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Differential Equations, Frobenius' Method

  1. Nov 22, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the indicial roots of the following Differential Equation: xy'' - y' + x3y = 0

    2. Relevant equations

    y = Ʃ[n=0 to infinity]cnxn+r
    y' = Ʃ[n=0 to infinity](n+r)cnxn+r-1
    y'' = Ʃ[n=0 to infinity](r+r)(n+r-1)cnxn+r-2

    3. The attempt at a solution

    Plugging these values into the differential equation, I got

    xr{Ʃ[n=0 to infinity](n+r)(n+r-1)cnxn-1 - Ʃ[n=0 to infinity](n+r)cnxn-1 + Ʃ[n=0 to infinity]4cnxn+3} = 0

    The three sums must produce the x to the same exponent, so I tried pulling out the first 4 terms of the first two sums, so the three sums would each output x3 as their first term [the first two sums starting from n=4]. However, this left me with the following equation:

    r(r-1)c0x-1 - rc0x-1 + r(r+1)c1 - (r+1)c1 + (r+1)(r+2)c2x - (r+2)c2x - (r+2)(r+3)c3x2 - (r+3)c3x2 + [remaining sums] = 0.

    How do I solve for r with this equation? I don't know how to find the roots.

    [the solution to the DE is y=c1cos(x2) + c2sin(x2)
     
    Last edited: Nov 22, 2011
  2. jcsd
  3. Nov 22, 2011 #2

    lanedance

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    Homework Helper

    So you have
    [tex] x^3y = \sum_{n=0}^{\infty} c_n x^{n+r+3}[/tex]
    [tex] y' = \sum_{n=0}^{\infty} c_n (n+r)x^{n+r-1}[/tex]
    [tex] xy'' = \sum_{n=0}^{\infty} c_n (n+r)(n+r-1)x^{n+r-1}[/tex]

    so lets shift the equations, so they have the same form for powers of x, as follows:

    lets leave the 2nd & 3rd leave unchanged, but replace the 1st so the terms match, first replace n with m
    [tex] x^3y = \sum_{m=0}^{\infty} c_m x^{m+r+3}[/tex]
    then let m=n-4, which when m=0 gives n=4, subtituting in gives
    [tex] x^3y =\sum_{n=4}^{\infty} c_{n-4} x^{n+r-1} [/tex]


    Putting this all together we get
    [tex] \sum_{n=4}^{\infty} c_{n-4} x^{n+r-1}- \sum_{n=0}^{\infty} c_n (n+r)x^{n+r-1}+ \sum_{n=0}^{\infty} c_n (n+r)(n+r-1)x^{n+r-1}=0[/tex]

    Now lets group terms
    [tex] \sum_{n=0}^{3} c_n(- (n+r)+ (n+r)(n+r-1))x^{n+r-1}+
    \sum_{n=4}^{\infty} (c_{n-4} - c_n (n+r)+ c_n (n+r)(n+r-1))x^{n+r-1}=0[/tex]

    Now the indicial equation is given by the lowest power of x, this occurs when n=0
     
    Last edited: Nov 22, 2011
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