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How to use the Laplace Transform on a DE with a time coefficent

  1. Feb 5, 2017 #1
    1. The problem statement, all variables and given/known data
    ty'' + y' = 2t2, y(0) = 0


    2. Relevant equations
    laplace(f''(t)) = s2laplace(f(t)) -sf(0) - f'(0)
    (-1) (d/ds) (F(s))


    3. The attempt at a solution
    I know how to solve the problem except for the ty'' part. I tried using the equation and I got -d/ds(s2Y(s) - 0 - f'(0)) which becomes -2sY(s). But this is wrong because it does not lead to the correct answer y = (2/3)t3+c1t2
     
  2. jcsd
  3. Feb 5, 2017 #2

    Ray Vickson

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    Your equation for ##Y(s)## is a first-order differential equation, since it involves both ##Y(s)## and ##dY(s)/ds##.
     
  4. Feb 5, 2017 #3
    So in the problem when I get the dY(s)/ds after using the product rule what am I supposed to do? Is it the same as having the laplace of y'(t)?
     
  5. Feb 5, 2017 #4

    LCKurtz

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    No. ##\frac d {ds} (s^2Y(s) \ne -2sY(s))##. You need the product rule.

    No, it isn't the same as the LaPlace of ##y'(t)##. You already know the LaPlace of that is ##sY(s)-y(0)##. Assuming you do correctly use the product rule, as has already been pointed out to you, you will get a differential equation in ##Y(s)## which you have to solve for ##Y(s)## before you try to take the inverse.
     
  6. Feb 5, 2017 #5
    Thanks for the help I got the answer!
     
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