082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

622 Chapter 20 Ordinary differential equations II
Therefore,
y 1
=y 0
+0.1(x 0
+x 1
) = 1 +0.1(0 +0.1) = 1.01
y 2
=y 1
+0.1(x 1
+x 2
) = 1.01 +0.1(0.1 +0.2) = 1.04
and so on. The complete solution appears in Table 20.5. Check this for yourself. The
analytical solution is
∫
y = 2xdx=x 2 +c
Applyingtheconditiony(0) = 1givesc = 1,andsoy =x 2 +1.Valuesofthisfunction
arealsoshowninthetableandwenotethemarkedimprovementgivenbytheimproved
Euler method.
EXERCISES20.7
1 Apply the improved Eulermethod to Question 1 in
Exercises20.6.
2 Apply the improved Eulermethod to Question 2 in
Exercises20.6.
3 Apply the improved Eulermethod to Question 3 in
Exercises20.6.
Solutions
1
x i
y i (h = 0.5) y i (h = 0.25) y(exact)
3
t i v i (h = 0.005) v(exact)
2
2.00 1.0000 1.0000 1.0000
2.25 -- 1.3889 1.3901
2.50 1.8000 1.8057 1.8080
2.75 -- 2.2476 2.2509
3.00 2.7017 2.7123 2.7165
x i y i (h = 0.25) y(exact)
0 0.0000 0.0000
0.25 0.0313 0.0340
0.50 0.1416 0.1487
0 0.0000 0.0000
0.005 0.2500 0.2673
0.010 0.2813 0.3954
t i v i (h = 0.002) v(exact)
0 0.0000 0.0000
0.002 0.0588 0.0569
0.004 0.1903 0.1919
0.006 0.3273 0.3354
0.008 0.4032 0.4178
0.010 0.3777 0.3954
x i y i (h = 0.1) y(exact)
0 0.0000 0.0000
0.1 0.0050 0.0052
0.2 0.0210 0.0214
0.3 0.0492 0.0499
0.4 0.0909 0.0918
0.5 0.1474 0.1487