Exploring the Finite and Infinite Solutions of ODEs with k < 1/4

[djh101]

Consider [itex]\frac{d^{2}y}{dx^{2}}+\frac{k}{x^{2}}y = 0[/itex]. 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:

[itex]y = Asin(\sqrt{k}ln(x)) + Bcos(\sqrt{k}ln(x))[/itex]​

And setting y = 0 gives:

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

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 [itex]\frac{d^{2}y}{dx^{2}}+\frac{k}{x^{2}}y = 0[/itex]. 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:

[itex]y = Asin(\sqrt{k}ln(x)) + Bcos(\sqrt{k}ln(x))[/itex]​

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:

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

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.