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Nice Laplace Transform

  1. Jun 19, 2008 #1
    I hope somebody's up tonight, this is due in the morning and I'm so close.

    So I was assigned to solve this differential equation using laplace transforms and although I (think I) can solve it, I'm not getting the same answer that Maple spits out.

    The DE is:

    [tex]x'' + 2x' + 5x = 3e^{-t}cos(2t); x(0) = x'(0) = 1[/tex]

    Let L(x) = Laplace(x)

    So here's my work:
    Take the Laplace of everything
    [tex]L(x'')+2L(x')+5L(x) = 3L(e^{-t}cos(2t))[/tex]

    Becomes:
    [tex]s^2L(x)-s(1)-(1)+2sL(x)+2(1)+5L(x)=3L(e^{-t}cos(2t))[/tex]

    Let L(x) = X

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

    I solved for X, simplified and broke it into partial fractions to figure out the Inverse Laplase but got the wrong answer. Is there anywhere I messed up in what you can see?
     
  2. jcsd
  3. Jun 19, 2008 #2

    Defennder

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    It should be [tex]s^2L(x)-s(1)-(1)+2sL(x)-2(1)+5L(x)=3L(e^{-t}cos(2t))[/tex] instead. You missed the minus sign.
     
  4. Jun 19, 2008 #3
    A thank you, took a little while to figure that out

    .... dang algebra
     
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