Finding regular singular point

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BearY
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Homework Statement


Show that ##x^2y''+sin(x)y'-y = 0## has a regular singular point at ##x=0##, determine the indicial equation and it's roots.

Homework Equations


For a DE in the form of ##y''+p(x)y'+q(x)y=0##, if both ##p(x)## and ##q(x)## are not analytic at ##x=x_0##, and both ##(x-x_0)p(x)## and ## (x-x_0)^2q(x)## are analytic at ##x=x_0##,
##x_0## is a singular regular point of the DE.

The Attempt at a Solution


I can't justify why ##\frac{sin(x)}{x}## is analytic at 0. Basically stucked at first step, so I am afraid no attempt worth mentioning.

Nevermind, I just realized that ##\frac{sin(x)}{x}## can be written as power series expension.
 
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BearY said:

Homework Statement


Show that ##x^2y''+sin(x)y'-y = 0## has a regular singular point at ##x=0##, determine the indicial equation and it's roots.

Homework Equations


For a DE in the form of ##y''+p(x)y'+q(x)y=0##, if both ##p(x)## and ##q(x)## are not analytic at ##x=x_0##, and both ##(x-x_0)p(x)## and ## (x-x_0)^2q(x)## are analytic at ##x=x_0##,
##x_0## is a singular regular point of the DE.

The Attempt at a Solution


I can't justify why ##\frac{sin(x)}{x}## is analytic at 0. Basically stucked at first step, so I am afraid no attempt worth mentioning.

Look at the Taylor series of ##\sin(x)##. Divide it by ##x##.
 
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