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Simple characterisitic equation clarification

  1. Oct 5, 2011 #1
    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. jcsd
  3. Oct 5, 2011 #2


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    (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)?
  4. Oct 5, 2011 #3
    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 ??
  5. Oct 6, 2011 #4


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    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.
  6. Oct 6, 2011 #5
    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 ?

  7. Oct 6, 2011 #6
    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
  8. Oct 6, 2011 #7


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    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.
  9. Oct 6, 2011 #8
    yeah... that was a lousy rushed ending on my behalf, sorry about that. and thanks:)
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