# Trying to get exact solution for a nonlinear DE

1. May 30, 2012

### Paalfaal

Ok, this should be quite simple. I've been looking at this problem for quite some time now, and I'm tired.. Please help me!

The equation to solve is $r'$ = $r(1-r^2)$

The obvious thing to do is to do partial fraction expansion and integrate(from r$_{0}$ to r):

$∫ (\frac{1}{r} + \frac{1}{1-r^2})dr = t$

After some trig substitution: $ln[r\sqrt{\frac{r+1}{1-r}}] = t$ (evaluated from r$_{0}$ to r). Then take the exponential on both sides.

$r\sqrt{\frac{r+1}{1-r}} = r_{0}\sqrt{\frac{r_{0}+1}{1-r_{0}}}e^{t}$

This leads to kind of a nasty expression which can't be solved explicitly for $r(t)$ (at least it seems that it can't be solved explicitly).. So this is where I'm stuck.

The solution to this ploblem is: $r(t) = \frac{r_{0}e^{t}}{\sqrt{1+r_{0}(e^{2t}-1)}}$

2. May 30, 2012

### JJacquelin

$∫ (\frac{1}{r} + \frac{1}{1-r^2})dr = t$ is false. r is missing on the second fraction.

3. May 31, 2012

### Paalfaal

Thank you! Got it now :)

4. May 31, 2012

### HallsofIvy

That's an incorrect "partial fraction expansion". $1- x^2= (1- x)(1+ x)$ so
$$\frac{1}{r(1- r^2)}= \frac{1}{r}- \frac{1}{r- 1}- \frac{1}{r+ 1}$$
There is no "trig substitution" involved in integrating that. the integral is
$$ln(r)- ln(r-1)- ln(r+1)+ C= ln\left(\frac{r}{r^2-1}\right)+ C$$

You have integrated incorrectly.