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Inverting a Laplace Transform w/non-factorable denominator

  1. Oct 13, 2009 #1
    1. The problem statement, all variables and given/known data

    A 2nd order DE with a Delta Dirac forcing function, I have been asked to solve in my DE course.
    y'' + 2y' + 10y = [tex]\delta[/tex](t-1)

    Also given are IV's y'(0)=0, y(0)=0

    2. Relevant equations

    At this stage in my course I am being asked to only use the method of partial fractions to rewrite my Laplace equations into a recognisable invertable form I am not supposed to be using the integral definition to invert my solutions.

    3. The attempt at a solution

    The Transforms of each side I have so far obtained:

    [tex] Y(s).(s^2 + 2s + 10) = e^{-x} [/tex]

    This gives me a solution for Y(s) -->
    [tex] Y(s) = \frac{ e^{-x} }{s^2 + 2s + 10}[/tex]

    So I have [tex] Y(s) = e^{-x} . F(s)[/tex]
    [tex] e^{-x} [/tex] is no problem to invert but [tex] F(s)[/tex] has a quadratic in denominator which cannot be factored. I am at a loss as to how I can find any invertable form for it and for some reason I'm not aware of any other technique that is possible.
  2. jcsd
  3. Oct 13, 2009 #2


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    Homework Helper

    try completing the square for s2+2s+10
  4. Oct 13, 2009 #3
    Ok I feel stupid, I was thinking that completing the square would give me Y(s) terms on the RHS which I would have to divide out but I just tried it and now I don't know why I was thinking that.

    I have factors but now I have a denominator with a dangling 6 that I don't know what to do with.

    [tex] Y(s)s^2 + Y(s)2s + Y(s)4 = e^{-s} - Y(s)6[/tex]

    [tex]Y(s)((s+2)(s+2) + 6) = e^{-s} [/tex]
    So what now?
    Last edited: Oct 14, 2009
  5. Oct 14, 2009 #4


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    Homework Helper

    While (s+2)2+6 will give you back the original function, that is not completing the square.

    Remember you want s2+2s+10 =(s+1)2/+A, so what is A?
  6. Oct 14, 2009 #5

    Yeah, it's 9. Thanks for pointing out my mistake there, but also that's not the problem that I am having. THe thing is that the 9 has a factor of Y(s) so I have no choice but to return to the original equation as far as I can see. That's what is bothering me what can I do about it?

    [tex] Y(s)s^2 + Y(s)2s + Y(s) = e^{-s} - Y(s)9[/tex]

    [tex]Y(s)((s+1)^2 +9) = e^{-s} [/tex]
  7. Oct 14, 2009 #6
    Thanks for your help Rock. I have discovered a table which answers my question so just ignore the recent post, I'm okay now. :)
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