# How is this a 4th order ODE?

1. Jan 31, 2010

### bravelittlemu

I am not interested in the solution, but I am curious what makes: uiv = u, a 4th-order ordinary differential equation. 'i' is the square-root of -1, v is some element of the reals, and the differentiating variable is x.

cheers

2. Feb 1, 2010

### MaxL

Really interesting! Thanks for bringing this to my attention!

I'm not familiar with this problem, so I'm not offering an answer, but here's what I did:

First restate

u^i*v = e^(ln(u)*i*v)=u

If you take the x derivative of both sides, you get

i*e^(ln(u)*iv)*(u'v/u + v'ln(u))=u'

recognize that the first part is u by the original eqn, then divide both sides by u',

iv +iv' u/u' = 1

That guy is separable, so

v'/(1-iv) = u'/(u*ln(u)) = a set of constants

Looks like a first order equation to me.

But I'm wondering if maybe that i allows higher orders? Like the way that i^i = (-e^(i*2*pi*n)*i/2) = -e^(pi*n) for any integer value of n? But that trick only works because i has an absolute value of 1, which isn't necessarily true of u or v. Also, I have no idea why n would stop at 4.

Where did you hear it was a 4th order ODE?

3. Feb 1, 2010

### bravelittlemu

My text book. :D
Birkhoff, Rota, Ordinary Differnential Equations (4th), 1989, pg 73 Example 1

4. Feb 1, 2010

### hamster143

This is not a differential equation at all. It's simply

$$u^{iv-1} = 1$$

$$(iv-1) \ln (re^{i\phi}) = 0$$

7. Feb 1, 2010

### bravelittlemu

That is it. The book uses z = μ + iυ (nu) to represent a complex number and upon closer inspection (holding the book up to my face) the exponent is iv (in cursive, go figure.).

Thanks all!

8. Feb 1, 2010

### MaxL

Hahaha, oh man that's hilarious.

9. Feb 4, 2010

:rofl: