Simple characterisitic equation clarification

[ProPatto16]

Homework Statement

use characteristic equation to solve y⁽⁴⁾-y=0

Homework Equations

characteristic equation would be r⁴-1=0

The Attempt at a Solution

my question is related to the number of roots. with r^4 that generally means there will be 4 roots?
but there's only 3? 1,0,-1... my calculator doesn't solve polynomials higher then third order so I am just looking for clarification on this.

would there be 4 roots and just one of them is repeated? so then the solution would be something like y(x)=c₁e^r₁x+c₂e^r₂x+(c₃+c₄x)e^r₃x ...

or are there just 3 roots and its y(x)=c1e^r1x + c2e^r2x + c3e^r3x ??

 

[Dick]

Homework Statement

use characteristic equation to solve y⁽⁴⁾-y=0

Homework Equations

characteristic equation would be r⁴-1=0

The Attempt at a Solution

my question is related to the number of roots. with r^4 that generally means there will be 4 roots?
but there's only 3? 1,0,-1... my calculator doesn't solve polynomials higher then third order so I am just looking for clarification on this.

would there be 4 roots and just one of them is repeated? so then the solution would be something like y(x)=c₁e^r₁x+c₂e^r₂x+(c₃+c₄x)e^r₃x ...

or are there just 3 roots and its y(x)=c1e^r1x + c2e^r2x + c3e^r3x ??

(r^4-1)=(r^2-1)*(r^2+1)=0. -1 and 1 are roots of (r^2-1). 0 isn't a root at all. What are the roots of (r^2+1)?

 

[ProPatto16]

complex. of course. tunnel vision on my behalf.

y(x)= c1e^r1x + c2e^r2x + e^ax(c1cosbx+c2sinbx)

where r1 and r2 are real roots 1 and -1 and c3 and c4 are complex conjugate roots i and -i ??

 

[Dick]

complex. of course. tunnel vision on my behalf.

y(x)= c1e^r1x + c2e^r2x + e^ax(c1cosbx+c2sinbx)

where r1 and r2 are real roots 1 and -1 and c3 and c4 are complex conjugate roots i and -i ??

Start here. A general solution is c1*e^(x)+c2*e^(-x)+c3*e^(ix)+c4*e^(-ix). Since those are the four roots. Where c1, c2, c3 and c4 might all be complex. Now think about how to get a real solution out of that.

 

[ProPatto16]

i don't get it?

if the characteristic equation has unrepeated pair of complex comjugate roots a+-bi then corresponding part of general solution is e^ax(c₁cosbx+c₂sinbx)

so for the two roots that are complex, a=0 and b=1 so then general real solution is e⁰(c₁cosx+c₂sinx)

= c₁cosx+c₂sinx

how is that not a real solution ?

sorry...

 

[ProPatto16]

oh i just realized in the part where you quoted my other reply i said c3 and c4 are complex conjugate roots i and -i... ignore that. thats... i don't know what that is. haha

 

[Dick]

i don't get it?

if the characteristic equation has unrepeated pair of complex comjugate roots a+-bi then corresponding part of general solution is e^ax(c₁cosbx+c₂sinbx)

so for the two roots that are complex, a=0 and b=1 so then general real solution is e⁰(c₁cosx+c₂sinx)

= c₁cosx+c₂sinx

how is that not a real solution ?

sorry...

That's fine. You didn't say what 'a' was and you didn't put it equal to zero. I was confused by your presentation. And yes, saying c3 and c4 were conjugate roots didn't help either.

 

[ProPatto16]

yeah... that was a lousy rushed ending on my behalf, sorry about that. and thanks:)