= + =

These results, the results for subsequent time steps for I = 4.0 s to 10.0 s, and the solution for M = 1.0 s are presented in Table 7.5.

The errors presented in Table 7.5 for the second-order modified midpoint method for M = 1.0 s are approximately 15 times smaller than the errors presented in Table 7.3 for the first-order explicit Euler method. This illustrates the advantage of the second-order method. To achieve a factor of 15 decrease in the error for a first-order method requires a reduction of 15 in the step size, which increases the number of derivative function evaluations by a factor of 15. The same reduction in error was achieved with the secondorder method at the expense of twice as many derivative function evaluations. An error analysis at I = 10.0 s gives

Ratio - E(111 = 2.0) _ 8.855320 - 4 64 - E(111 = 1.0) - 1.908094 - .