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Leplace transform

  1. May 5, 2008 #1
    1. The problem statement, all variables and given/known data

    y"+4y=7t , y(0)=-1, y'(0)=3


    3. The attempt at a solution

    L{y} + 4L{y} = 7L{t}

    (s^2L{y} - sy(0) - y'(0)) + 4L{y} = 7L{t}

    s^2L{y} - s(-1) - 3 + 4L{y} = 7(0!/s^(0+1))

    L{y}(s^2 + 4) = 7(0!/s^(0+1)) - s + 3
    = ((-s+3)s + 7)/s
    =(-s^2 + 3s +7)/s

    L{y}=(-s^2 + 3s + 7)/(s(s^2+4))

    here is where I run into trouble, setting up the partial fractions

    L{y}=-s^2 + 3s +7 = A/s + B/(s^2+4)

    is that correct so far?
     
  2. jcsd
  3. May 5, 2008 #2
    You should check the laplace transform for 7t.

    L{t^n} = (n!)/(s^(n+1))
    therefore L{t} = L{t^1} = (1!/(s^(1+1))

    L{7t} should look more like 7*(1/s^2) or (7/s^2)
     
  4. May 5, 2008 #3
    but its t^1, so n=0, t^0+1=t^1 right?
     
  5. May 5, 2008 #4
    Its t^1, so n = 1 when you plug into the transform for t^n.
     
  6. May 5, 2008 #5
    Or if you want the Laplace transform of t^n in terms of gamma function:

    [tex] L\{t^n\}=\frac{\Gamma(n+1)}{s^{n+1}}[/tex]

    [tex]\Gamma(n+1)=n\Gamma(n)=.....=n!,n \in Z^+[/tex]
    [tex]\Gamma(1)=1[/tex]
     
    Last edited: May 5, 2008
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