1. Numerical Analysis
Lecture 36

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. Euler method is one of the oldest numerical methods used for integrating the ordinary differential equations. Though this method is not used in practice, its understanding will help us to gain insight into nature of predictor-corrector method

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

12. The integral of this equation is a curve in, ty-plane.
Here, we find successively y1, y2, …, ym; where ym is the value of y at t =tm = t0 +mh, m =1, 2,… and h being small.

13. Here we use a property that in a small interval, a curve is nearly a straight line. Thus at (t0, y0), we approximate the curve by a tangent at that point.
Therefore,
That is,

14. Hence, the value of y corresponding to t = t1 is given by

15. Similarly approximating the solution curve in the next interval (t1, t2) by a line through (t1, y1) having its slope f(t1, y1), we obtain

16. Thus, we obtain in general, the solution of the given differential equation in the form of a recurrence relation

17. Geometrically, this method has a very simple meaning
Geometrically, this method has a very simple meaning. The desired function curve is approximated by a polygon train, where the direction of each part is determined by the value of the function f (t, y) at its starting point.

18. Example
Given
with the initial condition y = 1 at t = 0. Using Euler method, find y approximately at x = 0.1, in five steps.

19. Solution
Since the number of steps are five, we shall proceed in steps of (0.1)/5 = 0.02.
Therefore, taking step size
h = 0.02, we shall compute the value of y at
t = 0.02, 0.04, 0.06, 0.08 and 0.1

20. Thus
where
Therefore,
Similarly,

21. Hence the value of y corresponding to t = 0.1 is 1.091

22. MODIFIED EULER’S METHOD

23. The modified Euler’s method gives greater improvement in accuracy over the original Euler’s method. Here, the core idea is that we use a line through (t0, y0) whose slope is the average of the slopes at (t0 ,y0 ) and (t1, y1(1))

24. Where y1(1)= y0 + hf (t0, y0)
is the value of y at t = t1 as obtained in Euler’s method, which approximates the curve in the interval (t0 , t1)

25. Figure

26. Geometrically, from Figure, if L1 is the tangent at (t0, y0), L2 is the line through (t1, y1(1)) of slope f(t1, y1(1)) and L- is the line through (t1, y1(1)) but with a slope equal to the average of f(t0,y0) and f(t1, y1(1)),….

27. the line L through (t0, y0) and parallel to is used to approximate the curve in the interval (t0, t1).
Thus, the ordinate of the point B will give the value of y1.

28. Now, the equation of the line AL is given by

29. Similarly proceeding, we arrive at the recurrence relation
This is the modified Euler’s method.

30. Example
Using modified Euler’s method, obtain the solution of the differential equation
with the initial condition
y0 = 1 at t0 = 0 for the range
in steps of 0.2

31. Solution
At first, we use Euler’s method to get

32. Then, we use modified Euler’s method to find

33. Similarly proceeding, we have from Euler’s method

34. Using modified Euler’s method, we get

35. Finally

36. Modified Euler’s method gives

37. Hence, the solution to the given problem is given by
0.2
0.4
0.6 y
1.2295
1.5225
1.8819

38. Numerical Analysis
Lecture 36