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Math Help - Using Laplace, getting stuck on decomposition

  1. #1
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    Using Laplace, getting stuck on decomposition

    I think I have a fair idea of how to solve my equations, using Laplace transform. But the partial fraction decomposition seems to elude me...

    I get X and Y as such.

    \displaystyle Y(s) = \frac{s^3+3s}{s^4+5s^2+4}
    \displaystyle X(s) = \frac{2}{s^4+5s^2+4}

    I get this:

    \displaystyle Y(s) = \frac{1}{s^2+1} -  \frac{1}{s^2+4}

    I pulled the handbrake here because wolfram alpha got something else...

    Wolfram Alpha, see: Partition fraction expansion

    If anyone could find it in their heart to ease me throw this one, so that I can solve my system of differential equations...
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Using Laplace, getting stuck on decomposition

    Your answer can't be right because:
    Y(s)=\frac{1}{s^2+1}-\frac{1}{s^2+4}=\frac{(s^2+4)-(s^2+1)}{(s^2+1)(s^2+4)}=\frac{3}{(s^2+1)\cdot(s^2  +4)}
    which is different from the original given Y(s).

    We have Y(s)=\frac{s^3+3s}{(s^2+1)(s^2+4)}=\frac{As+B}{s^2  +1}+\frac{Cs+D}{s^2+4}

    If we proceed:
    \frac{(As+B)(s^2+4)+(Cs+D)(s^2+1)}{(s^2+1)(s^2+4)}
    =\frac{As^3+Bs^2+4As+4B+Cs^3+Ds^2+Cs+D}{(s^2+1)(s^  2+4)}
    =\frac{(A+C)s^3+(B+D)s^2+(4A+C)s+(4B+D)}{(s^2+1)(s  ^2+4)}

    This means: A+C=1, B+D=0, 4A+C=3 and 4B+D=0

    Can you proceed?
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  3. #3
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    Re: Using Laplace, getting stuck on decomposition

    Thank you very much! I got it from here. I didn't write the proposed denominators correctly.
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