1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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]


    RIGHT SIDE
    [tex]=\left(\frac{1}{3}\right)(-3(s^{2}+6)e^{-\left(\frac{s^2}{6}\right)[/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+6}{s^3}\}[/tex]

    [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???
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook




Loading...