y'' + by' + 4y = 0; what effect does changing b have?

[BlazingFire]

Here's the full text of the question:

To see the effect of changing the parameter b in the initial value problem:

\(\displaystyle y'' + by' + 4y = 0; y(0) = 1, y'(0) = 0\)

Solve for \(\displaystyle b = 5\), \(\displaystyle b = 4\) and \(\displaystyle b = 2\) and sketch the solutions.

So basically, in short, we have the following IVPs:

1. \(\displaystyle y'' + 5y' + 4y = 0; y(0) = 1, y'(0) = 0\)
2. \(\displaystyle y'' + 4y' + 4y = 0; y(0) = 1, y'(0) = 0\)
3. \(\displaystyle y'' + 2y' + 4y = 0; y(0) = 1, y'(0) = 0\)

I already worked out all of them. For #1, I got \(\displaystyle y = -\frac{1}{3}e^{4x} + \frac{4}{3}e^{-x}\). For #2, I wound up with \(\displaystyle y = e^{-2x} + 2xe^{-2x}\). And lastly, for #3, I got \(\displaystyle y = e^{-x}cos(\sqrt{3}x) + \frac{\sqrt{3}}{3}e^{-x}sin(\sqrt{3}x)\). I graphed all of these functions on my calculator to look for any similarities (I swear I didn't make any mistakes in doing so; I was extremely careful with the notation), but couldn't find any.

My question is this: what effect does changing the coefficient of the first derivative have on the solution of a second-order, linear, homogeneous ODE with constant coefficients, especially in my particular case?

 

[Deleted member 4993]

Here's the full text of the question:

To see the effect of changing the parameter b in the initial value problem:

\(\displaystyle y'' + by' + 4y = 0; y(0) = 1, y'(0) = 0\)

Solve for \(\displaystyle b = 5\), \(\displaystyle b = 4\) and \(\displaystyle b = 2\) and sketch the solutions.

So basically, in short, we have the following IVPs:

1. \(\displaystyle y'' + 5y' + 4y = 0; y(0) = 1, y'(0) = 0\)
2. \(\displaystyle y'' + 4y' + 4y = 0; y(0) = 1, y'(0) = 0\)
3. \(\displaystyle y'' + 2y' + 4y = 0; y(0) = 1, y'(0) = 0\)

I already worked out all of them. For #1, I got \(\displaystyle y = -\frac{1}{3}e^{4x} + \frac{4}{3}e^{-x}\). For #2, I wound up with \(\displaystyle y = e^{-2x} + 2xe^{-2x}\). And lastly, for #3, I got \(\displaystyle y = e^{-x}cos(\sqrt{3}x) + \frac{\sqrt{3}}{3}e^{-x}sin(\sqrt{3}x)\). I graphed all of these functions on my calculator to look for any similarities (I swear I didn't make any mistakes in doing so; I was extremely careful with the notation), but couldn't find any.

My question is this: what effect does changing the coefficient of the first derivative have on the solution of a second-order, linear, homogeneous ODE with constant coefficients, especially in my particular case?

You have plotted the solutions - and you don't see the effect!!!!

In physics, this is the equation for "damped" vibrating system (x is replaced by t - time).

By changing the damping coefficient - you are changing the nature of vibration.

If you change b to 1 - what change do you see?