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

An equilibrium bounds only one side

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

A.There exists an $α \in (0, \infty)$ such that for every $b \in R$ with $b > α$ , the solution y_b is bounded above on I_b
B.There exists an $α \in (0, \infty)$ such that for every $b \in R$ with $b > α$ , the solution y_b is bounded below on I_b✓
C.There exists an $α \in (- \infty, 0)$ such that for every $b \in R$ with $b < α$ , the solution y_b is bounded above on I_b✓
D.There exists an $α \in (- \infty, 0)$ such that for every $b \in R$ with $b < α$ , 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.

See pricing

50 are analysed free — try those first.

Related counterexample: Linearisation determines stability at every equilibrium

An equilibrium bounds only one side

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