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4th order differential equation

  1. Nov 19, 2012 #1
    I'm trying to find the gen. solution to the equation y''''-8y'=0
    I found the characteristic polynomial by plugging in ert as a solution to y.
    I got,
    r^4-8r=0
    I simplified to get
    r*(r^3-8)
    Thus one root is 0, for the other 3 i must find the cubed root of 8.
    I know the answer is 2*e2m*pi*i/3 for m=0,1,2
    How do I arrive at that answer?
    I tried the following:
    Represent 8 as 8=8[cos(2*pi)+i*sin(2*pi)]=8*ei*pi
     
  2. jcsd
  3. Nov 19, 2012 #2

    haruspex

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    It should be 8 = 8ei*2πm, for any m, right?
     
  4. Nov 20, 2012 #3
    The roots are r=0,2*e(2*m*pi*i)/3
    then it says this is equivalent to
    r=0,2,-1+i*sqrt(3),-1-i*sqrt(3)

    Then the gen solution is
    y=c1+c2*e2*t+e-t*[c3*cos(t*sqrt(3))+c4*sin(t*sqrt(3))]

    I dont know how they arrive to this
     
  5. Nov 20, 2012 #4
    Obviously, the cubic root of 8 is 2.
    r^4-8r = r(r-2)(r²+2r+4)
    = r (r-2) [r+1+i sqrt(3)] [r+1-i sqrt(3)]
     
  6. Nov 20, 2012 #5

    haruspex

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    Suppose λ is a root of the polynomial. That is, the differential equation has a factor (D-λ), where D = d/dt, so the equation can be written P(D)(D-λ)y = 0, for some polynomial P.
    Try the solution y = eλt: P(D)(D-λ)eλt = P(D)(D(eλt - λeλt) = P(D)(λeλt - λeλt) = 0. So the general solution is a linear combination of such terms.
    A complication arises when there is a repeated root, i.e. a factor (D-λ)n. It's fairly easy to show that treλt is also a solution for r = 1, .. n-1.
     
  7. Nov 20, 2012 #6
    But where is the root e(2*m*pi*i)/3 derived from? and how is this equivalent to the roots:2,-1+i*sqrt(3),-1-i*sqrt(3) for m=1,2,3
     
  8. Nov 20, 2012 #7

    lurflurf

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    exp(2∏n i)=1
    so
    (exp(2∏n i/3))^3=1
    also
    exp(2∏n ix)=cos(2∏n x)+i sin(2∏n x)
    so
    exp(2∏n i/3)=cos(2∏n /3)+i sin(2∏n /3)
     
    Last edited: Nov 20, 2012
  9. Nov 20, 2012 #8

    haruspex

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    Small correction (too many i's):
    exp(2∏n ix)=cos(2∏n x) + i sin(2∏n x)
    so
    exp(2∏n i/3)=cos(2∏n/3) + i sin(2∏n/3)

    ... and cos(2∏/3) = -cos(∏/3) = -(√3)/2 etc.
     
  10. Nov 21, 2012 #9
    What you are getting confused about is the field of the reals vs. the field of complex numbers. Over the field of the reals there is only one solution to the equation r^3-8=0 namely r=2, but over the field of complex numbers there are 3 distinct solutions to the equation r^3-8=0.

    To see this understand that [tex]exp(i2m\pi)=1[/tex] for any [tex]m\epsilon\mathbb{Z}[/tex]. So if, [tex]r^3=8\cdot1=8exp(i2m\pi)[/tex] then
    [tex]r =2exp(\frac{i2m\pi}{3})[/tex] but you can see the only distinct ones are for m= 0,1,2 since for m beyond or below that you repeat your answers. To answer your other question you need to know about Euler's formula which says that: [tex]e^{i\phi}=cos(\phi)+isin(\phi)[/tex]

    So for example, [tex]2exp(i\frac{2\pi}{3})=2(cos(\frac{2\pi}{3})+i\sin({\frac{2\pi}{3}})=2(-\frac{1}{2}+i\frac{\sqrt{3}}{2})[/tex]
     
    Last edited: Nov 21, 2012
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