numerical solution of differential equations

[GeeTwo]

The key to efficiently numerically solving this sort of differential equation is to understand the size/scale of the various derivatives. If the second derivative is consistently small, a linear (simple Euler) integration can provide a decent prediction at a reasonably large range step. As the second derivative increases, the range step must be reduced and/or the second derivative must be included in the calculation for each range step; midpoint is one way to do this. If the third derivative is large, you need to make the steps even smaller! Fortunately, the importance of the nth derivative is scaled by 1/n!, that is, the reciprocal of n factorial (google Taylor Expansion if you aren't familiar with it).

If you want an extreme case to work with that has a simple analytic solution, try this one:

x''(t) = 2x + 2x³
Initial conditions: x(0) = 0, x'(0) = 1.

The analytic solution is simply:

x=tan t
x'=sec²t
x''=2 tant sec²t = 2tant + 2tan³t= 2x + 2x³

Using a fixed step size in t, and a finite number of terms in the Taylor expansion with Euler-like integration, the numeric solution will never reach infinity, whereas the analytic solution reaches infinity at pi/2. Try to find solutions and step sizes which suitably predict pi/2, based on x being greater than, say, 10¹⁰⁰.

[Ether]

Originally Posted by Hitchhiker 42

I tried it using the slope at the end of the interval instead of at the beginning.

It's not clear what the point of that spreadsheet is. You are using the analytical solution for x' to compute an Euler step.

You weren't intending to compare that result to the Midpoint method, were you?