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Laplace Transform/Diff eq question

  1. Sep 7, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the solution to the differential equation using laplace transform techniques

    y''(t) + 2y'(t) + 2y(t) = u'(t)
    u(t) = 1(t) (step function)
    y(0) = 1, y'(0) = 0


    2. Relevant equations
    L{y''(t)} = (s^2)Y(s) + (s)y(0) - y'(0)
    L{y'(t)} = (s)Y(s) - y(0)
    u'(t) = d(t) (dirac/delta function)
    L{d(t)} = 1

    3. The attempt at a solution

    L{y''(t) + 2y'(t) + 2y(t) = u'(t)} =
    (s^2)Y(s) + (s)y(0) - y'(0) - 2s*Y(s) - 2y(0) + 2Y(s) = 1
    (s^2)Y(s) - s + 2s*Y(s) - 2 + 2Y(s) = 1
    (s^2)Y(s) + 2s*Y(s) + 2Y(s) = 3+s
    Y(s) (s^2 + 2s + 2) = 3+s
    Y(s) = (3+s)/(s^2 + 2s + 2)

    Now, the problem I am facing is that any calculator that I use tells me the solution to this problem is (e^(-t))*(cos(t) + sin(t))
    example
    http://www.wolframalpha.com/input/?i=y''(t)+2y'(t)+2y(t)=+dirac(t),+y(0)+=+1,+y'(0)+=+0

    I realized this when doing my analysis of the system in MATLAB.

    But when I take the inverse laplace transform of the equation I ended up with, I get
    (e^(-t))*(cos(t) + 2sin(t))

    which is inconsistent with the initial conditions, since it gives a 2 at t = 0.
    Any ideas about what I could be doing wrong here?
     
  2. jcsd
  3. Sep 7, 2009 #2
    e^(-0)=1
    cos(0) =1
    2sin(0) =0

    so y(0)=1.

    Your working seems correct :)
     
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