Basis functions of a differential equation, given boundary conditions

[rdfloyd]

First off, I've never taken a differential equations class. This is for my Math Methods for Physicists class, and we are on the topic of DE. Unfortunately, we didn't cover this much, so most of what I am about to show you comes from the professor giving me tips and my own common sense. I'd really like to learn this stuff, but I have no idea what I am doing.

Homework Statement

Find the basis functions that result from the differential equation u''(x)=-k^2u(x) and these boundary conditions:

u(0)=u(L)
u'(0)=u'(L)

Homework Equations

The Attempt at a Solution

I don't know how to code in Latex yet, so forgive me for having to use a picture of my work thus far. Again, most of this is from other's help and common sense. I couldn't begin to explain what is going on here, or its uses.

Since the image is too big to be displayed, here is the link to it on my imgur account: http://imgur.com/D8B3U

Thanks!

 

[HallsofIvy]

Since $e^{ikx}= cos(kx)+ i sin(kx)$ and $e^{-ikx}= cos(kx)- i sin(kx)$, you can use as your "basis functions" cos(kx) and sin(kx) rather than $e^{ikx}$ and $e^{-ikx}$. (That is normally done when all your conditions are in terms of real numbers.)

 

[rdfloyd]

I was originally going to do that (and basically ended up doing that anyways), but he told me I should use complex-exponentials instead.

What I ended up with was:

$u(x)=A \times Cos[(\frac{n \pi}{L}) x]$

I wish I had followed my gut instinct instead...Thanks for the help. I don't see a +reputation button (like on other forums), but I appreciate the help.I only have one more question: What was the purpose in all of this? What's the end result? I understand that the boundary conditions essentially setup a periodic function (looks a lot like the spring equation), but what does $u(x)$ really mean? And why doesn't B matter?