Radius of convergence quick help

SoulofLoneWlf
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quick help on this i seem to be missing some logic or process
determine the lower bound the radius of convergence of series solutions about the given X0

(2+x^2)y''-xy'+4y=0

xo=0
 
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Well, we can't really help you unless you start the process. Do you have any idea where to go from here?

EDIT: If you've got absolutely no idea where to start, you should just Google "radius of convergence series solutions."
 
nickmai123 said:
Well, we can't really help you unless you start the process. Do you have any idea where to go from here?

EDIT: If you've got absolutely no idea where to start, you should just Google "radius of convergence series solutions."

well i was seeing that (2+x^2) has no discontinuity
so shouldn't the radius be infinity?
 
What does discontinuities of P(x) have to do with anything? A hint would be writing the DE in the form y^{''}+P(x)y^{'}+Q(x)y=0 and using a Fuchs' Theorem.
 
nickmai123 said:
What does discontinuities of P(x) have to do with anything? A hint would be writing the DE in the form y^{''}+P(x)y^{'}+Q(x)y=0 and using a Fuchs' Theorem.

thanks guys just hammered through it think i got my head around it :)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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