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Laplace transforms

  1. Oct 18, 2006 #1
    Can someone show me how to do these laplace transforms of these differentials?

    1) y""-4y"'+6y" -4y'+y=0
    y(0)=0, y'(0)=1, y"(0)=0, y"'(0)=1

    2) y"-2y'+4y=0
    y(0)=2, y'(0)=0

    3) y"'+2y'+y=4e^-t
    y(0)=2, y'(0)=-1

    4) y"-2y'+2y=cos(t)
    y(0)=1, y'(0)=0
    the Laplace transfrom that i got for this was
    s/(s^2+a^2) * 1/(s^2-2s+2) + (s-2)/(s^2-2s+2)=y
    I'm trying to find the inverse transforms of these but i have no idea how to do this because i can't factor the numerator by completing the square.
     
  2. jcsd
  3. Oct 19, 2006 #2

    HallsofIvy

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    Staff Emeritus
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    Are you trying to find the Laplace transform or the inverse transform?

    Surely, if you are doing problems like that, you must know that:
    L(y')= sL(y)- y(0),
    L(y")= s2L(y)- y(0)- y'(0), and
    L(y"')= s3L(y)- y(0)- y'(0)- y"(0).

    s2- 2x+ 2= s2-2x+ 1+ 1= (s-1)2+ 1. You can't factor that, of course (with real numbers), but you should know inverse transforms involving [itex]\frac{1}{s^2+ 1}[/itex].
     
  4. Oct 19, 2006 #3
    My mistake, I'm looking for the inverse Laplace transform.
     
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