MHB Are the Singular Points of the Chebyshev Equation Regular?

Poirot1
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I have computed the singular points of Chebyshev equation to be x= 1, -1. What is the best way to find whether they are regular? Thanks.
 
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The Chebisheff DE is...

$\displaystyle y^{\ ''} - \frac{x}{1-x^{2}}\ y^{\ '} + \frac{\alpha^{2}}{1-x^{2}}\ y= y^{\ ''} + p(x)\ y^{\ '} + q(x)\ y=0$ (1)

If $x_{0}$ is a singularity of p(x) and q(x) and both the limits...

$\displaystyle \lim_{x \rightarrow x_{0}} (x-x_{0})\ p(x)$

$\displaystyle \lim_{x \rightarrow x_{0}} (x-x_{0})^{2}\ q(x)$ (2)

... exist finite, then $x_{0}$ is a regular singular point. You can verify that $x_{0}=1$ and $x_{0}=-1$ are both regular singular points...

Kind regards

$\chi$ $\sigma$
 
Thanks
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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