# Second order DE

I've been trying to deterimine the path of a particle. And somewhere in the solution, I need to solve this:
$$\frac{d^2x}{dt^2} = -\frac{L}{x}$$
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.

Tom Mattson
Staff Emeritus
Gold Member

In case you haven't noticed, the variable on the right side is x and not t.

Tom Mattson
Staff Emeritus
Gold Member
Whoops.

I haven't finished solving it yet, but I have started by moving the $x$ 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.

Thanks for taking time :)

Tom Mattson
Staff Emeritus
Gold Member
The Laplace transform method doesn't work. $X(s)$ cancels out entirely.

I then tried substituting $v=\frac{dx}{dt}$, which gives the following:

$$\frac{d^2x}{dt^2}=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}[/itex] This works in principle because it gives you a seperable first order ODE in $v(t)$. But the integration to get from $v$ to $x$ looks nasty, and I don't know if it can be done in practice. I too have tried solving the equation using the same trick. [tex]\frac{dv}{dt} \frac{dx}{dx} = vdv\frac{1}{dx}$$
and
$$vdv = -L\frac{dx}{x}$$

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

(note: negative v won't make sense, so I just ignored it)

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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:
$$\frac{d^2x}{dt^2} = -\frac{L}{x}$$
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?

dextercioby
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.

Clausius2
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.

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:
$$\int \frac{dx}{\sqrt{-2L \ln(x) + C_0}} = e^{\frac{C_0}{2L}} \sqrt{\frac{\pi}{2L}} erf(\sqrt{\frac{-2L \ln(x) + C_0}{2L}})$$

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