What Are the Solutions of ODEs When k < 1/4?

[djh101]

Consider $\frac{d^{2}y}{dx^{2}}+\frac{k}{x^{2}}y = 0$. Show that every nontrivial solution has an infinite number of positive zeroes if k > 1/4 and a finite number if k ≤ 1/4.

Solving gives:

$y = Asin(\sqrt{k}ln(x)) + Bcos(\sqrt{k}ln(x))$​

And setting y = 0 gives:

$tan(\sqrt{k}ln(x)) = -\frac{A}{B} = c$
$ln(x) = \frac{2n+1}{\sqrt{k}}arctan(c)$​

So I've mainly just rearranged everything a few times. When k ≤ 1/4, the last equation should hold for only a finite number of ns. Why, though? How would one go about proving this? The main thing I would think to look for would be a square root turning negative, but that doesn't seem to happen anywhere. Any ideas?

 

[Mark44]

Consider $\frac{d^{2}y}{dx^{2}}+\frac{k}{x^{2}}y = 0$. Show that every nontrivial solution has an infinite number of positive zeroes if k > 1/4 and a finite number if k ≤ 1/4.

Solving gives:

$y = Asin(\sqrt{k}ln(x)) + Bcos(\sqrt{k}ln(x))$​

Show what you did to get this. I believe you might be tacitly making assumptions about k to get to this point.

And setting y = 0 gives:

$tan(\sqrt{k}ln(x)) = -\frac{A}{B} = c$
$ln(x) = \frac{2n+1}{\sqrt{k}}arctan(c)$​

So I've mainly just rearranged everything a few times. When k ≤ 1/4, the last equation should hold for only a finite number of ns. Why, though? How would one go about proving this? The main thing I would think to look for would be a square root turning negative, but that doesn't seem to happen anywhere. Any ideas?

 

[vela]

Your y(x) doesn't satisfy the differential equation.

 

[djh101]

Sorry, I did a sloppy substitution. x = e^z gives y'' - y' + ky = 0, which has the needed 1 - 4k discriminate. Problem solved. Thanks.