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When solving second-order differential equations by the Frobenius method at a regular singular point, you are supposed to use the two roots of the indicial equation to give you two independent solutions.

If there is only one root, it makes sense that you would need another method to get the second independent solution. However, many texts say that you also need to do this when the roots differ by an integer.

Why?

Is it that when the roots differ by an integer, the two matching solutions are not independent? If so, why must they be independent?

Is it that sometimes they will be independent and sometimes they won't? If so, when will they be independent and when won't they?

Is it that there is something that prevents calculating one of the solutions? If so, why?

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  • $\begingroup$ Apparently you end up with an unsolvable recurrence for computing the coefficient $a_N$. See math.creighton.edu/nielsen/DE_Fall_2010/Series%20Solutions/… $\endgroup$ – Ian Jul 23 '14 at 20:28
  • $\begingroup$ Basically, the problem is that $a_n = .../F(n+r)$ where $F$ is the indicial polynomial. But if $F(n+r)=0$ then this doesn't make sense unless the $...$ expression is zero (which it isn't, in general). $\endgroup$ – Ian Jul 23 '14 at 20:36
  • $\begingroup$ It also crucially matters whether you're doing a series expansion at a regular singular point or not. The differential equation satisfied by elliptic integrals is a good example, if I remember right. $\endgroup$ – Semiclassical Jul 23 '14 at 20:40
  • $\begingroup$ I'm happy with an explanation just looking at regular singular points. $\endgroup$ – DavidButlerUofA Jul 23 '14 at 21:00
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    $\begingroup$ You're correct, since for ordinary points there's no need to wrroy about the indicial equation. I was thinking of power series solutions in general. $\endgroup$ – Semiclassical Jul 23 '14 at 22:20
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If you look at http://math.creighton.edu/nielsen/DE_Fall_2010/Series%20Solutions/Series_Solutions_Beamer.pdf, they write the resulting recurrence for one of the solutions as $a_n F(n+r) = E$, where $F(r)$ is the indicial equation, and I'm writing $E$ to abbreviate a complicated expression which depends on a variety of things, including $n$. If $F(n+r)=0$ for some $n$ and $E \neq 0$, then this isn't solvable.

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Each root from the indical equations may not always only find one group of the linearly independent solutions, sometimes we can find more than one group of the linearly independent solutions at the same time.

For the examples see Find the form of a second linear independent solution when the two roots of indicial equation are different by a integer and Solving a 2nd order differential equation by the Frobenius method.

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