1. Not finding help here? Sign up for a free 30min 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!

Second order DE

  1. Mar 30, 2006 #1
    I've been trying to deterimine the path of a particle. And somewhere in the solution, I need to solve this:
    [tex]\frac{d^2x}{dt^2} = -\frac{L}{x}[/tex]
    where L is a positive number (actually length). And ideas how this can be solved? I was just reluctant to try series expansion (frob., etc) since the problem isn't finished with the solution of this, and it'd be better if I had finite terms to go on.
     
  2. jcsd
  3. Mar 30, 2006 #2

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    How about integrating twice?
     
  4. Mar 30, 2006 #3
    How about reading the question?

    In case you haven't noticed, the variable on the right side is x and not t.
     
  5. Mar 30, 2006 #4

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Whoops. :redface:

    I haven't finished solving it yet, but I have started by moving the [itex]x[/itex] over to the left side and taking the Laplace transform. It requires some clever integration by parts, but I think it is going to come out.
     
  6. Mar 30, 2006 #5
    Thanks for taking time :)
     
  7. Mar 30, 2006 #6

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The Laplace transform method doesn't work. [itex]X(s)[/itex] cancels out entirely.

    I then tried substituting [itex]v=\frac{dx}{dt}[/itex], which gives the following:

    [tex]\frac{d^2x}{dt^2}=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}[/itex]

    This works in principle because it gives you a seperable first order ODE in [itex]v(t)[/itex]. But the integration to get from [itex]v[/itex] to [itex]x[/itex] looks nasty, and I don't know if it can be done in practice.
     
  8. Mar 30, 2006 #7
    I too have tried solving the equation using the same trick.
    [tex]\frac{dv}{dt} \frac{dx}{dx} = vdv\frac{1}{dx}[/tex]
    and
    [tex]vdv = -L\frac{dx}{x}[/tex]

    [tex]\frac{v^2}{2} = -L \ln(x) + C_0[/tex]
    [tex]v = \frac{dx}{dt} = \sqrt{-2L \ln(x) + C_0}[/tex]
    [tex]\frac{dx}{\sqrt{-2L \ln(x) + C_0}} = dt[/tex]
    Which seemed to be worse...
    But at least, it's first order!

    (note: negative v won't make sense, so I just ignored it)
     
    Last edited: Mar 30, 2006
  9. Mar 31, 2006 #8
    I guess x and t are not position and time as then the equation is dimensionally incorrect. Or is there still some coefficient missing?
     
  10. Mar 31, 2006 #9

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Yes, it all means that x(t) is not expressible in terms of elementary functions, and possibly neither in terms of known special functions...

    Daniel.
     
  11. Mar 31, 2006 #10

    Clausius2

    User Avatar
    Science Advisor
    Gold Member

    aaahhh nonlinearity.....makes a simple equation to be a very hard one. If you assume a series solution sometimes you come out with the series expansion of an analytical function. It's only an idea.
     
  12. Mar 31, 2006 #11
    ^-^' actually, I've just merged them into L to keep it simple, but L is still positive.

    BTW, integrator has given this solution:
    [tex]\int \frac{dx}{\sqrt{-2L \ln(x) + C_0}} = e^{\frac{C_0}{2L}} \sqrt{\frac{\pi}{2L}} erf(\sqrt{\frac{-2L \ln(x) + C_0}{2L}})[/tex]
     
    Last edited: Mar 31, 2006
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Second order DE
  1. Second order DE (Replies: 2)

  2. Second order DE (Replies: 6)

  3. Second Order DE question (Replies: 10)

Loading...