This is an example in my book. It is for the following system:
\begin{align*} x'&=y+x(1-x^2-y^2)\\ y'&=-x+y(1-x^2-y^2) \end{align*}
So using polar coordinates we get the following system \begin{align*} r'&=r(1-r^2)\\ \theta'&=-1, \end{align*} and the solutions are $$r(t)=(1+ce^{-2t})^{\frac{-1}{2}}, \,\,\,\,\,\,\theta(t)=-(t-\alpha).$$
So my question is how did they solve $r'$. It's been a while since I've take differential equations and I need help how to solve it.
So $r'=r(1-r^2)$ which means $$\int\frac{dr}{r(1-r^2)}=\int dt$$ Using partial fraction on LHS, we obtain, $$\int \frac{1}{r}-\frac{1}{2(r+1)}-\frac{1}{2(r-1)} dr = t+c$$ $$\ln r - \frac{1}{2}\ln(r+1)- \frac{1}{2}\ln(r-1)=t+c$$ $$\ln\left(\frac{r}{\sqrt{r^2-1}}\right)=t+c$$ $$\frac{r}{\sqrt{r^2-1}}=Ae^t$$ where $A=e^c$ $$\frac{r^2}{{r^2-1}}=Be^{2t}$$ where $B=A^2$ $$1-\frac{1}{r^2}=Be^{-2t}$$
so $$r =(1-Be^{-2t})^{\frac{-1}{2}}$$