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Consider an ODE of the form $y'' + P(x)y' + Q(x)y = 0$ where $(x-x_0)P(x)$ and $(x-x_0)^2 Q(x)$ are analytic functions on $x_0$. The Frobenius Theorem asserts that there exist at least one solution of the form:

$y=(x - x_0)^r \cdot \sum_{k = 0}^\infty a_k (x - x_0)^k $

where $r$ is an incognit, there is a formula for finding the $r$?

In some books appears, when $x_0 = 0$ that changes something? If this is the case, the formula is given by $ r(r-1) + p_0 r + q_0 = 0 $ where $q_0, p_0$ are the first terms of $ \begin{align*} P(x) &= (x-x_0)^{-1} \cdot \sum_{k = 0}^\infty p_k (x-x_0)^k \\ Q(x) &= (x-x_0)^{-2} \cdot \sum_{k = 0}^\infty q_k (x-x_0)^k \end{align*} $ Thanks, I only want to know that.

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The indicial equation does not change when $x_0 \ne 0$. For a nice and simple derivation of the indicial equation see page A52-A54 of the following reference

Nakhle H. Asmar, Partial Differential Equations with Fourier Series and Boundary Value Problem, Second Edition, Pages A52-A54

Also a more thorough explanation about the Frobenius method can be found in sections 5.5 and 5.6 of this book

William E. Boyce, Richard C. Diprima, Elementary Differential Equations and Boundary Value Problems, Tenth Edition, Section 5.4-5-5

To conclude, whether $x_0$ is $0$ or not, the indicial equation is

$r(r-1) + p_0 r + q_0 = 0$