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Solution of the Cauchy-Euler equation

  1. Oct 19, 2008 #1
    1. The problem statement, all variables and given/known data

    The Cauchy-Euler equation is:
    x2y'+axy'+by=0
    And has a solution of the form:
    y1(x)=xm
    Use the method of Variation of Parameters to show that the second independent solution
    is:
    y2(x)=xmln(x)

    So that the overall solution is:
    y(x)=[A+Bln(x)]xm
    Hint: This equation suggests that the characteristic equation has two identical roots.

    Hint: Use the following transformation:
    x=et

    2. Relevant equations

    None

    3. The attempt at a solution
    I'm not really sure how to start this. Any help would be very much appreciated.
     
  2. jcsd
  3. Oct 19, 2008 #2

    Dick

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    Science Advisor
    Homework Helper

    You should start it by putting y=x^m into the original equation and then realizing you meant to say x^2*y''+a*x*y'+b*y=0. Then think about it again using the hints.
     
  4. Oct 19, 2008 #3
    Ok, so:
    m2+(a-1)m+b=0

    So using variation of parameters the equation can be represented as:

    [tex]{\lambda}^{2}+ \left( a-1 \right) \lambda+b=0[/tex]

    and

    [tex]\lambda=\mbox {{\tt `\&+-`}} \left( 1-a,\sqrt {1/2\, \left( a-1 \right) ^{2}-2\,b} \right)[/tex]


    ???

    I'm not sure if that latex stuff worked:

    L2+(a-1)L+b=0

    and

    L=(1-a)(+/-)(((a-1)^2-4b)/2)1/2
     
    Last edited: Oct 19, 2008
  5. Oct 19, 2008 #4
    In any event, lets say I now have:

    u(x)=[c1+c2ln|x|]xm

    or

    u(x)=[c1+c2ln|x|]x-(a-1)/2

    How do I use the variations of parameters to show the second independent solution:

    y2=xmln(x)
     
  6. Oct 20, 2008 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Do you know what "variation of parameters" IS?

    Since you already know that xm is a solution, try a solution of the form y= u(x)xm. Then y'= u' xm+ m u xm-1 and y"= u" xm+ 2m u' xm-1+ m(m-1)u xm. Put those into the equation and use the fact that xm itself satifies the equation.
     
  7. Oct 20, 2008 #6
    No not really. I have notes on it buts I can't make heads or tails of it.

    Where is the 'u' coming from?
     
    Last edited: Oct 20, 2008
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