
#1
Feb1313, 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? 



#2
Feb1313, 05:31 PM

Mentor
P: 10,809

Did you try the usual methods, like separation of variables or a direct integration?




#3
Feb1313, 06:05 PM

HW Helper
P: 6,189

(I do not believe there is a nice analytic solution.) 



#4
Feb1313, 10:55 PM

HW Helper
P: 2,151

How to solve a second order diff eq?
$$\int \! \dfrac{d^2x}{dt^2}\dfrac{dx}{dt} \, \mathrm{d}t=
\int \! \frac{x}{1+x^2}\dfrac{dx}{dt} \, \mathrm{d}t$$ 



#5
Feb1413, 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?! :( 



#6
Feb1413, 03:47 AM

P: 24

This is what wolfram alpha says: http://m.wolframalpha.com/input/?i=x...B1%29&x=10&y=2




#7
Feb1413, 07:55 AM

Mentor
P: 10,809

The first integral should give something like ln(1+x^2). 



#8
Feb1413, 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.




#9
Feb1413, 03:10 PM

P: 262

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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