Method of Frobenius for Solving ODE at x=0 | Step-by-Step Guide

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SUMMARY

The discussion focuses on applying the Method of Frobenius to solve the ordinary differential equation (ODE) given by (1-x) y'' + xy' - \frac{\alpha^2}{x^2} = 0 at x=0. The roots from the indicial equation are established as λ = \frac{1}{2}(1 ± \sqrt{4α^2 + 1}). The recursion relation for the coefficients is derived as c_n = \frac{c_{n-1}(\lambda+n-1)(\lambda+n-2) - c_{n-2}(\lambda+n-2)}{(\lambda+k)(\lambda+n-1) - \alpha^2}. The user expresses uncertainty about verifying the correctness of the series solution, particularly when substituting specific values for α.

PREREQUISITES
  • Understanding of ordinary differential equations (ODEs)
  • Familiarity with the Method of Frobenius
  • Knowledge of series expansions in mathematical analysis
  • Basic algebraic manipulation skills for recursion relations
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  • Study the derivation of the Method of Frobenius in detail
  • Learn how to analyze convergence of power series solutions
  • Explore examples of solving ODEs using specific values of α
  • Investigate the implications of the indicial equation in ODE solutions
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Mathematics students, educators, and researchers interested in solving ordinary differential equations, particularly those utilizing series methods like the Method of Frobenius.

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Homework Statement


Using method of frobenius about [itex]x=0[/itex] to solve:
[itex](1-x) y''+xy'-\frac{\alpha^2}{x^2}+=0[/itex]
[itex][/itex]

Homework Equations


N/A

The Attempt at a Solution


1. plug in series into the equation.
2. adjust the index off all the terms.
3. write the extra terms separately so that we have all series starting at the same point.

and I get...

that the roots from the indicial equation
[itex]\lambda=\frac{1}{2}(1\pm\sqrt{4a^2+1})[/itex]
and for the recursion I get
[itex]c_n=\frac{c_{n-1}(\lambda+n-1)(\lambda+n-2)-c_{n-2}(\lambda+n-2)}{(\lambda+k)(\lambda+n-1)-\alpha^2}[/itex]

I am not sure how to check if this is correct or not.
Using a number for [itex]\alpha[/itex] I don't get any nice series that has a function.
 
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