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Help on Solving this Eq.

  1. Aug 3, 2009 #1
    I hope my posting is not against the rules.
    The equation is not that difficult but I had being out of school for 20 years.

    y"+2y'+4y=0 Then I do
    r^2+2r+4=0
    then you can not do
    (r+2)(r+2)=0 because do not work. can I have some help plz?
     
  2. jcsd
  3. Aug 3, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi jrbigfish! Fishy welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    use good ol' (-b ± √(b2 - 4ac))/2a …

    in this case -1 ± i√3. :wink:
    we need more fish! :biggrin:
     
  4. Aug 4, 2009 #3

    HallsofIvy

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    Most polynomial equations cannot be factored (with integer coefficients). Complete the square or use the quadratic formula.

    Completing the square:
    [itex]r^2+ 2r+ 4= r^2+ 2r+ 1- 1+ 4= r^2+ 2r+ 1+ 3= (r+1)^2+ 3= 0[/itex]

    Quadratic formula:
    [tex]r= \frac{-b\pm\sqrt{b^2- 4ac}}{2a}[/tex]
    with a= 1, b= 2, c= 4.
     
  5. Aug 5, 2009 #4
    Ok the quadratic formula! Yes! But what is this roots of the characteristic equation are
    r1,2=-a+/-(a^2-b)^1/2 what is this and when I use it?
     
  6. Aug 5, 2009 #5

    tiny-tim

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    Do you mean r1,2 = -a ± √(a2 - b) ?

    hmm :rolleyes: … that looks like the roots of x2 +2ax + b = 0.

    I'd forget that formula if I were you.
     
  7. Aug 6, 2009 #6
    Ok thank you mag.
     
  8. Aug 6, 2009 #7
    Ok after I use the quadratic formula (-2+/-((12))^1/2)/2 which solves 0.732i and -2.73i. So the solution is of the form underdamped. So I use the equation y=e^(∝x) (C_1 cosβx+C_2 sinβx). then what?
     
  9. Aug 6, 2009 #8

    tiny-tim

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    (copy the ± and √ symbols)

    Nooo … that's i(-1 ± √3) … you should have got -1 ± i√3. :redface:
     
  10. Aug 7, 2009 #9
    Ok -1 ± i√3 then what I do?
     
  11. Aug 7, 2009 #10

    tiny-tim

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    ?? :confused:

    you solve y' = (-1 ± i√3)y.
     
  12. Aug 7, 2009 #11
    so we end with y=Ce^(1±√3)t ?
     
  13. Aug 7, 2009 #12

    tiny-tim

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    what happened to i? :confused:
     
  14. Aug 10, 2009 #13
    so we end with y=Ce^(1±i√3)t ?? Help me here it has being 12 years after the B.S. is this correct?
     
  15. Aug 10, 2009 #14

    tiny-tim

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    (try using the X2 tag just above the Reply box :wink:)
    Sort of, but you won't get many marks if you write it like that.

    e(1±i√3)t = ete±i√3t,

    so the general solution is … ? :smile:
     
  16. Sep 5, 2009 #15

    tiny-tim

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    Hi jrbigfish! Thanks for the PM. :smile:

    The general solution will be Aetei√3t + Bete-i√3t

    though it would be more usual (and much easier if the solutions are going to be real anyway) to write it in the form Aetcos√3t + Betsin√3t :wink:
     
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