# How to solve a second order diff eq?

by Binaryburst
Tags: differetial, equations
 P: 24 If have this equation: $\frac {d^2x}{dt^2}=-\frac{x}{1+x^2}$ How do I solve it?
 Mentor P: 12,037 Did you try the usual methods, like separation of variables or a direct integration?
HW Helper
P: 6,189
 Quote by Binaryburst If have this equation: $\frac {d^2x}{dt^2}=-\frac{x}{1+x^2}$ How do I solve it?
How about numerical methods, like they are applied here?
(I do not believe there is a nice analytic solution.)

 HW Helper P: 2,264 How to solve a second order diff eq? $$\int \! \dfrac{d^2x}{dt^2}\dfrac{dx}{dt} \, \mathrm{d}t=- \int \! \frac{x}{1+x^2}\dfrac{dx}{dt} \, \mathrm{d}t$$
 P: 24 The solution to this equation should be: $x(t) = sin(t)$ 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?! :(
 P: 24 This is what wolfram alpha says: http://m.wolframalpha.com/input/?i=x...B1%29&x=10&y=2
Mentor
P: 12,037
 Quote by Binaryburst The solution to this equation should be: $x(t) = sin(t)$
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).
 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.
 P: 298 introduce a new variable $p=\frac{dx}{dt}$ by the chain rule we have: $\frac{d^2x}{dt^2}=\frac{dp(x(t))}{dx}=\frac{dp}{dx}\frac{dx}{dt}=\frac{ dp}{dx}p$ This means that $pdp = \frac{-1}{1+x^2}dx$ integrate! $\int pdp = \int \frac{-1}{1+x^2}dx$ $\frac{1}{2}p^2 = -\frac{1}{2}\ln(1+x^2)+C_1$ and because p is the derivative dx/dt, we get $\frac{dx}{\sqrt{-\ln(1+x^2)+C_1}} = 1$ integrate! $\int\frac{1}{\sqrt{\ln(1+x^2)+C_1}}dx = t + C_2$ Well, not a very nice solution because it's implicit.
 HW Helper P: 6,189 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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