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!(adsbygoogle = window.adsbygoogle || []).push({});

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!

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# Trying to get exact solution for a nonlinear DE

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