Question

Match List-I with List-II. List-I:
A. Node,
B. Center,
C. Saddle point,
D. Spiral when eigenvalues are complex with negative real parts. List-II: I. Stable, II. Unstable, III. Asymptotic stable, IV. Asymptotic stable or unstable.

• A. A-IV, B-I, C-II, D-III
• B. A-III, B-I, C-II, D-IV
• C. A-II, B-III, C-IV, D-I
• D. A-IV, B-II, C-I, D-III

Hint:

Negative real parts of eigenvalues indicate asymptotic stability.

Answer 1

The Correct Option is A

Concept:
Stability of equilibrium points depends on the nature of eigenvalues of the linearized system.

Step 1: Node.
A node may be stable or unstable depending on whether the eigenvalues are negative or positive. \[ A\rightarrow IV \]

Step 2: Center.
A center gives closed trajectories around the equilibrium point. It is stable but not asymptotically stable. \[ B\rightarrow I \]

Step 3: Saddle point.
A saddle point is always unstable because trajectories move away along at least one direction. \[ C\rightarrow II \]

Step 4: Spiral with complex eigenvalues having negative real parts.
If eigenvalues are complex and their real parts are negative, trajectories spiral towards the equilibrium point. \[ D\rightarrow III \] Therefore: \[ A-IV,\ B-I,\ C-II,\ D-III \] \[ \therefore \text{Correct Answer is (A)} \]