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Laplace transform

  1. May 1, 2005 #1
    I have this laplace transform that I need to solve: y''-6y'+13y=0 y'(0)=2 y(0)=-3


    I figured out my Y(s)=(-3s+20)/(s^2-6s+12). All I need to do is take the inverse laplace of this but I can't figure it. I know I need to split it into two fractions, but after that I'm lost. I'd appreciate any help.
     
  2. jcsd
  3. May 1, 2005 #2

    Pyrrhus

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    Yes Mr Beagss, it's right.
     
    Last edited: May 1, 2005
  4. May 1, 2005 #3
    Try doing complete the square on the bottom... see what happens
     
  5. May 1, 2005 #4
    Ok I got -3s/((s-3)^2)+4 + 20/((s-3)^2)+4) after completing the square and splitting up the fraction.
     
  6. May 1, 2005 #5

    Pyrrhus

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    Ok now look at your Laplace Table of inverse and convert them.

    For example

    [tex] e^{at} \sin (bt) = \frac{b}{(s-a)^{2} + b^{2}} [/tex]

    [tex]10 \frac{2}{(s-3)^{2} + (2)^{2}} = 10 e^{3t} \sin (2t) [/tex]
     
    Last edited: May 1, 2005
  7. May 1, 2005 #6

    Pyrrhus

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    For the other Laplace inverse is:

    [tex] e^{at} \cos (bt) = \frac{s-a}{(s-a)^{2} + b^{2}} [/tex]

    [tex] \frac{-3s}{(s-3)^{2} + 4} = \frac{-3s + 9 - 9}{(s-3)^{2} + 4} [/tex]

    thus

    [tex] \frac{-3s + 9 - 9}{(s-3)^{2} + 4} = \frac{-3(s - 3) - 9}{(s-3)^{2} + 4} [/tex]

    [tex] \frac{-3(s - 3) - 9}{(s-3)^{2} + 4} = \frac{-3(s - 3)}{(s-3)^{2} + 4} + \frac{-9}{(s-3)^{2} + 4} [/tex]

    and finally

    [tex] \frac{-3(s - 3)}{(s-3)^{2} + 4} = -3 e^{3t} \cos(2t) [/tex]

    [tex] \frac{-9}{2} \frac{2}{(s-3)^{2} + (2)^{2}} = \frac{-9}{2} e^{3t} \sin (2t) [/tex]

    so for the end Laplace inverse of

    [tex] \frac{-3s}{(s-3)^{2} + 4} = -3 e^{3t} \cos(2t) + \frac{-9}{2} e^{3t} \sin (2t) [/tex]
     
    Last edited: May 1, 2005
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