# Simple characterisitic equation clarification

1. Oct 5, 2011

### ProPatto16

1. The problem statement, all variables and given/known data

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

2. Relevant equations

characteristic equation would be r4-1=0

3. 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 theres only 3? 1,0,-1.... my calculator doesnt solve polynomials higher then third order so im 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)=c1er1x+c2er2x+(c3+c4x)er3x ...

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

2. Oct 5, 2011

### Dick

(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)?

3. Oct 5, 2011

### 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 ??

4. Oct 6, 2011

### Dick

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.

5. Oct 6, 2011

### ProPatto16

i dont get it?

if the characteristic equation has unrepeated pair of complex comjugate roots a+-bi then corresponding part of general solution is eax(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...

6. Oct 6, 2011

### ProPatto16

oh i just realised 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 dont know what that is. haha

7. Oct 6, 2011

### Dick

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.

8. Oct 6, 2011

### ProPatto16

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