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    Hi my images sarey question and my , although I think I have gone wrong, can anyone help me out please?
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    Without beating through your working (which seems over-complex) we need to say that there is a solution to the Bessel equation such that C zero is non zero.

    Once we have that solution written as a polynomial each coefficient in the polynomial must be separately zero.
    i.e. if the solution is ax + bx^2 + cx^3 ... then a=0, b = 0, c=0 etc and each of a, b, c will be a polynomial in r.

    So work out the bits (I write C rather than C sub zero to save having to latex this)

    y=Cx^r, y'=Crx^(r-1) and y''=Cr(r-1)x^(r-2)

    when these are substituted into Bessel's equn we get for terms involving C zero (again just using C). We are only interested in the C zero term

    y(x) = Cx^r [ r (r-1) + r + (x^2 - m^2)] + other terms not involving C zero

    Now multiplying this out (and rejecting the term in x^(r+2) cos that is not an x^r term) we get the coefficient of x^r is C[r^2 - r + r -m^2]

    If C is non-zero then for the Solution to be valid the r term in the coefficient of x^r must be zero.

    i.e. [r^2 - r + r -m^2] must be zero which gives us r^2=m^2

    Hope that's clear
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    (Original post by nerak99)
    Without beating through your working (which seems over-complex) we need to say that there is a solution to the Bessel equation such that C zero is non zero.

    Once we have that solution written as a polynomial each coefficient in the polynomial must be separately zero.
    i.e. if the solution is ax + bx^2 + cx^3 ... then a=0, b = 0, c=0 etc and each of a, b, c will be a polynomial in r.

    So work out the bits (I write C rather than C sub zero to save having to latex this)

    y=Cx^r, y'=Crx^(r-1) and y''=Cr(r-1)x^(r-2)

    when these are substituted into Bessel's equn we get for terms involving C zero (again just using C). We are only interested in the C zero term

    y(x) = Cx^r [ r (r-1) + r + (x^2 - m^2)] + other terms not involving C zero

    Now multiplying this out (and rejecting the term in x^(r+2) cos that is not an x^r term) we get the coefficient of x^r is C[r^2 - r + r -m^2]

    If C is non-zero then for the Solution to be valid the r term in the coefficient of x^r must be zero.

    i.e. [r^2 - r + r -m^2] must be zero which gives us r^2=m^2

    Hope that's clear
    Thank you for this, I will give this a go
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    Suppose further that I wanted to do this;


    All I can think of doing is just replacing "y" instead of "m" into the Bessel equation. Although that doesn't really help much.
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    Well this falls out after some algebra.
    Realise that the coefficients if x^(r+n) mainly involve Cn but the x^2-m^2 gives an -m^2 Cn x^(r+n) but the x^2 picks out C(n-2) x^(r+n).

     Cn [(x^2-m^2)+(r+n) + (r+n)(r+n-1)]x^{r+n}+...


    When we multiply out the x^2 we end up with a term in x^(r+n+2) and so we have to pick out a term in Cn-2 to find the x^(r+n) term.

    Sorry if that is not clear. Having a terrible time with getting tex to cooperate with subscripts
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    (Original post by nerak99)
    Well this falls out after some algebra.
    Realise that the coefficients if x^(r+n) mainly involve Cn but the x^2-m^2 gives an -m^2 Cn x^(r+n) but the x^2 picks out C(n-2) x^(r+n).

     Cn [(x^2-m^2)+(r+n) + (r+n)(r+n-1)]x^{r+n}+...


    When we multiply out the x^2 we end up with a term in x^(r+n+2) and so we have to pick out a term in Cn-2 to find the x^(r+n) term.

    Sorry if that is not clear. Having a terrible time with getting tex to cooperate with subscripts
    Thank you I needed this tip will try and solve further later today.
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    (Original post by nomad609)
    Thank you I needed this tip will try and solve further later today.
    You might also find this helpful:

    http://mathworld.wolfram.com/FrobeniusMethod.html

    The example they use on the method is the Bessel equation - lucky for you - but don't just copy it out or you won't learn anything...
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    (Original post by DFranklin)
    You might also find this helpful:

    http://mathworld.wolfram.com/FrobeniusMethod.html

    The example they use on the method is the Bessel equation - lucky for you - but don't just copy it out or you won't learn anything...
    great thanks, yes I will definietely use this and practice with TeeEm resources.
 
 
 
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