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Ode Again!

  1. Aug 22, 2005 #1
    Here is another problem which is driving me insane!I just need a hint on how to proceed!

    To Solve:-
    p^7 + p^3 - p^2 + 1 =0

    here p = dy/dx
     
  2. jcsd
  3. Aug 22, 2005 #2

    HallsofIvy

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    Since that is a linear homogeneous d. e., it characteristic equation is just what you give: p7+ p3- p2+ 1= 0. There are no general ways of solving 7th degree polynomial equations so the best you could hope for is some simple equation. Because the leading and ending coefficients are both 1, the only possible rational solutions are p= 1 and p= -1- and neither of those satisfy the equation!

    My suggestion- go insane!
     
  4. Aug 22, 2005 #3
    Halls,You triggered in the right direction!
    I asked my tutor and he said that there will be atleast 1 real root
    => p=k
    or y=kx + c
    From here i am thinking ahead!
     
  5. Aug 22, 2005 #4

    saltydog

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    I have a question:

    So we look for solutions of the form y=mx+b, plug it in, get a polynomial in m and solve for the roots. So there's one real root. So we find it numerically (I ain't proud), and then we get the solution:

    [tex]y(x)=m_rx+b[/tex]

    So, how do we know if there are other solutions?
     
  6. Aug 22, 2005 #5
    The same thing i asked him!
    He said our aim is to get just 1 family of curves of solution!
    The other six can be complex,real who knows!
     
  7. Aug 22, 2005 #6

    HallsofIvy

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    heman, I interpreted "p^7" where p= dy/dx to mean the seventh derivative (operator notation). saltydog tells me that he interpreted it as the seventh power of dy/dx so that this is not a linear equation at all. Which is it?

    (Yes, the polynomial p^7 + p^3 - p^2 + 1 =0, by DesCartes' "rule of signs", has one negative real root and either 0 or 2 positive real roots. That is, the equation may have (1) 1 negative real root and 3 pair of conjugate complex roots or (2) 1 negative real root, 2 positive real roots, and 2 pair of conjugate comples roots. However, the real roots are not rational.)
     
  8. Aug 22, 2005 #7
    Ohhhh....I am the source of miscommunication. :surprised ...Halls,it's 7th power,i should have had clarified more to avoid this..! o:)

    Your Predictions!!
    (1) 1 negative real root and 3 pair of conjugate complex roots
    I understand that it can have 3 pair of conjugate complex roots-but how do you say that real root will be negative?

    (2) 1 negative real root, 2 positive real roots, and 2 pair of conjugate comples roots
    How are you able to tell the signs??

    I have taken the Complex Analysis Course last semester but i haven't come across any such theorem which tells about the sign!
     
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