Why are Runge Kutta's method and Euler's so different? I am solving a system of linear equations $\underline {\dot A}=\underline A\cdot \underline x$ numerically. I have done this in the popular methods of of Euler and Runge Kutta. I have noticed quite a difference between the two methods in accuracy to the analytic solution.
Is there any reason for this?
 A: Expanding on the comment of Did: A method of order $p$ has an error term of $O(h^p)$ for step size $p$. For "tame" ODE (in contrast to e.g. stiff ODE) one can assume that the constants of the error terms are of similar magnitude. For simplicity in the example below, let the constants all be $1$.
Another concern for numerical methods is that truncation and cancellation errors accumulate, in general linearly in the number of steps. Thus there is a lower bound for useful step sizes of magnitude $\sqrt[p+1]{\mu}$ with $\mu$ the machine constant where the two error contributions balance. (The actual bound again also depends on the scales and magnitudes of the ODE functions.)
So in all simplifications, if one strives to obtain a global error of 1e-8 over an interval of length 1 for a machine constant of μ=1e-16, then with a method of order 1 like the Euler method one needs a step size of h=1e-8 (which is at the lower bound) and correspondingly 1e8=100 000 000 function evaluations. 
An order 4 method like classical Runge-Kutta needs h=1e-2 (lower bound h>6e-4) for the same global error target, and thus only 400 function evaluations.
On the other hand, the Euler method with 400 function evaluations and thus steps has a step size and error magnitude of h=2.5e-3
