Question

In Problems 1-3, find the general solution, and describe the long-term behavior of y(t) : exponential growth, exponential decay, simple oscillations, damped oscillations, or growing oscillations. Hint: if there are multiple terms in the solution, the qualitative behavior is determined by the largest term, i.e. the term that wins out for large t. 1. y′′+y′−6y=0. 2. y′′−4y′+4y=0. 3. y′′−y′+y=0. 4. Consider a mass-spring system with mass m=1, spring constant k=2, and damping constant b=2. Set up the equation of motion, and find the general solution y(t). Describe the long-time behavior as in problems 1-3. Is this overdamping, critical damping, or underdamping?

Answer 1

The differential equations provided can represent exponential growth, exponential decay, simple oscillations, damped oscillations, or growing oscillations. The solutions and long-term behaviors are summarized as follows: 1. exponential growth and exponential decay; 2. simple oscillations; 3. simple oscillations; 4. critically damped.

Step-by-step explanation:

To find the general solution for each of the given differential equations and determine their long-term behavior, let's analyze each equation step by step:

1. y'' + y' - 6y = 0: This equation can be factored as (y - 2)(y + 3) = 0, which gives us two solutions y = 2 and y = -3. Since the largest term in the solution is e^rt, where the growth or decay rate r is determined by the coefficient of y', this equation represents exponential growth (y = Ce^3t) and exponential decay (y = De^(-2t)).
2. y'' - 4y' + 4y = 0: This equation can be factored as (y - 2)^2 = 0, which gives us a repeated root y = 2. Since there is no term with e^rt in the solution, this equation represents simple oscillations (y = (A + Bt)e^(2t)).
3. y'' - y' + y = 0: This equation has a characteristic equation of r^2 - r + 1 = 0, which gives us complex roots r = (1 ± i√3)/2. Since there is no real term in the solution, this equation represents simple oscillations (y = e^(t/2)(Ccos(√3t/2) + Dsin(√3t/2))).
4. Mass-spring system with m=1, k=2, and b=2: The equation of motion for this system is y'' + 2y' + 2y = 0. The damping constant b is equal to √4mk, which means the system is critically damped. In this case, the system does not oscillate, but asymptotically approaches the equilibrium condition as quickly as possible (y = (A + Bt)e^(-t)).