How to solve a second order diff eq?

[Binaryburst]

If have this equation:

\frac {d^2x}{dt^2}=-\frac{x}{1+x^2}

How do I solve it?

 

[mfb]

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

 

[I like Serena]

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.)

 

[lurflurf]

$$\int \! \dfrac{d^2x}{dt^2}\dfrac{dx}{dt} \, \mathrm{d}t=-
\int \! \frac{x}{1+x^2}\dfrac{dx}{dt} \, \mathrm{d}t$$

 

[Binaryburst]

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?! :(

 

[Binaryburst]

This is what wolfram alpha says: http://m.wolframalpha.com/input/?i=x%27%27%28t%29%3D-x%2F%28x%5E2%2B1%29&x=10&y=2

 

[mfb]

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).

 

[Binaryburst]

@ 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. https://www.physicsforums.com/showthread.php?t=668751&goto=newpost It has to be done without conservation of energy. (Just using forces) Thanks in advance.

 

[bigfooted]

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.

 

[I like Serena]

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!