Finding the roots of a high degree polynomial equation

[1MileCrash]

Homework Statement

y(6) - 3y(4) + 3y''-y = 0

Homework Equations

The Attempt at a Solution

The characteristic equation of that differential equation is:

r^6 - 3r^4 + 3r^2 - r = 0

But how am I expected to solve such a high degree polynomial (and thus the DE?)

 

[Bacle2]

First, notice you can rewrite you char. equation as:

r(r^5-3r^3+3r-1)=0

Then notice too, that, inside your parenthesis: 1-3+3-1=0. What does this tell you?

 

[tiny-tim]

Hi 1MileCrash!

(try using the X² button just above the Reply box )

r^6 - 3r^4 + 3r^2 - r = 0

nooo …

r⁶ - 3r⁴ + 3r² - 1 = 0 ​

 

[SammyS]

Homework Statement

y⁽⁶⁾ - 3y⁽⁴⁾ + 3y''-y = 0

Homework Equations

The Attempt at a Solution

The characteristic equation of that differential equation is:

r^6 - 3r^4 + 3r^2 - r = 0

But how am I expected to solve such a high degree polynomial (and thus the DE?)

Following tiny-tim's corrected characteristic equation:

r⁶ - 3r⁴ + 3r² - 1 = 0​

Expand (a - b)³ .

 

[epenguin]

I don't know at what point certain things are supposed to be familiar but I would say pretty early there is a rather familiar pattern to be discerned in that last formula.

 

[HallsofIvy]

r^6- 3r^4+ 3r^2- 1 has only even powers of r. Let x= r^2 and that becomes x^3- 3x^2+ 3x- 1. And, as SammyS suggests, that is (x- 1)^3.

 

[Bacle2]

Worse comes to worse and you cannot see this pattern, it always makes sense to try for the simple roots,like 0,1 and -1 . Checking for 1 as a root comes down to adding the coefficients and seeing if the sum is zero; similar for -1.

 

[HallsofIvy]

Let me point out that Bacle2 is not just choosing "simple roots" at random. Since the leading coefficient is 1 and the constant term is 1, by the "rational root theorem" the only possible rational number roots are 1 and -1. (I don't know why he mentions "0".)

 

[Bacle2]

Let me point out that Bacle2 is not just choosing "simple roots" at random. Since the leading coefficient is 1 and the constant term is 1, by the "rational root theorem" the only possible rational number roots are 1 and -1. (I don't know why he mentions "0".)

Right, my bad. I thought the characteristic equation had no constant term.

 

[LCKurtz]

All this help and the OP is nowhere in sight.