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Initial value problem (Laplace)

  1. Jul 11, 2008 #1
    1. The problem statement, all variables and given/known data

    Use the Laplace transform method to solve the differential equation

    y" -4y' +4y = e^2t subject to initial conditions y(0)=1, y'(0)=0

    2. Relevant equations



    3. The attempt at a solution

    s^2Y(s) -sy(0) - y'(0) - 4[ sY(s) - y(0)] + 4Y(s)= L (e^2t)

    Y(s) [s^2 -4s +4] - s -4 = 1/(s-2)

    Y(s) [ (s-2)^2 ] = 1/(s-2) + s +4

    Y(s) = 1/(s-2)^3 + s/(s-2)^2 + 4/(s-2)^2

    Is this right, how do I find out the inverse laplace of Y(s) like for example 1/(s-2)^3 seems impossible!

    Thanks.
     
  2. jcsd
  3. Jul 11, 2008 #2

    HallsofIvy

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    ??? Pretty nearly every inverse lapalce transform is done by looking it up in a table of transforms. But, as I pointed out in response to your other post, generally, you find the Laplace transform (and almost always the inverse transform) by looking it up in a table. Here is a good one:
    http://www.vibrationdata.com/Laplace.htm

    In particular, the inverse transform of 1/(s-2)3 is (1/2)t2e-2t.
     
    Last edited: Jul 11, 2008
  4. Jul 11, 2008 #3
    how about s/(s-2)^2. It isnt a cosine or sine laplace thing?

    Thanks.
     
  5. Jul 11, 2008 #4
    You can find [tex]\frac{1}{(s-2)^2}[/tex] and s in that table he linked.
     
  6. Jul 12, 2008 #5

    HallsofIvy

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    [tex]\frac{1}{(s-2)^3}[/tex]
    and
    [tex]\frac{1}{(s-2)^2}[/tex]
    are given by number 2.11 in that table.

    By partial fractions,
    [tex]\frac{s}{(s-2)^2}= \frac{1}{s-2}+ \frac{2}{(s-2)^2}[/tex]
     
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