Exercise · ZW-PHASE-PLANE-002 · ★☆☆ Basic
For \(x'=x-2y,\ y'=3x+y\), what is the velocity vector \((x',y')\) at the point \((2,1)\)?
Choose one option.
Show answer
- B. \((0,7)\)
Explanation
\(x'=2-2\cdot1=0\) and \(y'=3\cdot2+1=7\). The point moves vertically, upward.
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
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- ZW-PHASE-PLANE-003 What are the equilibria of x'=x(1-y),\ y'=y(x-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)?