Simple ODE Solvers - Error Behaviour

y(t0) = y0 Here f(t, y) is a given function, t0 is a given initial time and y0 is a given initial value for y. The unknown in the problem is the function y(t). Two obvious considerations in deciding whether or not a given algorithm is of any practical value are (a) the amount of computational effort required to execute the algorithm and (b) the accuracy that this computational effort yields. For algorithms like our simple ODE solvers, the bulk of the computational effort usually goes into evaluating the function f(t, y). Euler’s method uses one evaluation of f(t, y) for each step; the improved Euler’s method uses two evaluations of f per step; the Runge–Kutta algorithm uses four evaluations of f per step. So Runge–Kutta costs four times as much work per step as does Euler. But this fact is extremely deceptive because, as we shall see, you typically get the same accuracy with a few steps of Runge–Kutta as you do with hundreds of steps of Euler. To get a first impression of the error behaviour of these methods, we apply them to a problem that we know the answer to. The solution to the first order constant coefficient linear initial value problem y′(t) = y − 2t