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HELP! on Partial Fraction Decomposition Problem - Heavy Side Technique

  1. Feb 17, 2009 #1
    HELP! on Partial Fraction Decomposition Problem!! - Heavy Side Technique

    I am doing a partial fraction decomposition problem for my calc 2 class.
    We use the Heavy Side Technique, but I will take help either way!!!

    [tex]\frac{2x-1}{x(x^2+1)^2}[/tex]

    Thank you!!!
     
  2. jcsd
  3. Feb 17, 2009 #2

    gabbagabbahey

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    Re: HELP! on Partial Fraction Decomposition Problem!! - Heavy Side Technique

    Hi khatche4, welcome to PF!:smile:

    What have you tried?....You need to show some attempt at the problem before we will assist you.
     
  4. Feb 17, 2009 #3
    Re: HELP! on Partial Fraction Decomposition Problem!! - Heavy Side Technique

    I tried splitting it up as..

    [tex]\frac{2x-1}{x(x^2+1)^2}[/tex] = [tex]\frac{A}{x}[/tex] + [tex]\frac{Bx+C}{x^2+1}[/tex] + [tex]\frac{Dx+E}{x^2+1}[/tex]

    Then I used x=0, x=1, x=-1, x=2, and x=-2 to get...

    1= -4 + 2B + 2C + 2D + 2E
    3= 4 - 2B + 2C - 2D + 2E
    3= -25 + 20B + 10C + 20D + 10E
    5= 25 - 20B + 10C - 20D + 10E

    And then I got stuck....


    Should it be this, initially?:

    [tex]\frac{2x-1}{x(x^2+1)^2}[/tex] = [tex]\frac{A}{x}[/tex] + [tex]\frac{Bx+C}{x^2+1}[/tex] + [tex]\frac{Dx+E}{(x^2+1)^2}[/tex] ???
     
  5. Feb 17, 2009 #4

    gabbagabbahey

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    Re: HELP! on Partial Fraction Decomposition Problem!! - Heavy Side Technique

    Yes, it should.
     
  6. Feb 17, 2009 #5
    Re: HELP! on Partial Fraction Decomposition Problem!! - Heavy Side Technique

    Ok! thank you! If I need anymore help, I'll come back!!
    =]
     
  7. Feb 17, 2009 #6
    Re: HELP! on Partial Fraction Decomposition Problem!! - Heavy Side Technique

    Ok... I'm stuck again.

    I went along with
    [tex]\frac{2x-1}{x(x^2+1)^2}[/tex] = [tex]\frac{A}{x}[/tex] + [tex]\frac{Bx+C}{x^2+1}[/tex] + [tex]\frac{Dx+E}{(x^2+1)^2}[/tex]

    From there, I used
    x=0 to get A = -1
    x=1
    x=-1
    x=2
    x=-2
    (Am I supposed to use 4?? We've only done up to Bx+C, and there we use 2.. So I assume to use 4)..

    Then, from there, I got
    1 = -4 + 2B + 2C + D + E
    3 = 4 - 2B + 2C - D + E
    3 = -25 + 20B + 10C + 4D + 2E
    5 = 25 - 20B + 10C - 4D + 2E

    So that boils down to..

    1 = 2C + E
    3 = 2C + E
    3 = 10C + 2E
    5 = 10C + 2E
    right??

    Will I be using substitution for this one? If so, do I use 2 different substitutions or just one?

    Thanks, again!!
     
  8. Feb 17, 2009 #7

    gabbagabbahey

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    Re: HELP! on Partial Fraction Decomposition Problem!! - Heavy Side Technique

    good :approve:
    You can get away with using only 3 values, if you choose those values wisely. What made x=0 such a useful choice was that it made one of the factors of [itex]x(x^2+1)^2[/itex] zero.....The other factor is of course just [itex]x^2+1[/itex], which is zero at [itex]x=\pm i[/itex], so try using [itex]x=i[/itex] to find easy solutions for [itex]D[/itex] and [itex]E[/itex]; you'll have a much easier time finding [itex]B[/itex] and [itex]C[/itex] after that.
     
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