Results 1 to 9 of 9

Math Help - Roots of this auxiliary eqn?

  1. #1
    Member
    Joined
    Apr 2009
    Posts
    85

    Roots of this auxiliary eqn?

    I'm trying to find the auxiliary roots of this DE

    y'''+iy=0
    So the auxiliary eqn is
    m^3+i=0
    and m=i ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    Use De Moivre's formula to find the three roots.

    EDIT:  -i = e^{-i \pi /2} = \cos(-\pi/2) + i \sin(-\pi/2)

     (-i)^{1/3} = e^{i(-\pi/6 + 2k \pi/3)} = \cos (-\pi/6 + 2k \pi /3) + i \sin(-\pi/6 + 2k \pi /3) for k = 0, 1, and 2

    When k=0,  (-i)^{1/3} = \cos(-\pi/6) + i \sin(-\pi/6) = \frac{\sqrt{3}}{2} - \ \frac{i}{2}

    when k =1,   (-1)^{1/3} = \cos(\pi/2) + i \sin(\pi/2) = i

    and for the k=2,  (-1)^{1/3} = -\frac{\sqrt{3}}{2} - \ \frac{i}{2} \

    The complex roots are not conjugate pairs like in the other problem, so the general solution is  y(x) = C_{1}e^{(\frac{\sqrt{3}}{2} - \ \frac{i}{2})x} + C_{2}e^{ix} + C_{3}e^{(-\frac{\sqrt{3}}{2} - \ \frac{i}{2} )x}


    BTW, Wolfram Alpha (the web site associated with Mathematica) can solve differential equations like these.
    Last edited by Random Variable; March 21st 2010 at 01:36 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Apr 2009
    Posts
    85
    Ah that formula is new to me. Just to be clear the -i = e^{-\pi /2} Comes from Euler's formula and \cos(-\pi/2) + i \sin(-\pi/2) comes from DeMoivre...is it \frac{\pi}{2} because we're dealing with just i?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    Quote Originally Posted by mmattson07 View Post
    Ah that formula is new to me. Just to be clear the -i = e^{-\pi /2} Comes from Euler's formula and \cos(-\pi/2) + i \sin(-\pi/2) comes from DeMoivre...is it \frac{\pi}{2} because we're dealing with just i?

    It should be  -i=e^{-i\pi/2} = \cos (-\pi/2) + i \sin (-\pi /2)

    Then  (-i)^{1/3} = e^{i (-\pi/6 + 2k \pi/3)}

    But the rest is correct.


    I used De Moivre's formula to find the roots for the other problem, too.

     -1= e^{i \pi} = \cos \pi + i \sin \pi

    so  (-4)^{-1/4} = 4^{1/4}(-1)^{1/4} = \sqrt{2}e^{i(\pi /4 + 2k \pi /4)} = \sqrt{2} [(\cos(\pi /4 + 2k \pi /4) + i \sin (\pi /4 + 2k \pi /4)]

    Then let k =0,1,2,and 3 to find the four roots.


    http://en.wikipedia.org/wiki/De_Moivre's_formula
    Last edited by Random Variable; March 21st 2010 at 01:40 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Apr 2009
    Posts
    85
    Ok I see thanks for the link. Why is x =\frac{\pi}{2} in this case?
    -i=e^{-i\pi/2} = \cos (-\pi/2) + i \sin (-\pi /2)
    Last edited by mmattson07; March 21st 2010 at 08:42 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    Quote Originally Posted by mmattson07 View Post
    Ok I see thanks for the link. Why is x =\frac{\pi}{2} in this case?
    -i=e^{-i\pi/2} = \cos (-\pi/2) + i \sin (-\pi /2)
    It's because of the location of -i on the complex plane.

    I'll try to explain, but you should really study up a bit on complex numbers.


    All real numbers (a+0i) are located along the x axis, all purely imaginary numbers (0+bi) are located along the y axis, and all complex numbers (a+bi) are located everywhere else.

    -i  = 0-i is located on the negative y axis and can be represented as a vector of unit length (since \sqrt{0^{2}+1^{2}} = 1) with it's tail at the origin and it's head at y=-1 .

    The angle (going counterclockwise) that the vector makes with the positive horizontal axis is  3 \pi/2 . But you usually want the angle to be between  \pi and -\pi . The angle in that range that is equal to  3 \pi /2 is  =-\pi/2

    You can write any number (real, imaginary, or complex) in the form re^{i\theta} = r(cos \theta + i \sin \theta) where r is the distance of the point from the origin in the complex plane and \theta is the angle formed by the vector and the positive x axis.

    So  -i= 1e^{-i \pi /2} = 1(\cos -\pi/2 + i \sin -\pi/2) = 1(0-i) = -i

    For the other problem the angle is  \pi because -1 is located on the negative x axis.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Apr 2009
    Posts
    85
    Should it be i or -i? Because (i)^3=-i satisfying m^3+i=0 and (-i)^3=i which would give 2i=0
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    Quote Originally Posted by mmattson07 View Post
    Should it be i or -i? Because (i)^3=-i satisfying m^3+i=0 and (-i)^3=i which would give 2i=0
    You're correct.  -i is not a root. But I never said it was. We're looking for the three roots of (-i)^{1/3} . I need to first express  -i in complex polar form (which I did above) before I can apply De Moivre's formula to find the roots. One of those roots is  i.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Apr 2009
    Posts
    85
    OH I see you get (-i)^{1/3} from solving for m then apply DeMoivre. Ok thanks.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Auxiliary Circle
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: July 26th 2011, 10:52 AM
  2. Replies: 1
    Last Post: November 17th 2010, 11:41 PM
  3. Roots & Imaginary Roots
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: October 4th 2009, 09:24 AM
  4. Given 2 imaginary roots find other 2 roots
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: January 26th 2008, 09:24 PM
  5. Auxiliary Eq (Dif EQ)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 5th 2007, 12:37 AM

Search Tags


/mathhelpforum @mathhelpforum