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Laplace transform of f(x)= xcos(ax)

  1. Jan 30, 2008 #1
    1. The problem statement, all variables and given/known data

    Laplace transform from f(x) = x cos(ax)
    to ^f(s) = (s^2 - a^2) / (s^2 + a^2)^2

    how do you get the -a^2 term in the numerator,
    all i come up with is s^2?

    2. Relevant equations

    f(x) = x e^(ax) ----> ^f(s) = 1/(s-a)^2

    2 cos(ax) = e^(aix) + e^(-aix)


    3. The attempt at a solution

    f(s) = integral[ x/2 (e^-(s-ai)x + e^-(s+ai)x) ]dx
    = 1/2[1/(s-ai)^2 + 1/(s+ai)^2]
    =1/2[ (s+ai)^2 + (s-ai)^2] / (s^2+a^2)^2
    =1/2(s^2+a^2)^2 [ s^2 + 2asi - a^2 + s^2 - 2asi +a^2]
    =(s^2 )/ (s^2 + a^2)^2

    but the given solution is:
    (s^2 - a^2) / (s^2 + a^2)^2

    so where am i missing the -a^2 term?
     
  2. jcsd
  3. Jan 30, 2008 #2

    Avodyne

    User Avatar
    Science Advisor

    =1/2[ (s+ai)^2 + (s-ai)^2] / (s^2+a^2)^2
    =1/2(s^2+a^2)^2 [ s^2 + 2asi - a^2 + s^2 - 2asi +a^2]
    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~|wrong sign!
     
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