1 = b0 + ∆t kb11 + 2(kb12 + kb13) + kb14 6 0, 36 + 2(0, 3597 + 0, 3597) + 0, 3594 = 1, 2 + 0, 01 · = 1, 2036 6 r1 = b0 + ∆t kr11 + 2(kr12 + kr13) + kr14 6 0, 14 + 2(0, 1414 + 0, 1414) + 0, 1428 = 0.7 + 0, 01 · = 0, 7014 6 Iteration 2: Nu bestemmes værdierne kb21 . . . kb24 og kr21 . . . kr24: kb21 = f(t0, b0) = (A − Br0)b0 = (1 − 1 · 0.7014)1, 2036 = 0.3594 kr21 = f(t0, b0) = (Cb0 − D)r0 = (1 · 1, 2036 − 1)0, 7014 = 0, 1428 kb22 = f t0 + ∆t 2 , b0 + ∆t 2 kb21 = (A − Br0)(b0 + ∆t 2 kb21) = 0, 01 (1 − 1 · 0, 7014)(1, 2036 + · 0, 3594) = 0, 3591 2 kr22 = f t0 + ∆t 2 , r0 + ∆t 2 kr21 = (Cb0 − D)(r0 + ∆t 2 kr21) = (1 · 1, 2036 − 1)(0, 7014 + 0, 01 2 · 0, 1428) = 0, 1442 kb23 = f t0 + ∆t 2 , b0 + ∆t 2 kb22 = (A − Br0)(b0 + ∆t 2 kb22) = 0, 01 (1 − 1 · 0, 7014)(1, 2036 + · 0, 3591) = 0, 3590 2 kr23 = f t0 + ∆t 2 , r0 + ∆t 2 kr22 = (Cb0 − D)(r0 + ∆t 2 kr22) = (1 · 1, 2036 − 1)(0, 7014 + 0, 01 2 · 0, 1442) = 0, 1442 kb24 = f (t0 + ∆, x0 + ∆t · kb23) = (A − Br0)(b0 + ∆t · kb23) = (1 − 1 · 0, 7014)(1, 2036 + 0, 01 ∗ 0, 3590) = 0, 3587 kr24 = f (t0 + ∆t, x0 + ∆t · kr23) = (Cb0 − D)(r0 + ∆t · kr23) = (1 · 1, 2036 − 1)(0, 7014 + 0, 01 · 0, 1442) = 0, 1456
2 = b1 + ∆t kb21 + 2(kb22 + kb23) + kb24 6 0, 3594 + 2(0, 3591 + 0, 3590) + 0, 3587 = 1, 2036 + 0, 01 · = 1, 2072 6 r2 = r1 + ∆t kr21 + 2(kr22 + kr23) + kr24 6 0, 1428 + 2(0, 1442 + 0, 1442) + 0, 1456 = 0.7014 + 0, 01 · = 0, 7029 6