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Laplace Transform

  1. Jul 30, 2009 #1
    1. The problem statement, all variables and given/known data
    Use the Laplace Transform to solve: y"+2y'+2y=t y(0)=y'(0)=1

    2. Relevant equations
    L{y(t)} = Y(s)
    L{y'(t)} = sY(s)-y(s)
    L{y"(t)} = s2Y(s)-sy(0)-y'(0)
    using the laplace transform table: tn = n!/(sn+1) where n=1

    3. The attempt at a solution
    Take laplace on both sides:
    L{y"(t)} + 2L{y'(t)} + 2L{y(t)} = L{t}
    s2Y(s) - sy(0) - y'(0) + 2[sY(s)-y(0)] + 2Y(s) = 1/s2

    after plugging in the initial conditions I get:
    s2Y(s) - s - 3 + 2sY(s) + 2Y(s) = 1/s2

    isolate Y(s):
    Y(s)[s2+2s+2] = 1/s2 + s + 3
    Y(s) = (1/s2 + s + 3) / (s2+2s+2)
    I completed the square in the denominator to get: (m+1)2+1

    Y(s) = s / [(s2+1)(s+3)2-8]}

    Take the Laplace inverse
    L-1{ s / [(s2+1)(s+3)2-8]}

    I add and subtract 3 in the numerator to get:
    L-1{ (s+3)-3 / [(s2+1)(s+3)2-8]}

    Use linearity property of inverse transform to get from L-1{Y(s)} to y(t):
    L-1{ (s+3) / [(s2+1)(s+3)2-8]} - 3L-1{ 1 / [(s2+1)(s+3)2-8}

    How do I apply partial fraction decomposition to get y(t)?
    Last edited: Jul 30, 2009
  2. jcsd
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