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Solve Using Frobenius Method
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[QUOTE="vanhees71, post: 4528225, member: 260864"] Yes, that's all correct, and you have to figure out your [itex]k[/itex] from the equation. Since you've found the solution now, I think it's ok to redo the full problem in a systematic way to get the last uncertainties solved. I hope I don't violate the rules of the homework forum with this, but I'm convinced that at this point of understanding of the OP it helps most to give a systematic application of the Frobenius-Fuchs theorem for this nice example. The equation to solve was (written in standard form) [tex](1-x)y''(x)-\frac{2}{x} y'(x)+\frac{2}{x}y(x)=0.[/tex] This indeed fulfills the conditions of the theorem for the singular point [itex]x=0[/itex] (note that there is another singular point at [itex]x=1[/itex] here!). So here we have the simplifying case that we don't need series expansions for the coefficient functions as stated already in the OP. So the Frobenius ansatz must lead to at least one solution of the form [tex]y_1(x)=x^s \sum_{k=0}^{\infty} a_k x^k.[/tex] To find [itex]s[/itex] and the recursion for the [itex]a_k[/itex] we plug in the ansatz into the ODE, finding after rearrangement of the summation index and dividing by [itex]x^{s-2}[/itex] [tex]s(s-3)+\sum_{k=1}^{\infty} (s+k)(s+k-3)[a_k - a_{k-1}] x^{k}=0.[/tex] It's clear that this implies the two solutions [itex]s=0[/itex] and [itex]s=3[/itex] for the leading behavior of the solutions. So both solutions are analytic in [itex]x=0[/itex], i.e., the weak singularity is integrable here. Further we are lucky in this case and can circumvent the somewhat cumbersome case that we need [itex]k \neq 0[/itex], because in this case for both values for [itex]s[/itex] the coefficients can be set [itex]a_k \equiv 1[/itex] to solve the recursion equation, which gives the two solutions [tex]y_1(x)=x^3 \sum_{k=0}^{\infty} x^k=\frac{x^3}{1-x}, \quad y_2(x) = \sum_{k=0}^{\infty} x^k = \frac{1}{1-x}.[/tex] Since these are obviously linearly independent functions, because [itex]y_1(x)/y_2(x) = x^3 \neq \text{const}[/itex] the general solution of the ODE is [tex]y(x)=(a_1+a_2 x^3) \frac{1}{1-x}.[/tex] In the general case for [itex]m=s_1-s_2 \in \mathbb{N}_0[/itex] only the first solution with the larger [itex]s_1[/itex] is given by the Frobenius ansatz, while the second solution must be found by the ansatz given in your last posting. For the case [itex]s_1=s_2[/itex] you can set [itex]k=1[/itex], and then the [itex]b_k[/itex] are uniquely determined with [itex]b_0=0[/itex], which is always a possible choice for the initial condition for the recursion for the [itex]b_k[/itex] in this case. For [itex]m \in \mathbb{N}[/itex] the [itex]k[/itex] must be determined from the ODE, and it is always uniquely defined (given the choice of the solution [itex]y_1[/itex]), and you can always set [itex]b_m=0[/itex], and then all coefficients [itex]b_k[/itex] are uniquely determined by further setting [itex]b_0=1[/itex]. [/QUOTE]
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