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Math Help - help with partial fractions

  1. #46
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    I let s = -1:

    1 + 2(-1) = -3*(-1-2)(-1)^2 + 5/4*(-1-1)(-1)^2 + C*(-1-1)(-1-2)(-1)^2 + 1/2*(-1-1)(-1-2)

    -1 = 9 -5/2 + 3 + 6C
    -21/2 = 6C
    -7/4 = C
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  2. #47
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    Hmm. You should have this equation:

    1+2s=A(s-2)s^{2}+B(s-1)s^{2}+C(s-1)(s-2)s+D(s-1)(s-2).

    You can see that there is only an s multiplying C, not a s^{2}.

    Carry this correction all the way through to see how that changes things.
    Last edited by Ackbeet; November 8th 2010 at 05:38 AM. Reason: Correct grammar.
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  3. #48
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    I used \dfrac{1+2s}{(s-1)(s-2)s^{2}}=\dfrac{A}{s-1}+\dfrac{B}{s-2}+\dfrac{C}{s}+\dfrac{D}{s^{2}}.

    Is this not the way?

    Yeah the answers are now correct. Nice one.
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  4. #49
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    I got s^2 at the end of C using     \dfrac{1+2s}{(s-1)(s-2)(s)^2}=\dfrac{A}{s-1}\,\dfrac{(s-2)(s)^2}{(s-2)(s)^2}+\dfrac{B}{s-2}\,\dfrac{(s-1)(s)^2}{x(s-1)(s)^2}<br /> <br />
+\dfrac{C}{s}\,\dfrac{(s-1)(s-2)(s)^2}{(s-1)(s-2)(s)^2}+\dfrac{D}{s^2}\,\dfrac{(s-1)(s-2)}{(s-1)(s-2)}.
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  5. #50
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    No. The C term doesn't require both s's, since it already has one in its denominator.

    Glad you got required result using the other method. Any more questions?
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  6. #51
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    Given that \dfrac{3s}{(s-2)(s+1)(s^2+4)}=\dfrac{A}{s-2}\,\dfrac{(s+1)(s^2+4)}{(s+1)(s^2+4)}+\dfrac{B}{s  +1}\,\dfrac{(s-2)(2^2+4)}{(s-2)(2^2+4)}<br /> <br />
+\dfrac{Cs+D}{s^2+4}\,\dfrac{(Cs+D)(s-2)(s+1)}{(Cs+D)(s-2)(s+1)}

    Can you confirm that C = -17/10 and D = -3/10
    the answers say C = -9/20 D = -3/20
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  7. #52
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    You should have

    \dfrac{3s}{(s-2)(s+1)(s^2+4)}=\dfrac{A}{s-2}\,\dfrac{(s+1)(s^2+4)}{(s+1)(s^2+4)}+\dfrac{B}{s  +1}\,\dfrac{(s-2)(s^2+4)}{(s-2)(s^2+4)}<br />
+\dfrac{Cs+D}{s^2+4}\,\dfrac{(s-2)(s+1)}{(s-2)(s+1)};

    You'll see that there should be an s in a coupled of places, not a 2. Also, no need to multiply by Cs+D top and bottom, since that's in the numerator, and plays no role in determining the common denominator.

    What did you get for A and B?
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  8. #53
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    Yeah sorry the extra (Cs+D) shouldnt of been there. For A = 1/4 and B = 1/5
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  9. #54
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    That's correct. D you can get by setting s = 0. What do you get for D?
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  10. #55
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    Yeah I got when S = 0

    3(0) = 1/4*(0+1)((0)^2+4) + 1/5*(0-2)((0)^2+4) + (C(0) + D)*(0-2)(0+1)

    0 = 1 - 8/5 - 2D

    3/5 = -2D

    -3/10 = D

    What did you get?
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  11. #56
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    Your D is correct. So, what do you get for C? I'd try s = 1 to get your C.
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  12. #57
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    yeah I used s = 1 and got C = -17/10
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  13. #58
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    Let me see your work.
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  14. #59
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    s = 1

    3(1) = 1/4*(1+1)((1)^2+4) + 1/5*(1-2)((1)^2+4) + (C(1) -3/10)*(1-2)(1+1)

    3 = 5/2 -1 -2C + 3/5

    9/10 = -2C

    -9/20 = C
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  15. #60
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    There you go. So you got D correct, and the book got C correct.
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