractions will get rid of the s in the numerator for you, and you will get a 2.9 form and a 2.10 form. Yes??
     
  11. Nov 22, 2006 #10
    i can see the first relationship with 2.10 but what do i need to do to s/(s-2)^2 ? use partial fractions? Thats what i tried at the start but how do i find A and B if s = A(s-2) + B(s-2) ?
     
  12. Nov 22, 2006 #11

    OlderDan

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    You cannot use that form when the denominator is a perfect square. Use
    (s + 1)/(s - 2)² = A/(s - 2) + B/(s - 2)²
     
  13. Nov 22, 2006 #12
    how does that change anything........still can't solve A and B??
     
  14. Nov 22, 2006 #13

    OlderDan

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    Sure you can.

    (s + 1)/(s - 2)² = A/(s - 2) + B/(s - 2)²

    s + 1 = A*(s - 2) + B

    s + 1 = As - 2A + B
    A = 1

    1 = -2A + B

    B = 1 + 2A = 3
     
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