Solutions to initial value problems of first order using Runge kutta method

1. Runge Kutta Method by Ch.M.Verriyya.Naidu is licensed under a Creative Commons Attribution 4.0 International
License.

7. RUNGE-KUTTA’S METHOD
1. Second order
𝑦 𝑛+1 = 𝑦𝑛 +
1
2
𝑘1 + 𝑘2
where 𝑘1 = ℎ𝑓 𝑥 𝑛, 𝑦𝑛
𝑘2 = ℎ𝑓 𝑥 𝑛 + ℎ, 𝑦𝑛 + 𝑘1
2. Fourth order
𝑦 𝑛+1 = 𝑦𝑛 +
1
6
𝑘1 + 2𝑘2 + 2𝑘3 + 𝑘4
where 𝑘1 = ℎ𝑓 𝑥 𝑛, 𝑦𝑛
𝑘2 = ℎ𝑓 𝑥 𝑛 +
ℎ
2
, 𝑦𝑛 +
𝑘1
2
𝑘3 = ℎ𝑓 𝑥 𝑛 +
ℎ
2
, 𝑦𝑛 +
𝑘2
2
𝑘4 = ℎ𝑓 𝑥 𝑛 + ℎ, 𝑦𝑛 + 𝑘3

8. Problem 1
Given
𝑑𝑦
𝑑𝑥
= 𝑥 + 𝑦, with initial conditions 𝑦 0 = 1. Choose ℎ = 0.
𝑦 0.1 , 𝑦 0.2 and 𝑦 0.3 using Runge-Kutta’s method of fourth o

9. Problem 1
Given
𝑑𝑦
𝑑𝑥
= 𝑥 + 𝑦, with initial conditions 𝑦 0 = 1. Choose ℎ = 0.
𝑦 0.1 , 𝑦 0.2 and 𝑦 0.3 using Runge-Kutta’s method of fourth o
Putting 𝑛 = 0 in Runge-Kutta’s formula for fourth order, we get

10. Problem 1
Given
𝑑𝑦
𝑑𝑥
= 𝑥 + 𝑦, with initial conditions 𝑦 0 = 1. Choose ℎ = 0.
𝑦 0.1 , 𝑦 0.2 and 𝑦 0.3 using Runge-Kutta’s method of fourth o
Putting 𝑛 = 0 in Runge-Kutta’s formula for fourth order, we get
𝑦1 = 𝑦0 +
1
6
𝑘1 + 2𝑘2 + 2𝑘3 + 𝑘4

11. Problem 1
Given
𝑑𝑦
𝑑𝑥
= 𝑥 + 𝑦, with initial conditions 𝑦 0 = 1. Choose ℎ = 0.
𝑦 0.1 , 𝑦 0.2 and 𝑦 0.3 using Runge-Kutta’s method of fourth o
Putting 𝑛 = 0 in Runge-Kutta’s formula for fourth order, we get
𝑦1 = 𝑦0 +
1
6
𝑘1 + 2𝑘2 + 2𝑘3 + 𝑘4
where 𝑘1 = ℎ𝑓 𝑥0, 𝑦0 = 0.1 0 + 1 = 0.1

12. Problem 1
Given
𝑑𝑦
𝑑𝑥
= 𝑥 + 𝑦, with initial conditions 𝑦 0 = 1. Choose ℎ = 0.
𝑦 0.1 , 𝑦 0.2 and 𝑦 0.3 using Runge-Kutta’s method of fourth o
Putting 𝑛 = 0 in Runge-Kutta’s formula for fourth order, we get
𝑦1 = 𝑦0 +
1
6
𝑘1 + 2𝑘2 + 2𝑘3 + 𝑘4
where 𝑘1 = ℎ𝑓 𝑥0, 𝑦0 = 0.1 0 + 1 = 0.1
𝑘2 = ℎ𝑓 𝑥0 +
ℎ
2
, 𝑦0 +
𝑘1
2
= 0.1 0.05 + 1.05

13. Problem 1
Given
𝑑𝑦
𝑑𝑥
= 𝑥 + 𝑦, with initial conditions 𝑦 0 = 1. Choose ℎ = 0.
𝑦 0.1 , 𝑦 0.2 and 𝑦 0.3 using Runge-Kutta’s method of fourth o
Putting 𝑛 = 0 in Runge-Kutta’s formula for fourth order, we get
𝑦1 = 𝑦0 +
1
6
𝑘1 + 2𝑘2 + 2𝑘3 + 𝑘4
where 𝑘1 = ℎ𝑓 𝑥0, 𝑦0 = 0.1 0 + 1 = 0.1
𝑘2 = ℎ𝑓 𝑥0 +
ℎ
2
, 𝑦0 +
𝑘1
2
= 0.1 0.05 + 1.05
𝑘3 = ℎ𝑓 𝑥0 +
ℎ
2
, 𝑦0 +
𝑘2
2
= 0.1 0.05 + 1.05

14. Problem 1
Given
𝑑𝑦
𝑑𝑥
= 𝑥 + 𝑦, with initial conditions 𝑦 0 = 1. Choose ℎ = 0.
𝑦 0.1 , 𝑦 0.2 and 𝑦 0.3 using Runge-Kutta’s method of fourth o
Putting 𝑛 = 0 in Runge-Kutta’s formula for fourth order, we get
𝑦1 = 𝑦0 +
1
6
𝑘1 + 2𝑘2 + 2𝑘3 + 𝑘4
where 𝑘1 = ℎ𝑓 𝑥0, 𝑦0 = 0.1 0 + 1 = 0.1
𝑘2 = ℎ𝑓 𝑥0 +
ℎ
2
, 𝑦0 +
𝑘1
2
= 0.1 0.05 + 1.05
𝑘3 = ℎ𝑓 𝑥0 +
ℎ
2
, 𝑦0 +
𝑘2
2
= 0.1 0.05 + 1.05
𝑘4 = ℎ𝑓 𝑥0 + ℎ, 𝑦0 + 𝑘3 = 0.1 0.1 + 1.110

