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Part CCSIR NET December 2024an-equilibrium-bounds-only-one-side

An equilibrium bounds only one side

For b∈Rb \in \mathbb{R}, let y_b = y_b(x) be the unique solution of the initial value problem dy/dx =y5+y4+y3+y2+y+1,y(0)=b= y^{5} + y^{4} + y^{3} + y^{2} + y + 1, y(0) = b defined on its maximal interval of existence I_b. Then which of the following statements are true?

  1. A.There exists an α∈(0,∞)\alpha \in (0, \infty) such that for every b∈Rb \in \mathbb{R} with b>αb > \alpha, the solution y_b is bounded above on I_b
  2. B.There exists an α∈(0,∞)\alpha \in (0, \infty) such that for every b∈Rb \in \mathbb{R} with b>αb > \alpha, the solution y_b is bounded below on I_b✓ Correct
  3. C.There exists an α∈(−∞,0)\alpha \in (-\infty, 0) such that for every b∈Rb \in \mathbb{R} with b<αb < \alpha, the solution y_b is bounded above on I_b✓ Correct
  4. D.There exists an α∈(−∞,0)\alpha \in (-\infty, 0) such that for every b∈Rb \in \mathbb{R} with b<αb < \alpha, the solution y_b is bounded below on I_b

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests boundary and endpoint.

See pricing

50 are analysed free — try those first.

The trap it tests

Boundary and endpoint

The statement turns at the edge of the interval, the domain, or the parameter range.

Drill statements like this

Related counterexample: Linearisation determines stability at every equilibrium

More on this topic

The chapter behind this: Stability and phase portraits — free to read

From Ordinary Differential Equations › Stability and phase portraits

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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