Consider the embedded methods where x = x(to) +=(F1 + F2+4F3) 11 (F + F + 4F 3 ) 6 1 x = x(to))+(F + F) (F + F) If Fhf (to, x(to)) F2 = hf (to+h, x(to) + F) F3 = hf(to+h/2, x(to) + (F1 + F2)/4). x(t) = x(t) and x(to). = x0 compute 1 and x as well as |x1 x(to+h) and |x1 = x(to + h)| where x(to + h) is the solution to the initial value problem evaluated at to + h.