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Differential equation with Laplace transform

  1. Feb 3, 2010 #1
    1. The problem statement, all variables and given/known data
    [tex]t y'' + (3 t-1) y' + 3 y = 6 e^{-3t}[/tex]
    y(0) = 1
    y(5) = 2

    2. Relevant equations

    3. The attempt at a solution
    I tried applying the Laplace transform to the equation but I was having a little trouble...
    L [t·y''] = -dL[y'']/ds = s2·Y' + 2·s·Y - y(0) = s2·Y' + 2·s·Y - 1
    L [t·y'] = -dL[y']/ds = s·Y' + Y
    Therefore:
    [tex]s^2 Y' + 2 s Y - 1 + 3 s Y' +3 Y - s Y + 1+ 3 Y = \frac{6}{s+3}[/tex]
    [tex](s^2+3s)Y'+(s+6)Y=\frac{6}{s+3}[/tex]

    And then solve the Y(s) and then use the inverse transform.
    But it doesn't yield the correct result. I did the rest on the computer so I guess it's when I apply the transform where I make a mistake.

    Thanks for your help.
     
  2. jcsd
  3. Feb 3, 2010 #2

    vela

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    I think you dropped the negative sign when calculating L[t y''] and L[t y'].
     
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