I want to draw the integral curves of the differential equation in polar coordinates $(\theta, \rho)$
$\frac{d\rho}{d\theta}= \rho^3-6\rho^2+8\rho$
At first I thought it would suffice to analyse where does the polinomial is zero and where is it negative and positive to obtain where the function $\rho(\theta)$ is monotone, etc.
But since $\theta$ is the free variable here, I'm confused. I feel there's no way I can know for which values of $\theta$, $\rho$ falls in a certain interval.
Can someone help me understand how I should draw the integral curves in the phase space $(\theta, \rho)$ if possible, without solving the equation.
You can do this exactly how you would do it normally. Factor the RHS:
$\frac{d\rho}{d\theta} = (\rho)(\rho - 2)(\rho - 4)$
Which means you get equilibria at $\rho = 0,2,4$, which look like circles of radius 2 and 4, and then the constant curve at 0. Between $0$ and $2$ the RHS is positive, so the integral curves look like spirals which approach the $\rho = 2$ equilibrium circle. Between 2 and 4, the RHS is negative and so the integral curves spiral inward. And then outside $4$, the curves spiral outward again.