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Solving polynomial using matrix methods

  1. May 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Let y(x) = a0 + a1x + a2x2 + a3x3 + a4x4

    Represent the differential equation

    (d/dx)[ (1-x2)(dy/dx) ] + λy = 0

    in matrix form. Find the values of λ for which there is a solution to the matrix equation, and find the solutions for the smallest and largest values of λ. Normalize these solutions with the condition y(1) = 1.


    2. Relevant equations



    3. The attempt at a solution

    I got as far as finding the values of λ = 0, 2, 6, 12, 20. Am i supposed to plug in the values of λ into the differential equation and solve it by finding the characteristic equation by assuming y = Aeβx? But the question gives y as a polynomial: y(x) = a0 + a1x + a2x2 + a3x3 + a4x4

    6nz8nl.png
     
  2. jcsd
  3. May 27, 2013 #2

    Dick

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    Check your matrix. You are missing a λ in the first row. That will give you another value for λ. And you are supposed to substitute the values of λ into your matrix and find corresponding a's.
     
  4. May 28, 2013 #3
    Oops i wrote my matrix wrongly from my working on paper. The smallest value of λ = 0.

    I don't understand what you mean by "finding the corresponding a's".

    Say i sub in λ = 0. All i get is: BA = 0, where B is the matrix with the subbed value of λ and A is the column with all the a0, a1, ....

    For example, let's consider [5,5].

    (-20)(a4) = 0.

    This doesn't mean anything other than a4 = 0. Then it would mean a0 = a1 = a2 = a3 = a4 = 0
     
  5. May 28, 2013 #4

    Dick

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    You are jumping to conclusions. Just because a4=0 doesn't mean ALL of they a's have to be zero. They don't. Which one doesn't?
     
  6. May 28, 2013 #5
    Taking λ = 0,

    Let's start with the bottom row.

    -20a4 = 0,

    => a4 = 0

    -12a3 = 0

    => a3 = 0

    -6a2 + 12a4 = 0

    => a2 = 0

    -2a1 + 6a3 = 0

    => a1 = 0

    and finally 2a2 = 0
     
  7. May 28, 2013 #6

    Dick

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    You forgot a0.
     
  8. May 28, 2013 #7
    I put in λ = 0 into the matrix at [1,1]. This gives a0 = 0 as well..
     
  9. May 28, 2013 #8

    vela

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    No, it doesn't.
     
  10. May 28, 2013 #9
    0 * a0 + 2a1 = 0

    does this mean a0 can be any real number?
     
  11. May 28, 2013 #10

    vela

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    To satisfy that equation, yes. But it needs to be non-zero because you're looking for a non-trivial solution to the system of equations.
     
  12. May 28, 2013 #11
    Then when i sub in λ = 20,

    I have a4 = any real number,

    a3 = 0,

    a2 = -(6/7)a4

    a1 = 0

    a0 = 0

    is this right?
     
  13. May 28, 2013 #12

    vela

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    You shouldn't get ##a_0=0##. The rest are right.
     
  14. May 29, 2013 #13
    For λ = 0

    -20a4 = 0,

    => a4 = 0

    -12a3 = 0

    => a3 = 0

    -6a2 + 12a4 = 0

    => a2 = 0

    -2a1 + 6a3 = 0

    a1*0 = 0

    a1 [itex]\subset[/itex] ℝ


    is it because from the condition:

    y(1) = 1

    a0 + a1 + a2 + a3 + a4 = 1

    So, a0 = 1


    For λ = 20

    => a4 [itex]\subset[/itex] ℝ,

    => a3 = 0

    14a2 + 12a4 = 0
    => a2 = (-6/7) a4

    18a1 + 6a3 = 0
    => a1 = 0

    20a0 + 2a2 = 0
    => a0 = (-1/10)a2 = (3/35)a4

    Using a0 + a1 + a2 + a3 + a4 = 1



    a0 = 3/8
    a1 = 0
    a2 = -48/245
    a3 = 0
    a4 = 35/8
     
    Last edited: May 29, 2013
  15. May 29, 2013 #14

    Dick

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    All correct, except your final a's for λ=20 don't sum to 1.
     
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