Trying to get exact solution for a nonlinear DE

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 [itex]r'[/itex] = [itex]r(1-r^2)[/itex]

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

[itex] ∫ (\frac{1}{r} + \frac{1}{1-r^2})dr = t[/itex]

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

[itex]r\sqrt{\frac{r+1}{1-r}} = r_{0}\sqrt{\frac{r_{0}+1}{1-r_{0}}}e^{t} [/itex]

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

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

Again, this should be fairly simple, but I'm stuck.. Please help me!

JJacquelin

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

Paalfaal

Thank you! Got it now :)

HallsofIvy

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

You have integrated incorrectly.