Exercise · ZW-PHASE-PLANE-003 · ★☆☆ Basic
What are the equilibria of \(x'=x(1-y),\ y'=y(x-1)\)?
Choose one option.
Show answer
- A. \((0,0)\) and \((1,1)\)
Explanation
If \(x=0\), the second equation gives \(y'=-y=0\), so \(y=0\). If \(y=1\), the second requires \(x=1\). There are two: \((0,0)\) and \((1,1)\); at \((1,0)\) we have \(x'=1\neq0\).
Review the concept: Phase plane
Coordinate plane \((x,y)\) of an autonomous system whose velocities \(f\) and \(g\) do not depend on \(t\); each solution is drawn there as a curve, its trajectory.
\[x'=f(x,y),\quad y'=g(x,y)\;\Longrightarrow\;\dfrac{dy}{dx}=\dfrac{g(x,y)}{f(x,y)}\quad(f\neq0)\]
More practice problems: phase plane
- ZW-PHASE-PLANE-001 For x'=x+y,\ y'=x-y, what is the slope dy/dx of the trajectory through (2,1)?
- ZW-PHASE-PLANE-002 For x'=x-2y,\ y'=3x+y, what is the velocity vector (x',y') at the point (2,1)?
- ZW-PHASE-PLANE-004 The trajectories of x'=2x,\ y'=y are the curves:
- ZW-PHASE-PLANE-005 In the system x'=y,\ y'=-x, which expression stays constant along each trajectory?
- ZW-PHASE-PLANE-006 For x'=y,\ y'=x, which point lies on the same trajectory as (1,0)?
- ZW-PHASE-PLANE-007 For x'=-x,\ y'=-2y, what happens as t→∞ to the solution starting at (1,1)?
- ZW-PHASE-PLANE-008 In an autonomous system with smooth f and g, can two different trajectories cross at a point that i…
- ZW-PHASE-PLANE-009 For x'=y,\ y'=-4x, which of these points lies on the trajectory through (1,0)?