Consider the equation
𝑑
𝑦
𝑑
𝑡
=
𝑦
2
(
25
−
𝑦
2
)
. Which of the following are true?

(1)
𝑦
=
0
is an (asymptotically) unstable equilibrium point.

(2)
𝑦
=
0
is a (asymptotically) stable equilibrium point.

(3)
𝑦
=
0
is a (asymptotically) semistable equilibrium point.

(4)
𝑦
=
−
5
is an (asymptotically) unstable equilibrium point.

(5)
𝑦
=
5
is a (asymptotically) stable equilibrium point.

Question

Consider the equation 
𝑑
𝑦
𝑑
𝑡
=
𝑦
2
(
25
−
𝑦
2
)
.   Which of the following are true?

(1) 
𝑦
=
0
  is an (asymptotically) unstable equilibrium point.

(2)  
𝑦
=
0
  is a (asymptotically) stable equilibrium point.

(3) 
𝑦
=
0
 is a (asymptotically) semistable equilibrium point.

(4) 
𝑦
=
−
5
 is an (asymptotically) unstable equilibrium point.

(5) 
𝑦
=
5
  is a (asymptotically) stable equilibrium point.

BlackTom AI · Accepted Answer

We begin by identifying the equilibria of dy/dt = y^2(25 − y^2). Setting the right-hand side to zero gives y = 0 and y = ±5 as equilibrium points. Option (1): y = 0 as an asymptotically unstable equilibrium. To judge stability, consider small perturbations around 0. For small y ≠ 0, dy/dt ≈ 25 y^2 > 0, so any nonzero y near 0 tends to increase. If y > 0, it moves away from 0 toward larger positive values; if y < 0, dy/dt is still positive and y increases toward 0, but does not cross and continue to stay near or move away depending on the trajectory. The net behavior indicates that zero is not asymptotically unstable in the typical sense, and in fact the analysis shows a semi-stable behavior rather than simple asymptotic instability. Therefore this statement is not correct. Option (2): y = 0 as a (asymptotically) stable equilibrium. Since small positive y grows (dy/dt > 0) and tends to move away from 0, whereas small negative y increases toward 0 but does not settle there for all directions, the equilibrium is not asymptotically stable from all perturbations. Hence this statement is false. Option (3): y = 0 is a semistable equilibrium point. As discussed, for tiny positive perturbations, the solution moves away from 0 toward the positive side (toward +5), whereas for tiny negative perturbations, the solution increases toward 0 but from the negative side it tends to approach 0 in the short term and then moves away in the long term depending on the trajectory. This asymmetric behavior characterizes semistability. Thus this statement is true. Option (4): y = −5 is an unstable equilibrium point. Evaluate stability via f'(y) at the equilibrium: f(y) = y^2(25 − y^2); f'(y) = 2y(25 − y^2) − 2y^3 = 50y − 4y^3. At y = −5, f'(−5) = 50(−5) − 4(−125) = −250 + 500 = 250 > 0, indicating instability. Therefore this statement is true. Option (5): y = 5 is a stable equilibrium point. Using the same derivative, f'(5) = 50(5) − 4(125) = 250 − 500 = −250 < 0, which corresponds to asymptotic stability. Hence this statement is true. In summary, the true statements are those asserting semistability at 0 and instability at −5 and stability at 5, i.e., options (3), (4), and (5) are correct. The remaining statements (1) and (2) misclassify the nature of the equilibrium at 0.

2251-1_15MATH2073005 Quiz 2 (2025SS)

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