$\displaystyle (x^2+2)y'' + 2xy' = 0$

Verify $\displaystyle x=0$ is an ordinary point. Then find a lower bound for the radius of convergence of power series solutions about $\displaystyle x=0$

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- Jan 19th 2010, 10:03 AMCreebeEasy Question
$\displaystyle (x^2+2)y'' + 2xy' = 0$

Verify $\displaystyle x=0$ is an ordinary point. Then find a lower bound for the radius of convergence of power series solutions about $\displaystyle x=0$ - Jan 19th 2010, 10:31 AMTheEmptySet
If you put it in standard form you get

$\displaystyle y'' +\frac{2x}{x^2+2}y'=0$

Since the limit as $\displaystyle \lim_{x \to 0}\frac{2x}{x^2+2}=0$ is finite it is an ordinary point.

The power series will have radius of convergence of at least the distance to the nearest singularity in the complex plane.

$\displaystyle x^2+2=0 \iff x \pm i\sqrt{2}$

So the radius of convergence of at least $\displaystyle R=\sqrt{2}$