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Using Frobenius Method

  1. Nov 17, 2013 #1
    1. The problem statement, all variables and given/known data
    I want to find two linearly independent solutions of
    $$
    x^{2}y''-2x^{2}y'+(4x-2)y=0.
    $$

    3. The attempt at a solution
    The roots to the indicial polynomial are ##r_{1}=2## and ##r_{2}=-1##.
    I found one solution which was ##x^{2}## and I am having trouble finding the second solution.
    After all of the arithmetic I came up with the equation
    $$
    [r(r-1)-2]c_{0}+\sum\limits_{k=1}^{\infty}\left[(k+r)(k+r-1)-2\right]c_{k}x^{k}+4c_{k-1}x^{k}-2c_{k-1}(k+r-1)x^{k}.
    $$
    I let ##r=-1## so that I get
    $$
    \left[(k-1)(k-2)-2\right]c_{k}+4c_{k-1}-2c_{k-1}(k-2)=0
    $$
    and so
    $$
    c_{k}=\frac{2(k-4)c_{k-1}}{k(k-3)}.
    $$
    The recurrence relation becomes undefined for k=3 and onwards. How do I deal with this? The solution skips ##c_{3}## and seemingly magically obtains a formula for ##c_{4}## and onwards.
     
  2. jcsd
  3. Nov 17, 2013 #2

    vanhees71

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    First of all I guess the equation should be
    [tex]x^{2}y''-2x^{2}y'-(4x+2)y=0,[/tex]
    because otherwitse [itex]y(x)=x^2[/itex] isn't a solution, and your indical roots are not [itex]2[/itex] and [itex]-1[/itex].

    Assuming the above ODE, you get the indical equation to be yours,
    [tex]r(r-1)-2=0 \; \Rightarrow r_1=2, \quad r_2=-1,[/tex]
    and the recursion equation reads (after some algebra)
    [tex]C_k k(k+2r-1)-2 C_{k-1}(k+r-3)=0.[/tex]

    Now here you have the special case of a Fuchs-class ODE, where the difference of the indical roots is natural and not 0. In this case you always have this "problem" with the coefficient of [itex]c_{r_1-r_2}[/itex] becoming [itex]0[/itex] for the solution for the smaller indical root, [itex]r<2[/itex]. For the solution [itex]y_1[/itex] referring to the larger indical root, there is no problem with the recursion. Given this solution, the correct ansatz for the 2nd linearly independent solution is
    [tex]y_2(x)=\gamma y_1(x) \ln x+x^{r_2} \sum_{k=0}^{\infty} C_k x^k.[/tex]
    Of course you assume [itex]C_0 \neq 0[/itex]. Then for [itex]k<r_1-r_2[/itex] the recursion can be used to get the coefficients for these [itex]k[/itex]. For [itex]k=r_1-r_2[/itex] you find that you can set [itex]C_{r_1-r_2}=0[/itex] and then get [itex]\gamma[/itex] uniquely, and for [itex]k>r_1-r_2[/itex] you get another recursion relation which is well-defined for the [itex]C_k[/itex] with [itex]k>r_1-r_2[/itex].
     
  4. Nov 17, 2013 #3
    The equation is ##x^{2}y''-2x^{2}y'+(4x-2)y=0##. Unless I'm completely crazy, ##y(x)=x^{2}## is a solution.
     
  5. Nov 17, 2013 #4

    vanhees71

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    Argh. When copying your equation to a piece of paper, I wrote [itex]+2x^2y'[/itex] for the middle term. Of course, you are right, for
    [tex]x^2 y''-2x^2 y'+(4x-2)y=0[/tex]
    one solution is [itex]y_1=x^2[/itex].

    The rest of my posting is, however correct. For [itex]r=-1[/itex] you make the ansatz
    [tex]y(x)=\gamma x^2 \ln x+z(x), \quad z(x)=\frac{1}{x} \sum_{k=0}^{\infty} d_k x^k.[/tex]
    This plugged into the ODE you get after some algebra the equation
    [tex]\sum_{k=1}^{\infty} [d_k k(k-3)+d_{k-1} (8-2k)] x^k=(-3x^3+2x^4)\gamma.[/tex]
    Setting [itex]d_0=1[/itex] you get recursively all the [itex]d_k[/itex] by comparing coefficients. [itex]d_3[/itex] is of course arbitrary. You can set it to [itex]0[/itex].
     
  6. Nov 17, 2013 #5
    Ah! Thanks! I see what I was doing wrong now. I was plugging in ##y(x)=z(x)## to try to get the second solution instead of the correct ##y(x)=cx^{2}\ln x+z(x)##. Thanks for your help!
     
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