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Partial fractions

  1. Aug 16, 2008 #1
    got an exam coming up in a few days and half way through my question i ran into a partial fractions question instead of having the standard (1/(y+c)(y+d))= A/(y+c) + B(y+d) and multiplying out i had a double root so (1/(y+c)(y+c)) does this change the way i go about the question and are there any other similar situations that i am likely to run into?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 16, 2008 #2

    Hootenanny

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    Although it shouldn't affect the method you use, it will affect the form of your solution. In this case your solution will be of the form:

    [tex]\frac{1}{\left(y+c\right)^2} = \frac{A}{y+c\right} + \frac{B}{\left(y+c\right)^2}[/tex]

    i.e. a linear denominator followed by a quadratic denominator. You should then follow your usual technique (e.g. equating coefficients) to solve for A and B.
     
  4. Aug 16, 2008 #3
    are you sure this is right cos this gives me A=0 and B=1 giving me 1/(y+c)^2=1/(y+c)^2 which doesnt achive anything at all
     
  5. Aug 16, 2008 #4

    tiny-tim

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    Hi franky2727! :smile:

    (since Hootenanny is offline …)

    This is really for fractions like (P + Qy)/(y + c)2,

    and it gives you partial fractions of the form A/(y + c) + B/(y + c)2, without any y on top.

    In this case, you started without any y on top, so indeed you didn't achieve anything at all, because 1/(y + c)2 was already the simplest answer. :wink:
     
  6. Aug 16, 2008 #5
    Ahh.. ok so i just end up with Ln|(y+c)^2|?
     
  7. Aug 16, 2008 #6

    tiny-tim

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    erm … if you started with ∫dy/y2 (you didn't say), then nooo … :frown:

    ln(y + c)2 is just 2 ln(y + c), isn't it? :wink:
     
  8. Aug 16, 2008 #7
    i feal so stupid :P log laws a bit rusty to say the least
     
  9. Aug 17, 2008 #8

    HallsofIvy

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    the log law is not the problem here.
    [tex]\int \frac{1}{y^2}dy= \int y^{-2} dy[/tex]
    is NOT a logarithm.
     
  10. Aug 17, 2008 #9
    no i started with dy/(v+1)2
    so am i right to get ln |(v+1)2| and does that go to 2ln|V+1| i thought it wouldnt because the squared bit is inside the modulus not outside or does that not matter because squaring something gets rid of the need for a modulus anyway
     
  11. Aug 17, 2008 #10
    so what does this end up as when i e-it-out (V+1)^2 or 2(V+1) or something different
     
  12. Aug 17, 2008 #11

    tiny-tim

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    Nooo …

    Hint: what do you have to differentiate to get -1/x2? :smile:
    (V+1)^2 :smile:
     
  13. Aug 17, 2008 #12
    is this even ment to be in logs ? 1/((V+1)^2 can be (V+1)^-2 which goes to (-1/V+1)?????
     
  14. Aug 17, 2008 #13

    tiny-tim

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    Bingo! :smile:
     
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