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Differential Eq: Inverse Laplace Transforms

  1. May 10, 2010 #1
    1. The problem statement, all variables and given/known data

    I have a couple of problems that im stuck on. The following:

    y'' - 6y' + 9y = t
    y(0) = 0
    y'(0) = 1

    and

    y'' - 6y' + 13y = 0
    y(0) = 0
    y'(0) = -3

    2. Relevant equations

    y'' = s2Y(s) - sf(0) - f'(0)
    y' = sY(s) - f(0)
    y = Y(s)


    3. The attempt at a solution

    ===================================
    for the first one:

    once I substitute into the original equation, I can move things around and I came out with the following:
    Y(s){s2-6s+9} = (1 +s2)/s2

    So I get Y(s) = (1+s2)/s2(s-3)2

    IIRC for partial fractions it should be the following: A/(s-3) + B/(s-3)2 + (Cs+D)/s2

    I dont know if this is where i messed up but I got:
    A=-2/27
    B=-1/9
    C=2/27
    D=1/9

    as a final result I get:
    -2/27e^3t - 1/92e^3t and the other part (Cs + D)/s^2 . . .I cant find any way to transform that

    the back of the book says:
    1/9t + 2/27 - 2/27e^3t + 10/9te^3t

    =============================

    The second one:

    Starting by substitution, plugging in the values and solving for Y(s)

    Y(s) = -3/(s2-6s + 13)

    and well. . Im lost from this point onwards. . .I dont remember how to do partial fractions if you cant factorize that denominator. And the quadratic formula (A=1 B = 6 C=13) gives imaginary numbers

    =============================

    Thank you!
     
    Last edited: May 10, 2010
  2. jcsd
  3. May 10, 2010 #2

    vela

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    You just have to separate it into individual terms.

    [tex]\frac{Cs+D}{s^2} = \frac{C}{s} + \frac{D}{s^2}[/tex]

    Also, according to Mathematica, you should have B=10/9. The other coefficients you found are correct.
     
    Last edited: May 10, 2010
  4. May 10, 2010 #3

    vela

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    Complete the square in the bottom so you get something that looks like

    [tex]Y(s) = -\frac{3}{(s-a)^2+b^2}[/tex]

    You should be able to find the inverse of that using the tables and the properties of Laplace transforms.
     
  5. May 10, 2010 #4
    wow. . .I cant believe I missed that. . .how does really old math comes back and haunt you huh?. . .thanks a lot
     
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