15. Problem 1
Given
𝑑𝑦
𝑑𝑥
= 𝑥 + 𝑦, with initial conditions 𝑦 0 = 1. Choose ℎ = 0.
𝑦 0.1 , 𝑦 0.2 and 𝑦 0.3 using Runge-Kutta’s method of fourth o
Putting 𝑛 = 0 in Runge-Kutta’s formula for fourth order, we get
𝑦1 = 𝑦0 +
1
6
𝑘1 + 2𝑘2 + 2𝑘3 + 𝑘4
where 𝑘1 = ℎ𝑓 𝑥0, 𝑦0 = 0.1 0 + 1 = 0.1
𝑘2 = ℎ𝑓 𝑥0 +
ℎ
2
, 𝑦0 +
𝑘1
2
= 0.1 0.05 + 1.05
𝑘3 = ℎ𝑓 𝑥0 +
ℎ
2
, 𝑦0 +
𝑘2
2
= 0.1 0.05 + 1.05
𝑘4 = ℎ𝑓 𝑥0 + ℎ, 𝑦0 + 𝑘3 = 0.1 0.1 + 1.110
𝑦 0.1 = 𝑦1 = 1 +
1
6
0.1 + 0.22 + 0.221 + 0.12105 =1.11034

16. Putting 𝑛 = 1 in Runge-Kutta’s formula for fourth order, we get
𝑦2 = 𝑦1 +
1
6
𝑘1 + 2𝑘2 + 2𝑘3 + 𝑘4

17. Putting 𝑛 = 1 in Runge-Kutta’s formula for fourth order, we get
𝑦2 = 𝑦1 +
1
6
𝑘1 + 2𝑘2 + 2𝑘3 + 𝑘4
where 𝑘1 = ℎ𝑓 𝑥1, 𝑦1 = 0.1 0.1 + 1.11034 = 0.12

18. Putting 𝑛 = 1 in Runge-Kutta’s formula for fourth order, we get
𝑦2 = 𝑦1 +
1
6
𝑘1 + 2𝑘2 + 2𝑘3 + 𝑘4
where 𝑘1 = ℎ𝑓 𝑥1, 𝑦1 = 0.1 0.1 + 1.11034 = 0.12
𝑘2 = ℎ𝑓 𝑥1 +
ℎ
2
, 𝑦1 +
𝑘1
2
= 0.13208

19. Putting 𝑛 = 1 in Runge-Kutta’s formula for fourth order, we get
𝑦2 = 𝑦1 +
1
6
𝑘1 + 2𝑘2 + 2𝑘3 + 𝑘4
where 𝑘1 = ℎ𝑓 𝑥1, 𝑦1 = 0.1 0.1 + 1.11034 = 0.12
𝑘2 = ℎ𝑓 𝑥1 +
ℎ
2
, 𝑦1 +
𝑘1
2
= 0.13208
𝑘3 = ℎ𝑓 𝑥1 +
ℎ
2
, 𝑦1 +
𝑘2
2
= 0.13263
𝑘4 = ℎ𝑓 𝑥1 + ℎ, 𝑦1 + 𝑘3 = 0.14429

20. Putting 𝑛 = 1 in Runge-Kutta’s formula for fourth order, we get
𝑦2 = 𝑦1 +
1
6
𝑘1 + 2𝑘2 + 2𝑘3 + 𝑘4
where 𝑘1 = ℎ𝑓 𝑥1, 𝑦1 = 0.1 0.1 + 1.11034 = 0.12
𝑘2 = ℎ𝑓 𝑥1 +
ℎ
2
, 𝑦1 +
𝑘1
2
= 0.13208
𝑘3 = ℎ𝑓 𝑥1 +
ℎ
2
, 𝑦1 +
𝑘2
2
= 0.13263
𝑘4 = ℎ𝑓 𝑥1 + ℎ, 𝑦1 + 𝑘3 = 0.14429
𝑦 0.2 = 𝑦2 = 1.2428

21. Putting 𝑛 = 1 in Runge-Kutta’s formula for fourth order, we get
𝑦2 = 𝑦1 +
1
6
𝑘1 + 2𝑘2 + 2𝑘3 + 𝑘4
where 𝑘1 = ℎ𝑓 𝑥1, 𝑦1 = 0.1 0.1 + 1.11034 = 0.12
𝑘2 = ℎ𝑓 𝑥1 +
ℎ
2
, 𝑦1 +
𝑘1
2
= 0.13208
𝑘3 = ℎ𝑓 𝑥1 +
ℎ
2
, 𝑦1 +
𝑘2
2
= 0.13263
𝑘4 = ℎ𝑓 𝑥1 + ℎ, 𝑦1 + 𝑘3 = 0.14429
𝑦 0.2 = 𝑦2 = 1.2428
𝑛 = 2 will give 𝑦 0.3 = 1.399711
𝑘1 = 0.14428, 𝑘2 = 0.156494, 𝑘3 = 0.157105, 𝑘4 = 0.169990