Answer:
overdamped = (-∞,√20) u (√20,∞)
underdamped = (-√20,√20)
critically damped = [√20,√20] u [-√20,-√20]
Step-by-step explanation:
Hi!
In order to answer this question we must first write the charcteristic equation, which is obtained replacing a new variable, lets say ω by the derivatives of s, and putting it to the power of the same order of the derivative, e.g
s''' -> ω^3
The characteristic equation is:
ω^2 + bω + 5 = 0
We can now solve for ω using the quadratic formula:
[tex]\omega =\frac{-b \pm \sqrt(b^2 - 20)}{2}[/tex]
If b=0, the solution for ω will be a pure complex number, and that means an undamped oscialtor.
The critically damped solution is given when the determinant of ω is zero, that is
b^2 = 20
|b| = √20
The underdamped is given when we have a non-zero imaginary part for ω
that is
b^2<20
|b| < √20
The overdamped is given when the determinant is positive:
b^2>20
|b| > √20
The solution is critically damped at [tex]|b| = \sqrt{20[/tex], the solution is underdamped at [tex]|b| < \sqrt{20[/tex], and the solution is overdamped at [tex]|b| > \sqrt{20[/tex]
The differential equation is given as:
[tex]s"+bs'+5s=0[/tex]
Rewrite the equation as follows:
[tex]x^2+bx+5s=0[/tex]
Where:
x represents s
Using the quadratic formula, we have the value of x to be:
[tex]x = \frac{-b \pm \sqrt{b^2 - 4 * 1 * 5}}{2 * 1}[/tex]
So, we have:
[tex]x = \frac{-b \pm \sqrt{b^2 - 20}}{2}[/tex]
In the above equation, the determinant (d) is:
[tex]d = b^2 - 20[/tex]
When the solution is overdamped, then:
d > 0
So, we have:
[tex]b^2 - 20 > 0[/tex]
Add 20 to both sides
[tex]b^2 > 20[/tex]
Take the square root of both sides
[tex]|b| > \sqrt{20[/tex]
When the solution is underdamped, we have:
d < 0
So, we have:
[tex]|b| < \sqrt{20[/tex]
When the solution is critically damped, we have:
[tex]|b| = \sqrt{20[/tex]
Read more about differential equations at:
https://brainly.com/question/18760518
