Differential equation mass spring oscillator

[max]

I need help explaining what's happening instead of solving this problem.
The question is "Provide a mass spring oscillator analogy to determine what happens to the differential equation solutions as t goes to infinity of
y''+2y'/(t+1)+4y= 0."

I know that my''+by'+ky= external force, where m =mass, b=damping, and k=spring stiffness.
For this problem, spring stiffness is 4 and solutions will oscillate at a frequency that will get a little faster with amplitude decreasing over time because the damping approach zero as the lim as t goes to infinity of 2/(t+1) goes to zero which means there would be less and less friction involved until there is no more friction.

Did I explain this correctly?
I picture it as oscillating with amplitude decreasing until there is a straight line.

 

[Deleted member 4993]

I need help explaining what's happening instead of solving this problem.
The question is "Provide a mass spring oscillator analogy to determine what happens to the differential equation solutions as t goes to infinity of
y''+2y'/(t+1)+4y= 0."

I know that my''+by'+ky= external force, where m =mass, b=damping, and k=spring stiffness.
For this problem, spring stiffness is 4 and solutions will oscillate at a frequency that will get a little faster with amplitude decreasing over time because the damping approach zero as the lim as t goes to infinity of 2/(t+1) goes to zero which means there would be less and less friction involved until there is no more friction.

Did I explain this correctly?
I picture it as oscillating with amplitude decreasing until there is a straight line.

I think you need to explain this into two regions - over-damped and under-damped regions.

For the under damped case - the dampening factor will decrease with time and the oscillating-frequency will ramp upto the natural frequency and will be in steady state at that frequency and amplitude (as you have guessed).

For the overdamped case (w = imaginary), I cannot be sure about the steady state solution without looking at the actual solution of the DE. Intuitively, I think the dampening factor will reduce to "critical dampening factor" at steady state (w² = 0) and will remain there.

 

[max]

I don't understand your overdamped case. Are you letting t go to negative infinity?
I can't picture it.