Derivation of finite difference schemes for a boundary value problem. I have had no experience with differential equations before I was presented with this problem on a homework. The equation is:
$ -u'' + \beta u' = 0 $
$ u(0) = 0  $
$ u(1) = 1 $
I have found the general solution to be $ u = \frac{-1}{e^\beta - 1} + \frac{e^{\beta x}}{e^{\beta} - 1}$ and am now asked to derive the central, forward, and backward finite difference schemes (I am only worried about understanding central for now though).
Bear with me now, like I said I have not had any experience with differential equations. From my understanding, the central finite difference yields:
$ u''(x) \approx \frac{2u(x) - u(x+h) - u(x-h)}{h^2} $
and therefore $ -u'' + \beta u' = 0 $ becomes:
$ \frac{2u(x) - u(x+h) - u(x-h)}{h^2} + \beta u'(x) = 0 $
$ 2u(x) + u(x + h) - u(x - h) = -\beta u'(x)h^2 $
Am I on the right track so far? I am not sure what a complete "finite difference scheme" even is, so when should I stop? I have seen examples where the end result is a large matrix equation - is that required to answer this question?
 A: It looks like your overall approach is correct. Your second derivative central difference seems to be off by a sign though, the numerator should be $u(x+h)-2u(x)+u(x-h)$, see http://en.wikipedia.org/wiki/Finite_difference . 
You still need to expand the first derivative of $u$ as a finite difference. If the problem is asking for first order central differences as well, you would write $u'(x)\approx \frac{u(x+h/2)-u(x-h/2)}{h}$, or most likely for this problem is wanted $u'(x)\approx \frac{u(x+h)-u(x-h)}{2h}$. 
Notice that once you write all derivatives out, you get an equation that involves only $u(x)$, $u(x+h)$ and $u(x-h)$. This is effectively your "scheme" or rule for updating your $u$. If say your $u$ is a function of space, then you can write out a matrix equation. Here's how: if you are on the interval $[0,1]$, then you can write down the vector $[u(0),u(h),u(2h),\ldots,u(1-h),u(1)]$. Since your $u$ satisfies a linear equation involving $u(x)$, $u(x+h)$ and $u(x-h)$, you can think of this equation as an appropriate matrix multiplying the aforementioned vector. 
