# Finite difference schemes for an ODE

If we are given an equation $-y''+b y'=0$ how would I construct a forward and backward difference for it. Would I write out the general Taylor series and then rearrange terms to get the needed equation on the LHS?

Thank you.

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You are aware of how derivatives are replaced with finite differences, no? Doesn't your book discuss the usual replacements done? – J. M. Apr 15 '12 at 15:40

For 1st order finite difference you don't have to go all the way to Taylor series, start with

$f'(x) = \frac{f(x+h) - f(x)}{h}$ which is of course just the definition of a derivative.

Now if you have a gridsize of L, and you want to divide it evenly in n parts, then we have $h = L/n$, and we can write

$f'(i) = \frac{f(i+h) - f(i)}{h}$

where $i = 1, 2, 3, ...n$. So you pretty much just replace $f'(i)$ with $\frac{f(i+h) - f(i)}{h}$ and you've got a forward finite difference scheme. Backward would be $\frac{f(i) - f(i-h)}{h}$, and centered would be $\frac{f(i+h) - f(i-h)}{h}$.

Now to move onto what the second order replacements should look like, we can again go back to the definition of a derivative. Since $f''(x) = \frac{d (f'(x))}{dx}$.

So $f''(x) = \frac{ f'(x+h) - f'(x)}{h}$. Now use the 1st order replacement to get

$f''(x) = \frac{(f(x+h +h) - f(x+h))/h - (f(x+h) + f(x))/h}{h}$. Then

$f''(i) = \frac{f(x+2h) - 2f(x+h) + f(x)}{h^2}$ is the 2nd order forward finite difference scheme. You can play the same game and get the centered and backward 2nd order scheme.

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