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Laplace Transofrm using IVP

  1. May 8, 2006 #1
    Okay, I know this is alot... but I am stuck, so here goes...

    Use the method of Laplace transform to solve the initial value problem

    [tex]y''+3ty'-6y=0, y(0) = 1, y'(0) = 0[/tex]
    [tex]L\{y'' + 3ty' - 6y\} = L\{0\}[/tex]
    [tex]s^{2}Y(s) - sy(0) - y'(0) + 3L\{ty'\} - 6Y(s) = 0[/tex]
    [tex]s^{2}Y(s) - s(1) - 0 - \frac{d}{ds}\left(3 L\{ty'\}\right) -6Y(s) = 0[/tex]

    Now to resolve the [tex]- \frac{d}{ds}\left(3 L\{ty'\}\right)[/tex]
    [tex]= - \frac{d}{ds}\left(3 L\{ty'\}\right)[/tex]

    [tex]= - \frac{d}{ds}3 \left(sY(s) - y(0)\right)[/tex]

    [tex]= -3sY'(s) - 3Y(s)[/tex]

    Plugging it back into the eq we now have

    [tex]s^{2}Y(s) - s - 3sY'(s) - 3Y(s) - 6Y(s) = 0[/tex]

    [tex]-3sY'(s) + (s^{2}-9)Y(s) - s = 0[/tex]

    [tex]Y'(s) + \left(-\frac{s}{3} + \frac{3}{s}\right)Y(s) = -\frac{1}{3}[/tex]

    [tex]\mu = e^{\int\left(-\frac{s}{3} + \frac{3}{s}\right)ds}[/tex]

    [tex]\mu = e^{\left(-\frac{s^{2}}{6} + ln(s^{3})\right)[/tex]

    [tex]\mu = s^{3} e^{\left(-\frac{s^{2}}{6}\right)[/tex]

    [tex]\int\left(\frac{d}{ds}(s^{3}e^{-\left(\frac{s^2}{6}\right)}Y(s)\right) = \int-\left(\frac{1}{3}\right)s^{3}e^{-\left(\frac{s^2}{6}\right)} ds[/tex]

    [tex]s^{3}e^{-\left(\frac{s^2}{6}\right)}Y(s) = \int-\left(\frac{1}{3}\right)s^{3}e^{-\left(\frac{s^2}{6}\right)} ds[/tex]


    [tex]=(s^2+6)e^{-\left(\frac{s^2}{6}\right) + A[/tex]

    [tex]Y(s)=\frac{(s^2+6)}{s^{3}} + \frac{A e^ \frac{s^2{6}{s^{3}}[/tex]

    [tex] Limit...as... s \rightarrow \infty........Y(s) = 0...therefore A = 0[/tex]

    [tex]Y(s) = \frac{s^2+6}{s^3}[/tex]

    Break down the Inverse Laplace

    [tex]=L^{-1}\{\frac{s^2}{s^3}\} + L^{-1}{\frac{6}{s^3}\}[/tex]

    [tex]=L^{-1}\{\frac{1}{s}\} + L^{-1}{\frac{6}{s^3}\}[/tex]

    [tex]= 1 + ?????? [/tex]

    This is where I get lost.... I don't know how to do the other side... Please help.
    Last edited: May 9, 2006
  2. jcsd
  3. May 8, 2006 #2
    Why does my tex look all ugly and bad ???
  4. May 9, 2006 #3


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    What other side? I hope it is not calculating the antitransform of
    [tex]\frac{6}{s^3}[/tex] because it's pretty easy, it is in any table.

  5. May 9, 2006 #4

    Yes this is where I was lost because I didn't have that in the table my professor gave me. His said:


    I am lost because I only see this problem as:

    [tex]= t^2[/tex]...... i know this isn't right.

    wait... is this what I think it is? Am I overthinking this. Is the answer:

    [tex]2t^2 [/tex] ? A 2 multiplied into the answer would make the numerator a 6. Am I right ?
  6. May 9, 2006 #5
    can someone verify if [tex]2t^2 [/tex]is right ?
  7. May 9, 2006 #6
    I have looked all over the internet and in my book and I cannot find the above mentioned Laplace Transform formula anywhere. Did you mean (n!) in the numerator vice the (n+1) ? In that case, that would make my answer [tex]3t^2[/tex]
  8. May 9, 2006 #7


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    For n a positive integer:

  9. May 9, 2006 #8
    I appreciate the help :smile:

    thanks. I can't believe I didn't see that the antitransform was so simple...

    [tex] y(t) = 1 + 3t^2 [/tex]
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