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No-homo second order DE

  1. Apr 29, 2013 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations



    3. The attempt at a solution

    Again.. I find it to be really easy, then get it wrong. My answer is not even close. Applying the properties of the laplace transform in the usual way,

    s^2L{y} - 4 - 2L{y} + 2L{y} = L{e^-t}

    =>

    [itex]L{y} = (\frac{1}{s+1})(\frac{1}{s^2 - 2s + 2}) + (\frac{4}{s^2-2s+2})[/itex]

    Now, I split the first term into partial fractions and factored the denomenator of the second. I'll spare you all of that tedious algebra:

    [itex]-\frac{1}{4}(\frac{1}{s+1}) + \frac{1}{4}(\frac{1}{s-1}) - \frac{1}{2}(\frac{1}{(s-1)^{2}}) + 4(\frac{1}{(s-1)^{2}})[/itex]


    The inverse laplace of that, I find to be

    -(1/4)e^-t + (1/4)e^t - (1/2)te^t + 4te^t

    The answer given and the one given by wolfram involves trig functions, what am I doing wrong??

    Thanks!
     
  2. jcsd
  3. Apr 29, 2013 #2

    Mark44

    Staff: Mentor

    For starters, s2 - 2s + 2 ≠ (s - 1)2.
     
  4. Apr 29, 2013 #3


    I think I need a nap.
     
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