Thread: Power Series Solutions of linear DE

1. Power Series Solutions of linear DE

Without actually solving the DE, find a lower bound for the radius of convergence of power series solutions about the ordinary points x=0 and x=1

DE: (x^2+25)y''+2xy'+y=0

Is there a quicker way to solve this other than substituting y= Σcnx^n into the DE and then combine them to one power series and then using the ratio test to find the radius???

2. A power series solution about a regular point will converge for all x up to the nearest singular point and so the radius of convergence is the absolute vaule of the difference of the two points. Since $\displaystyle x^2+ 25$ is never 0 this equation has no singular points and a power series solution,about any point, converges for all x.

3. Hi. Seems to me the equation $\displaystyle (x^2+25)y''+2xy'+y=0$ has singular points at $\displaystyle \pm 5 i$ and therefore the power series solution centered at the origin would have a radius of convergence of 5 and the series centered at the point $\displaystyle x=1$ would have a convergence radius of $\displaystyle \sqrt{26}$, the distance to $\displaystyle 5i$. Or no?

4. Originally Posted by shawsend
Hi. Seems to me the equation $\displaystyle (x^2+25)y''+2xy'+y=0$ has singular points at $\displaystyle \pm 5 i$ and therefore the power series solution centered at the origin would have a radius of convergence of 5 and the series centered at the point $\displaystyle x=1$ would have a convergence radius of $\displaystyle \sqrt{26}$, the distance to $\displaystyle 5i$. Or no?
Hmm so I looked up the answers for the question and for x=0 its 5 but for x=1 its 4...any idea how to explain that?

Another same question with different DE: (x^2-2x+10)y''+xy'-4y=0
Using your method the solutions are x=1 +-3i So for x=0 radius= sqrt(10) and for x=1 the radius is 3?

Hmm so I looked up the answers for the question and for x=0 its 5 but for x=1 its 4...any idea how to explain that?

Another same question with different DE: (x^2-2x+10)y''+xy'-4y=0
Using your method the solutions are x=1 +-3i So for x=0 radius= sqrt(10) and for x=1 the radius is 3?
Alright, this is what I'm gonna' do. First off, Hall is better at this than me so that's why I was asking. But let's consider the power series for $\displaystyle (x^2+25)y''+2xy'+y=0$ centered at x=1. We then let $\displaystyle v=x-1$ and obtain the new DE:

$\displaystyle [(v+1)^2+25]\frac{d^2y}{dv^2}+2(v+1)\frac{dy}{dv}+y=0$ and now solve that one at $\displaystyle v=0$ and we obtain a solution:

$\displaystyle y(v)=\sum_{n=0}^{\infty}a_nv^n$

which to me seems like it would converge in v up to the nearest singular point which is $\displaystyle v=-1\pm 5i$ so $\displaystyle R=\sqrt{26}$ in v. Now we substitute $\displaystyle v=x-1$:

$\displaystyle y(x)=\sum_{n=0}^{\infty}a_n(x-1)^n$

and that series, again to me, would now converge for $\displaystyle |x-1|<\sqrt{26}$ or $\displaystyle R=\sqrt{26}-1\approx 4.1$

Tell you what, that what I'm puttin' on the test and if push comes to shove, we could actually solve the DE, obtain the power series and confirm it's radius of convergence either directly or empirically.