# How to solve a second order diff eq?

1. Feb 13, 2013

### Binaryburst

If have this equation:

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

How do I solve it?

2. Feb 13, 2013

### Staff: Mentor

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

3. Feb 13, 2013

### I like Serena

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

4. Feb 13, 2013

### lurflurf

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

5. Feb 14, 2013

### 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?! :(

6. Feb 14, 2013

### Binaryburst

7. Feb 14, 2013

### Staff: Mentor

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).

8. Feb 14, 2013

### 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.

9. Feb 14, 2013

### 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.

10. Feb 14, 2013

### 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!

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