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Reduction od order question

  1. Aug 17, 2010 #1
    1. The problem statement, all variables and given/known data

    find the first and second derivative of first solution.

    2. Relevant equations

    [itex]y(x)=m(x)y_1(x)[/itex]

    [itex]y'(x)=m'(x)y_1(x)+m(x)y_1'(x)[/itex]




    3. The attempt at a solution

    I have been given [itex]y_1=\frac{1}{x^n}[/itex]

    Which part is the m(x) and which is [itex]y_1(x)[/itex]
    I'm not sure how to do the substitution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 17, 2010 #2

    hunt_mat

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    Can you give us a little more of the question.
     
  4. Aug 17, 2010 #3
    Sure the original question is Differential equation.

    It asks find a pair of fundamentle soultions to the DE


    [itex]x^2y''-3xy'+y=0[/itex]


    So I'm trying to find the two solutions I'm taking [itex]y_1=\frac{1}{x^n}[/itex]
    as one solution.
     
  5. Aug 17, 2010 #4

    hunt_mat

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    This is known as Eulers equation, you are correct (partially) in looking for solutions of the form:
    [tex]
    y=x^{n}
    [/tex]
    Insert this into your ODE and you will obtain a quadratic equation for [tex]n[/tex], this will have two solutions which correspond to the two solutions.
     
  6. Aug 17, 2010 #5
    So does

    [itex]x^2y''-3xy'+y=0[/itex]

    become


    [itex]x^2(n^2-2)x^{n-2} -3xnx^{n-1} + x^n=0[/itex]
     
  7. Aug 17, 2010 #6
    I seem to be having trouble getting a quadratic for n

    is this the right method?

    I have

    [itex]y=x^n[/itex]
    than
    [itex]y'=nx^{n-1}[/itex]
    then
    [itex]y''=(n^2-n-1)x^{n-2}[/itex]


    so do you substitute into

    [itex]x^2y''-3xy'+y=0[/itex] to give

    [itex]x^2((n^2-n-1)x^{n-2})- 3x(nx^{n-1})+x^n=0[/itex]

    I'm getting

    [itex](n^2-4n+1)x^n[/itex]

    I can see a characteristic equation there but not sure about the coefficient [itex]x^n[/itex]?
     
  8. Aug 18, 2010 #7

    hunt_mat

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    Almost:
    [tex]
    y''=n(n-1)x^{n-2}
    [/tex]
    So
    [tex]
    x^{2}(n(n-1)x^{n-2}-3xnx^{n-1}+x^{n}=0
    [/tex]
    which yields
    [tex]
    (n^{2}-4n+1)x^{n}=0
    [/tex]
    So
    [tex]
    n^{2}-4n+1=0
    [/tex]
    Can you calculate n from the above?
     
  9. Aug 18, 2010 #8
    Thanks,

    I solve the quadratic [tex]n^{2}-4n+1=0[/tex]
    and get:

    [tex]n_1=-\sqrt{3}-2[/tex] and [tex]n_2=\sqrt{3}+2[/tex]


    So are the two solutions....?

    [tex]y_1=c_1x^{-\sqrt{3}-2}[/tex] and [tex]y_2=c_2x^{\sqrt{3}+2}[/tex]
     
    Last edited: Aug 18, 2010
  10. Aug 18, 2010 #9

    vela

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    You flipped a sign in n1, but otherwise your solutions are correct.
     
  11. Aug 18, 2010 #10
    Sorry,

    [tex]
    y_1=c_1x^{-\sqrt{3}+2}
    [/tex]

    thanks for your help guys
     
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