Processing math: 100%

Vector field for the system $\dot{x} = 1-x^2+y,\ \dot{y}=bx$ with different value of $b$

Move the slider below to change the value of

$b$ (shown on the slider). The red dot, located at

$(0,-1)$ , is the only fixed point; The green dot shows the movement under the vector field (click your mouse to change the starting point).

b = 0

Vector field for the system ˙x=1−x2+y, ˙y=bx\dot{x} = 1-x^2+y,\ \dot{y}=bx with different value of bb