Questions: Autonomous Equations and Equilibrium Solutions

Question 1

For the autonomous equation dy/dt = y(1 − y), which of the equilibrium solutions is stable?

• A. y = 0 is stable — it is the trivial solution and autonomous systems return to zero
• B. y = 1 is stable — solutions above it are pushed down and solutions below it are pushed up
• C. Both y = 0 and y = 1 are stable — all equilibria of autonomous equations are stable
• D. Neither is stable — stability cannot be determined without solving the ODE

Answer: (B)

Evaluate f(y) = y(1−y) on either side of each equilibrium. Near y = 1: for y slightly below 1 (say y = 0.9), f(0.9) = 0.9(0.1) > 0, so solutions increase toward y = 1. For y slightly above 1 (say y = 1.1), f(1.1) = 1.1(−0.1) < 0, so solutions decrease toward y = 1. Both sides push toward y = 1 — it is stable. Near y = 0: for y slightly below 0, f(y) < 0 (solutions move away); for y slightly above 0, f(y) > 0 (solutions move away). Both sides push away from y = 0 — it is unstable.

Question 2

For dy/dt = f(y), suppose f(y) > 0 for 0 < y < 2 and f(y) < 0 for y > 2, with f(2) = 0. What can you conclude?

• A. y = 2 is an unstable equilibrium — the sign of f(y) changes there
• B. y = 2 is a stable equilibrium — solutions below it increase toward it and solutions above it decrease toward it
• C. y = 2 is a semi-stable equilibrium — it attracts from above but repels from below
• D. Nothing can be concluded without solving the ODE explicitly

Answer: (B)

The phase-line analysis is direct: f(y) > 0 below y = 2 means solutions there are increasing (arrows point up, toward y = 2). f(y) < 0 above y = 2 means solutions there are decreasing (arrows point down, toward y = 2). Both sides funnel toward y = 2 — it is a stable equilibrium (a sink/attractor). This determination required no integration, only evaluating the sign of f(y) on either side of the equilibrium.

Question 3

An equilibrium solution y = c of an autonomous ODE is typically stable.

• T. True
• F. False

Answer: False

False. Equilibria can be stable (attractors/sinks), unstable (repellers/sources), or semi-stable. For dy/dt = y(1−y), y = 1 is stable but y = 0 is unstable — solutions near y = 0 move away from it. Stability depends on the sign of f(y) on either side of the equilibrium: if f changes from positive to negative through the equilibrium, it is stable; if from negative to positive, unstable. Nothing about being an equilibrium (f(c) = 0) guarantees attraction.

Question 4

The long-run behavior of any solution to an autonomous ODE dy/dt = f(y) can be determined from the phase line without explicitly solving the equation.

• T. True
• F. False

Answer: True

True. The phase line encodes the sign of f(y) on each interval between equilibria and at each equilibrium. This is sufficient to determine where every solution goes as t → ∞: solutions follow the arrows on the phase line — moving in the direction f(y) points — until they reach a stable equilibrium or diverge to ±∞. No integration is required because the qualitative structure (which equilibria exist and which are stable) completely determines long-run behavior from any initial condition.

Question 5

Why does an autonomous equation dy/dx = f(y) — where only y appears on the right, not x — make it especially amenable to qualitative (phase-line) analysis?

Model answer: Because the rate of change depends only on the current value of y, the behavior is the same regardless of when or where you start. The phase line captures all possible behaviors: at each value of y, f(y) tells you whether solutions increase or decrease, and equilibria (where f(y) = 0) are fixed points that cannot be crossed. Since f(y) is fixed across all x, the arrows on the phase line never change — you can trace every solution's long-run fate without solving the equation.

In contrast, a non-autonomous equation dy/dx = g(x, y) has a rate of change that varies with x, so the same y-value produces different behavior depending on where you are in x. A phase line would need to change as x changes, making qualitative analysis far harder. Autonomy removes this complication: the system's 'rules' are the same at every x, making the phase line a complete and time-independent map of all solution behaviors.