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Laplace Transform With Initial Values

  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.
  2. jcsd
  3. May 8, 2006 #2
    Why does my LaTex look all funny???
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