Question

The critical point of the differential equation \[ \frac{d^2y}{dt^2} + 2 \alpha \frac{dy}{dt} + \beta^2 y = 0, \alpha > 0, \, \beta > 0, \] is a

• A. node and is asymptotically stable
• B. spiral point and is asymptotically stable
• C. node and is unstable
• D. saddle point and is unstable

Hint:

For second-order linear differential equations with real and distinct roots, the system exhibits a node, and the stability depends on the sign of the roots. If both roots are negative, the node is asymptotically stable.

Answer 1

The Correct Option is A

The given differential equation is a second-order linear differential equation with constant coefficients. To analyze the stability, we first find the characteristic equation of the differential equation: \[ r^2 + 2 \alpha r + \beta^2 = 0. \] The roots of this quadratic equation are given by: \[ r = \frac{-2 \alpha \pm \sqrt{(2 \alpha)^2 - 4 \beta^2}}{2}. \] Simplifying this: \[ r = -\alpha \pm \sqrt{\alpha^2 - \beta^2}. \] Since \( \alpha > 0 \) and \( \beta > 0 \), the discriminant \( \alpha^2 - \beta^2 \) is positive, meaning the roots are real and distinct. Therefore, the critical point is a node. Since the roots are negative, the solution decays to zero as \( t \to \infty \), which indicates that the node is asymptotically stable. Final Answer: (A) node and is asymptotically stable