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 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)}}

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

 

[JJacquelin]

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

 

[Paalfaal]

Thank you! Got it now :)

 

[HallsofIvy]

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

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

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)}}

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

You have integrated incorrectly.