MY NO Consider the initial-value problem y' = 3y, y(0) = 1. The analytic solution is y(x) = e3x,
(a) Approximate y(0.1) using one step and the fourth-order RK4 method. (Round your answer to six decimal places.)
1.349838
(b) Find a bound for the local truncation error in y. (Round your answer to nine decimal places.)
0.000020256
(c) Compare the error in y, with your error bound.
The actual error, rounded to nine decimal places, is 0.000021308
, which is less than
the local truncatic
(d) Approximate y(0.1) using two steps and the RK4 method. (Round your answer to six decimal places.)
1.368076
(e) Verify that the global truncation error for the RK4 method is ocn4) by comparing the errors in parts (a) and (d).
The actual error for the answer in (d), rounded to nine decimal places, is 0.202524005 X . With global truncation e
h = 0.05 to be approximately one-sixteenth the error when h = 0.1. Comparing the two errors we see that thi