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Math Help - rearranging an equation

  1. #1
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    rearranging an equation

    hi
    im trying to rearrange an equation to make y the subject to then use partial fraction expansion and laplace inverses to solve.
    so far i have simplified to give

    y(S^4 + 3S^2 -12) = 104/s+3 -24 +12S^3 - 3S^2 + 25S -15

    i know i need to divide by the bracketed terms on the left but i dont know how to do this. i.e what is (104/s+3)/(S^4 + 3S^2 -12)

    do i need a common denominator

    any help would be appreciated
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  2. #2
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    Re: rearranging an equation

    Solution :- Given y(S^4 + 3S^2 -12) = (104/s)+3 -24 +12S^3 - 3S^2 + 25S -15
    s(S^4+3S^2-12)Y=104+s(12S^3-3S^2+25S-36)
    because s(S^4+35^2-12)?0
    Y=(12sS^3-3sS^2+25sS-36s+104)/s(S^4+3S^2-12)
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  3. #3
    Super Member ebaines's Avatar
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    Re: rearranging an equation

    Please clarify the right hand side of yuor equation. What you wrote is this:

     \frac {104} {S} + 3 -24 +12S^3-3S^2 + 25S - 15

    But I suspect what you meant is this:

     \frac {104} {S + 3} -24 +12S^3-3S^2 + 25S - 15

    Why didn't you combine the -24 and -15 terms?

    One option is to multiply both sides through by (S+3), then divide both sides by the Y coeffeicient:


    \small y(S^4 + 3S^2 -12)(S+3) = (\frac {104}{s+3} -24 +12S^3 - 3S^2 + 25S -15)(s+3)

    \small y(S^5+3S^4+3S^3+9S^2-12S-36) = 104 -39(S+3)+12S^3(S+3)-3S^2(S+3)+25S(S+3)


    \small y= \frac {(12S^4+33S^3+16S^2+10S+36S+179)}{(S^5+3S^4+3S^3+9  S^2-12S-36)}
    Last edited by ebaines; August 14th 2013 at 09:47 AM.
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