1. Numerical Analysis
Lecture 38

2. Chapter 7 Ordinary Differential Equations

3. Introduction Taylor Series Euler Method Runge-Kutta Method Predictor Corrector Method

4. TAYLOR’S SERIES
METHOD

5. We considered an initial value problem described by

6. We expanded y (t ) by Taylor’s series about the point t = t0 and obtain

7. Noting that f is an implicit function of y, we have

8. Similarly

9. EULER METHOD

10. Consider the differential equation of first order with the initial condition y(t0) = y0.

11. The value of y corresponding to t = t1

12. Similarly

13. Then, we obtained the solution in the form of a recurrence relation

14. MODIFIED EULER’S METHOD

15. The recurrence relation
is the modified Euler’s method.

16. RUNGE – KUTTA METHODS

17. We considered the IVP
We also defined
and took the weighted average of k1 and k2 and added to yn to get yn+1

18. We obtained
Implying

19. We considered two cases,
Case I We choose W2 = 1/3,
then W1 = 2/3 and

20. Case II: We considered
W2 = ½, then W1 = ½ and
Then

21. Finally, we obtain
The expression can further be simplified to

22. Therefore, the expression for local truncation error is given by

23. The fourth-order R-K method was described as

24. where

25. Example
Use the following second order Runge-Kutta method described by

26. where
and
and find the numerical solution of the initial value problem described as
at x = 0.4 and taking h = 0.2.

27. Solution
Here
We calculate

28. Now, using the given R-K method, we get
Now, taking x1 = 0.2,
y1 = 1.24, we calculate

29. Again using the given R-K method, we obtain

30. Example Solve the following differential equation
with the initial condition y(0) = 1, using fourth- order Runge-Kutta method from t = 0 to t = 0.4 taking h = 0.1

31. Solution
The fourth-order Runge-Kutta method is described as
(1)
where

32. In this problem,
As a first step, we calculate

33. Now, we compute from
Therefore y(0.1) = y1=1.1103
In the second step, we have to find y2 = y(0.2)

34. We compute

35. From Equation (1), we see that
Similarly we calculate,

36. Using equation (1), we compute
Finally, we calculate

37. Using them in equation (1), we get
which is the required result.

38. RUNGE – KUTTA METHOD FOR SOLVING A SYSTEM OF EQUATIONS

39. The fourth-order Runge-Kutta method can be extended to numerically solve the higher-order ordinary differential equations- linear or non-linear

40. For illustration, let us consider a second order ordinary differential equation of the form

41. Using the substitution
this equation can be reduced to two first-order simultaneous differential equations, as given by

42. Now, we can directly write down the Runge-Kutta fourth-order formulae for solving the system.
Let the initial conditions of the above system be given by

43. Then, we define

44. Now, using the initial conditions yn, pn and 4th -order R-K formula, we compute

45. This method can be extended on similar lines to solve system of n first order differential equations.

46. Example Solve the following equation
Using 4th order Runge-Kutta method for x = 0.2, with the initial values y(0) = 1, y’(0)=0

47. Solution
Let
Then
Thus, the given equation reduced to two first-order equations.

48. In the present problem, we are given
x0 = 0, y0 = 1, p0 =yo’ =0
Taking h = 0.2, we compute

51. Now, y (0.2) = y1 is given by

52. Therefore, the required solution is

53. Numerical Analysis
Lecture 38