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Solving an IVP using Laplace Transforms.

  1. Sep 3, 2012 #1
    Solving an IVP using Laplace Transforms. HELP!

    Ok I'm supposed to Solve this problem using Laplace Transforms.

    [tex] \frac{d^2 x}{dt^2}+2\frac{dx}{dt}+x = 5e^{-2t} + t [/tex]

    Initial Conditions

    x (0) = 2 ; [tex] \frac{dx}{dt} (0) = -3 [/tex]

    so I transformed the the IVP and it looks like this

    [tex] s^2 x(s) - s x(s) - x (0) + 2 x(s) - x(0) + x(s) = \frac{5}{s+2} + \frac {1}{s^2} [/tex]

    Then I plugged in my initial conditions


    [tex] s^2 x(s) + 3s - 2 + 2 x(s) - 2 + x(s) = \frac{5}{s+2} + \frac {1}{s^2} [/tex]


    Then I factored the x(s) on the left side, and found a common denominator on the right side

    [tex] s^2 x(s) + 2 x(s) + x(s) + 3s - 4 = \frac{5}{s+2} + \frac {1}{s^2} [/tex]


    [tex] s^2 x(s) + 2 x(s) + x(s) + 3s - 4 = \frac{5}{s+2}*\frac {s^2}{s^2} + \frac {1}{s^2} *\frac {s+2}{s+2} [/tex]

    [tex] s^2 x(s) + 2 x(s) + x(s) + 3s - 4 = \frac{5s^2}{(s+2)s^2} + \frac {s+2}{(s^2)(s+2)}[/tex]

    [tex] s^2 x(s) + 2 x(s) + x(s) + 3s - 4 = \frac{5s^2 + s + 2}{(s+2)s^2} [/tex]

    [tex] (s^2 + 2s + 1) x(s) = \frac{5s^2 + s + 2}{(s+2)s^2} - 3s +4 [/tex]

    Then I found x(s)

    [tex] (s+1)(s+1) x(s) = \frac{5s^2 + s + 2}{(s+2)s^2} (- 3s +4)*\frac{(s+2)s^2}{(s+2)s^2} [/tex]

    [tex](-3s +4)*(s^3 + 2s^2) = -3s^4 - 6s^3 + 4s^3 + 8s^2 [/tex]

    [tex] x(s) = \frac{5s^2 + s + 2 -3s^4 - 6s^3 + 4s^3 + 8s^2 }{(s+2)(s^2)(s+1)^2} [/tex]

    [tex] x(s) = \frac{13s^2 + s + 2 -3s^4 -2s^3 }{(s+2)(s^2)(s+1)^2} [/tex]

    Now I used the heaviside theorem to find the residues.

    [tex] x(s) = \frac{13s^2 + s + 2 -3s^4 -2s^3 }{(s+2)(s^2)(s+1)^2} = \frac{R1}{s} + \frac{R2}{s^2} + \frac{R3}{s+2} + \frac{R4}{(s+1)^2} + \frac{R5}{s+1} [/tex]

    Ok so here is where I'm stuck. Say I try to solve for R3.

    [tex] R3 = (s+2) x(s) |s = -2 , \frac{13s^2 + s + 2 -3s^4 -2s^3 }{(s^2)(s+1)^2} | s = -2 ] = 5 [/tex]

    Ok so I got the correct residue for R3, but now I'm stuck what do I do after this?
     
    Last edited: Sep 3, 2012
  2. jcsd
  3. Sep 3, 2012 #2
    Hi !

    OK. R3=5
    Why not doing the same for the other coefficients ?
    multiply by (s+1)² to compute R4
    multiply by s² to obtain R2
     
  4. Sep 3, 2012 #3

    Mute

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    Homework Helper

    Re: Solving an IVP using Laplace Transforms. HELP!

    Before you get too deep into your calculation, you might want to double check this result. The Laplace transform of the second derivative is

    $$\mathcal L[\ddot{x}(t)](s) = s^2 X(s) - s x(t=0) - \dot{x}(t=0),$$
    whereas you seem to have written

    $$s^2 X(s) - sX(s) - x(0).$$
    (I'm writing ##X(s)## for the transform to make it easier to distinguish from x(t)).

    When you transform the first derivative term, it also looks like you didn't factor through the 2 : you wrote ##2X(s)-x(t=0)## instead of ##2X(s) - 2x(t=0)##.
     
  5. Sep 3, 2012 #4
    Re: Solving an IVP using Laplace Transforms. HELP!

    Oh I see it now that's why I've been messing up. Thanks.
     
    Last edited: Sep 3, 2012
  6. Sep 3, 2012 #5
    Re: Solving an IVP using Laplace Transforms. HELP!

    So it ends up becoming

    [tex] s^2 X(s) - s X(0) - \dot{x}(0) [/tex]

    Which in turn becomes

    [tex] s^2 X(s) - 2s +3 [/tex]
     
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