Find explicit solution of the differential equation $\dot{r} = r(1-r^{2})$

Question

We have $$ \dot{r} = r(1-r^{2}) $$ and I try to derive the explicit solution. The solution is already given; $$ r(t,r_{0}) = \frac{ r_{0} } { \sqrt{r_{0}^{2} + (1-r_{0}^{2})e^{-2t} } }, $$ but I can not see why.

Normally I would use the seperation of variables and hence write $$ \frac{dr}{dt} = r(1-r^{2})) \Rightarrow \int\frac{1}{r-r^{3}}dr = \int dt, $$ but I do not know how to proceed.

Any suggestions are more than welcome.

Answer 1

Under the assumption that $r(t_o)\notin\{0,1\}$ within some interval you can rewrite the equation as $$\frac{\dot{r}}{r(1-r^2)}=1$$ Then integrate from $t_0$ to say $t$. I'll write more details if you still need help. On the left hand side you may need to decompose the expression as partial fractions.

$$ \int^t_{t_0}\frac{r'(s)}{r(s)(1-r^2(s))}\,dt = t- t_0 $$ Substitution $u=r(t)$ gives $$ \int^{r(t)}_{r(t_0)}\frac{du}{u(1-u^2)}=t-t_0 $$

Find $A$, $B$, $C$ such that

$\frac{A}{u}+\frac{B}{1-u}+\frac{C}{1+u}=\frac{1}{u(1-u)(1+u)}$

Answer 2

$$\dot{r} = r(1-r^{2})$$ $$\int_{r_0}^r \dfrac {ds} {s(1-s^{2})}=\int_{0}^t dt$$ $$\int_{r_0}^r \dfrac {2sds} {s^2(1-s^{2})}=2\int_{0}^t dt$$ Note that $ds^2=2sds$: $$\int_{r_0}^r\dfrac {ds^2} {s^2(1-s^{2})}=2t$$ Decompose with fraction method: $$\int_{r_0}^r\dfrac {ds^2} {(1-s^{2})}+\int_{r_0}^r \dfrac {ds^2} {s^2}=2 t$$ Evaluate the integrals. $$\ln r^2-\ln |r_0^2|-\ln|r^2-1|+\ln |r_0^2-1|=2t$$ $$\ln \left | \dfrac {r^2}{r^2-1} \right |= \ln \left | \dfrac {r_0^2}{r_0^2-1} \right | +\ln (e^{ 2t})$$ $$ \dfrac {r^2}{r^2-1} = \dfrac {r_0^2e^{ 2t}}{r_0^2-1} $$ You can surely finish.