time traveled between two points on phase portrait

Question

Given the dynamical system:

$\dot x = 1-x^2-2y^2-xy$

$\dot y = 2x^2+y^2+xy-1$

If the system starts at (1,0), show that it goes to(0,1) and find the time t at this point.

Attempt:

The Phase portrait of the system with x and y nullclines will show a trajectory (blue color) resembling a unit circle (solution with initial condition (1,0):

phase portrait

To find the time, is it plausible to say that the solution with the given initial condition of the system is $x^2+y^2=1$? How can I find the time traveled since the velocity is changing along this trajectory (i.e. not uniform circular motion)?

Answer 1

Attempt:

using $dy/dt = 2x^2+y^2+xy−1$

$ dt = dy/(2x^2+y^2+xy−1)$

$ \int_0^T(dt) = \int_0^1 {dy \over (2x^2+y^2+xy−1)}$

Substituting $x = \sqrt {1-y^2}$

$T = \int_0^1 {dy \over (2(1-y^2)+y^2+ \sqrt {1-y^2}y−1)}$

Using a program to evaluate integral gives T = 1.25