Second order DE

  • Thread starter gulsen
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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.
 

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  • #2
Tom Mattson
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How about integrating twice?
 
  • #3
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How about reading the question?

In case you haven't noticed, the variable on the right side is x and not t.
 
  • #4
Tom Mattson
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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.
 
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Thanks for taking time :)
 
  • #6
Tom Mattson
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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.
 
  • #7
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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:
  • #8
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gulsen said:
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.

I guess x and t are not position and time as then the equation is dimensionally incorrect. Or is there still some coefficient missing?
 
  • #9
dextercioby
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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.
 
  • #10
Clausius2
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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.
 
  • #11
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da_willem said:
I guess x and t are not position and time as then the equation is dimensionally incorrect. Or is there still some coefficient missing?

^-^' 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]
 
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