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Laplace Transformation Question

  1. Oct 8, 2011 #1
    1. The problem statement, all variables and given/known data

    Use Laplace transformations to solve:
    Y''+tY'+2Y=0, Y(0)=0, Y'(0)=1

    2. Relevant equations

    Y''==>[itex]s^{2}[/itex]y-sY(0)-Y'(0)=[itex]s^{2}[/itex]y-1
    tY'==>-(y+sy')
    2Y==>2y

    3. The attempt at a solution

    The question felt pretty straightforward till I hit a rock.


    [itex]s^{2}[/itex]y-1+y-sy'=0

    y'=[itex]\frac{1}{s}[/itex](y([itex]s^{2}[/itex]+1)-1)

    My problem is that I cant integrate this because of the -1. What to do? Am I on the right track?

    (First time using these forums and symbols so give me a sec to fix them up. Not sure why the S is coming out weird.)
     
    Last edited: Oct 8, 2011
  2. jcsd
  3. Oct 8, 2011 #2
    Hi Applejacks,

    I verified your calculations, and I think you are on the right track. The [itex]-1[/itex] does indeed preclude the use of separation of variables to solve the ODE for [itex]y(s)[/itex]. If I put that ODE into Maple I get a complicated expression involving the error function, so it does not look like there's an easy way; maybe you could look up the integral in, for example, Abramowitz and Stegun.

    Cheers,
    Kurt

    P.S. According to Maple the solution of your ODE with boundary conditions is [tex]Y(t)=t \exp{(-t^2/2)}.[/tex]
     
  4. Oct 8, 2011 #3

    vela

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    Rearranging terms a little, you get
    [tex]sy' - (s^2+1)y = -1[/tex]which is a first-order differential equation. I'd try solving it using an integrating factor.
     
  5. Oct 8, 2011 #4
    Hi,

    Indeed, vela, I had forgotten about that technique. Upon investigation I find that it does not make the problem much easier, however. By using the integrating factor technique, I recast the first-order ODE for [itex]y(s)[/itex] into the following form: [tex]\frac{d}{ds}\left(\frac{1}{s}\exp{\left(-\frac{s^2}{2}\right)}y(s)\right)=-\frac{1}{s}\exp{\left(-\frac{s^2}{2}\right)}.[/tex] Integrating this equation requires integration of the Gaussian function on the right hand side, which leads to an expression involving an error function (in agreement with the Maple output I mentioned earlier).

    Cheers,
    Kurt

    P.S. That the expression for [itex]y(s)[/itex] contains an error function can also be seen by taking the Laplace transform of the known solution [itex]Y(t)=t\exp(-t^2/2)[/itex] that I posted earlier.
     
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