Exercise · ZW-PHASE-PLANE-008 · ★★☆ Intermediate
In an autonomous system with smooth \(f\) and \(g\), can two different trajectories cross at a point that is not an equilibrium?
Choose one option.
A No: through each non-equilibrium point passes exactly one trajectory B Yes, they always end up crossing C Yes, if they traverse the curve in opposite senses D Only if the system is linear Show answer A. No: through each non-equilibrium point passes exactly one trajectory← Back to the question
Explanation The velocity at each point is determined by \((f,g)\) and does not depend on \(t\); by uniqueness, two solutions that meet at a point are the same trajectory traversed with a time shift.
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 .
\[x'=f(x,y),\quad y'=g(x,y)\;\Longrightarrow\;\dfrac{dy}{dx}=\dfrac{g(x,y)}{f(x,y)}\quad(f\neq0)\]
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