How to solve a second order diff eq?


by Binaryburst
Tags: differetial, equations
Binaryburst
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#1
Feb13-13, 04:35 PM
P: 24
If have this equation:

[itex] \frac {d^2x}{dt^2}=-\frac{x}{1+x^2} [/itex]

How do I solve it?
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mfb
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#2
Feb13-13, 05:31 PM
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Did you try the usual methods, like separation of variables or a direct integration?
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#3
Feb13-13, 06:05 PM
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Quote Quote by Binaryburst View Post
If have this equation:

[itex] \frac {d^2x}{dt^2}=-\frac{x}{1+x^2} [/itex]

How do I solve it?
How about numerical methods, like they are applied here?
(I do not believe there is a nice analytic solution.)

lurflurf
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#4
Feb13-13, 10:55 PM
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How to solve a second order diff eq?


$$\int \! \dfrac{d^2x}{dt^2}\dfrac{dx}{dt} \, \mathrm{d}t=-
\int \! \frac{x}{1+x^2}\dfrac{dx}{dt} \, \mathrm{d}t$$
Binaryburst
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#5
Feb14-13, 03:08 AM
P: 24
The solution to this equation should be:

[itex] x(t) = sin(t) [/itex]

I simply don't know how to get there ...

I have integrated but I get something like integ of 1/sqr(-ln(1+x^2)) dx?! :(
Binaryburst
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#6
Feb14-13, 03:47 AM
P: 24
This is what wolfram alpha says: http://m.wolframalpha.com/input/?i=x...B1%29&x=10&y=2
mfb
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#7
Feb14-13, 07:55 AM
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Quote Quote by Binaryburst View Post
The solution to this equation should be:

[itex] x(t) = sin(t) [/itex]
This would give ##x''=-x##, and ##1+x^2 \neq 1## (in general), so it is not a solution.

The first integral should give something like ln(1+x^2).
Binaryburst
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#8
Feb14-13, 08:12 AM
P: 24
@ mfb I am out of ideas. I have a problem that I have been struggling with for half a year and still haven't solved it. I posted it once but didn't get an exact complete solution. http://www.physicsforums.com/showthr...1&goto=newpost It has to be done without conservation of energy. (Just using forces) Thanks in advance.
bigfooted
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#9
Feb14-13, 03:10 PM
P: 263
introduce a new variable
[itex]p=\frac{dx}{dt}[/itex]

by the chain rule we have:
[itex]\frac{d^2x}{dt^2}=\frac{dp(x(t))}{dx}=\frac{dp}{dx}\frac{dx}{dt}=\frac{ dp}{dx}p[/itex]
This means that
[itex]pdp = \frac{-1}{1+x^2}dx[/itex]

integrate!

[itex]\int pdp = \int \frac{-1}{1+x^2}dx[/itex]
[itex]\frac{1}{2}p^2 = -\frac{1}{2}\ln(1+x^2)+C_1[/itex]

and because p is the derivative dx/dt, we get

[itex]\frac{dx}{\sqrt{-\ln(1+x^2)+C_1}} = 1[/itex]

integrate!

[itex]\int\frac{1}{\sqrt{\ln(1+x^2)+C_1}}dx = t + C_2[/itex]


Well, not a very nice solution because it's implicit.
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#10
Feb14-13, 03:32 PM
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Let us denote ##\dot x## to be the same as ##dx \over dt##.

Then:
$$\ddot x = - \frac{x}{1+x^2}$$
$$2 \ddot x \dot x = -\frac{1}{1+x^2} \cdot 2x \cdot \dot x$$
$$\dot x^2 = -\ln(1+x^2) + C$$

And there it ends!
No nice integration after this!


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