Determine the lower bound of the radius of convergence of each of the following equations:
$$(2+x^2)\ddot y- x\dot y + 4y = 0, \space\space x_0 = 0 $$
$$(3 – x^2)\ddot y – 3x\dot y – y = 0, \space\space x_0 = 0 $$
If these equations are in the form:
$$P(x)\ddot y + Q(x)\dot y + R(x)y = 0$$
and for Equation 1:
$$p(x) = Q(x) / P(x) = \frac{-x}{2+x^2}$$
$$q(x) = R(x) / P(x) = \frac{4}{2+x^2}$$
Equation 2:
$$p(x) = Q(x) / P(x) = \frac{-3x}{3-x^2}$$
$$q(x) = R(x) / P(x) = \frac{-1}{3-x^2}$$
Is the minimum radius of convergence as simple as determining the distance to the poles for $p(x)$ and $q(x)$ ?
So the answers would be: $\sqrt 2$ and $\sqrt 3$ respectively?
Or do I have to apply something like the ratio test for absolute convergence? i.e.
$$\left|x-x_0\right| \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n} \right |$$
I have solved for the power series solutions, so the above is also easy to compute
Thanks!
Best Answer
Yes, that is totally correct. While the first equation has solutions with the whole of $\Bbb R$ as domain, for the power series also the complex roots of the leading coefficient count for limiting the radius of convergence.