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Finding the G.S.(for equal roots) of the Euler-Cauchy diff. equation

  1. Oct 11, 2007 #1

    rock.freak667

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    Homework Helper

    1. The problem statement, all variables and given/known data

    [tex]x^2\frac{d^2y}{dx^2}+ax\frac{dy}{dx}+by=0[/tex]

    show that if there is one real double root of the aux. eq'n show that the G.S. is given by

    [tex]y=c_1x^{n_1}+c_2x^{n_1}ln(x)[/tex]

    2. Relevant equations

    Assume the trial solution [itex]y=x^n[/itex]

    3. The attempt at a solution
    [tex] y=x^n
    \frac{dy}{dx}=nx^{n-1}
    \frac{d^2y}{dx^2}=n(n-1)x^{n-2}

    => n(n-1)x^n + anx^n +bx^n =n^2-(a-1)n+b=0 (Aux. Eq'n)[/tex]

    for equal roots [tex](a-1)^2-4b=0[/tex]

    therefore [tex]n=\frac{1-a}{2}[/tex]
    hence one solution is [tex]y=x^\frac{1-a}{2}[/tex]

    so the other solution is [tex]y=vx^\frac{1-a}{2}

    \frac{dy}{dx}=\frac{dv}[dx}x^\frac{1-a}+ \frac{1-a}{2}Vx^\frac{-a-1}{2}
    \frac{d^2y}{dx^2}=\frac{d^2v}{dx^2}x^\frac{-1-a}{2} + (1-a)\frac{dv}{dx}x^\frac{-1-a}{2} -\frac{1-a^2}{4}vx^\frac{-a-3}{2}[/tex]

    sub. into the diff. eq'n....(this is where I think I am wrong)
    (This is what I get)
    [tex]\frac{d^2v}{dx^2}x^\frac{3-a}{2} + \frac{dv}{dx}x^\frac{3-a}{2} +\frac{vx^\frac{1-a}{2}((a-1)^2+4b)}{4}[/tex]

    Now I was hoping to get [itex](a-1)^2-4b[/itex] in the last term there so that it would be zero but I apparently did something wrong..where did I go wrong?
     
  2. jcsd
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