Runge Kutta Method
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The document describes the Runge-Kutta method for numerically solving differential equations. It presents the formulas for the second-order and fourth-order Runge-Kutta methods. It then works through an example problem of solving the differential equation dy/dx = x + y with initial condition y(0) = 1, calculating the solutions y(0.1), y(0.2), and y(0.3) using the fourth-order Runge-Kutta method.
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