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DE using Laplace transform

  1. Mar 2, 2010 #1
    1. The problem statement, all variables and given/known data

    y'' -6y' + 9y = f(t)

    f(t)=
    0, 0<t<1
    1, 1<t<3
    0, t>3

    2. Relevant equations



    3. The attempt at a solution

    Turning it to heaviside functions I get:
    y'' -6y' + 9y = u1 - u3

    and I solve.

    in the answers it should be:
    y(t) = 2tet + u1(1/9 - 1/9e3(t-1)+ 1/3(t-1)e3(t-1)).

    where did u3 go in the answers? why doesn't it appear? the u1 I can get just like that, but I can't seem to figure out where the u3 went and why there is a 2tet in there.
     
  2. jcsd
  3. Mar 2, 2010 #2
    It looks like we're missing some information. Were there additional conditions given? (e.g. y(0) = ? and y'(0) = ?)

    Also, what did the DE look like after you applied the Laplace transform?
     
  4. Mar 2, 2010 #3
    it doesn't look as though you changed anything by putting it in unit step form.
     
  5. Mar 2, 2010 #4
    Sorry, forgot.
    y(0) = 1, y'(0) = 2.
     
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