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Math Help - partial fraction problem

  1. #1
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    Cool partial fraction problem

    i amstruggling with the partial fraction:
    8/s^2*(s^2+4)) = A/s^2 + B/s + Cs+D/(s^2+4)

    i get A=2 but cannot find a decent s value to give B,C and D
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  2. #2
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    Quote Originally Posted by ben.mahoney@tesco.net View Post
    i amstruggling with the partial fraction:
    8/s^2*(s^2+4)) = A/s^2 + B/s + Cs+D/(s^2+4)

    i get A=2 but cannot find a decent s value to give B,C and D
    Multiplying on both sides by s^2(s^2+ 4) gives
    8= A(s^2+4)+ Bs(s^2+ 4)+ (Cs+D)s^2. Setting s= 0 gives 8= 4A so A= 2. I presume that is what you have already done.

    Although s= 0 is the only value that makes the equation that simple, any value of s give an equation.

    If s= 1, 8= 5A+ 5B+ C+ D and you already know that A= 2: 5B+ C+ D= -2.

    If s= -1, 8= 5A- 5B- C+ D so 5B+ C- D= 2.
    Adding those two equations eliminates D immediately.

    If s= 2, 8= 8A+ 16B+ 8C+ 4D so 16B+ 8C+ 4D= -8 or 4B+ 2C+ D= -2.
    Adding that to 5B+ C+ D= -2 again eliminates D leaving two equations to solve for B and C.

    Another method for problems like this is to combine like powers of s.

    From 8= A(s^2+4)+ Bs(s^2+ 4)+ (Cs+D)s^2, 8= As^2+ 4A+ Bs^3+ 4Bs+ Cs^3+ Ds^2 or, combining like powers, 8= (B+ C)s^3+ (A+ D)s^2+ (4B)s+ 4A. For that to be true for all x, we must have coefficients of the same powers equal on both sides: B+ C= 0, A+ D= 0, 4B= 0, 4A= 8.

    This method is often harder that just choosing value for s but here is much easier.
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