Presentation on Numerical Solution of Ordinary Differential Equations

1. NUMERICAL METHODS
(06CT42)
Dr. N. Meenakshi Sundaram
Asst. Prof. of Physics
Vivekananda College
Tiruvedakam West - Madurai

2. NUMERICAL SOLUTION OF ORDINARY DIFFERENTIAL EQUATIONS
CONTENTS
Power Series approximations
Solution by Taylor series (Type 1)
Euler’s Method
Runge-Kutta Method

6. Substituting these derivatives and truncating the series in (1) give the approximate solution at x.

7. Euler’s Method
In solving a first order differential equation by numerical methods, we come across two types of solutions:
• A series solution of y in terms of x, which will yield the value of y at a particular value of x by direct substitution in
the series solution.
• Values of y at specified values of x.

8. Runge-Kutta Method
The use of the previous methods to solve the differential equation numerically is restricted due to either slow
convergence or due to labour involved, especially in Taylor series method. But, in Runge-Kutta methods, the derivatives
of higher order are not required and we require only the given function values at different points. Since the derivation of
fourth order Runge-Kutta method is tedious, we will derive Runge-kutta method of second order.
Second order Runge-Kutta method (for first order O.D.E.)
By Taylor series

9. where a, b and m are constants to be determined to get the better accuracy of
where ℎ =∆𝑥.
Second order R.K. algorithm

10. Since the derivation of third and fourth order Runge-Kutta algorithms are tedious, we state them below for use.
The third order Runge-Kutta method algorithm is given below:
Third order
R.K. algorithm
The fourth order Runge-Kutta method algorithm is mostly used in problem unless other mentioned. It is

11. By Taylor Series Method,

13. By Taylor Series Method,

14. =0.3486875+(0.1)(4.012887)+0.01/2 (11.341286)+(0.001/6)(25.99808)+⋯
=0.8110156≈0.811 (three digits)
The exact value of y (0.1) = 0.3486955
and y (0.2) = 0.8112658.
Solution. h=0.2, ,
,
By Euler algorithm

20. Solution.
=(0.1)f(0.1,0.9005)=-0.0982.

23. By Taylor series method,

24. =0.1+0.01+0.00033+0.00000833+0.000000166+⋯
y(1.1)=0.11033847
By Taylor series

25. =0.11033847+0.121033847+2.21033847(0.005+0.0016666+⋯)
=0.24280160.

26. By Taylor series.

27. =0.194752003 + 0.18441984 - 0.0151668737 - 0.00451341243 + 0.00039360239
=0.35988515

28. =1 + (0.01) (-1) = 1-0.01=0.99.
=0.99 +(0.01) (-0.99) = 0.9801

29. x 0 0.01 0.02 0.03 0.04
Y 1 0.9900 0.9801 0.9703 0.9606
Exact y 1 0.9900 0.9802 0.9704 0.9608
Tabular values (step values) are
since y=e-x is the exact solution.
By fourth order Runge-Kutta method,

30. =(0.1) f (0.05,1.055)
=(0.1)(0.05+1.055)=0.1105
=(0.1) f (0.1,1.1105) = 0.12105

31. Now starting from
=0.144298048

32. =1.110342+1/6(0.794781008)=1.2428055
y(0.2)=1.2428055
Correct to four decimal places, y (0.2) =1.2428.

33. =(0.1)f(0.65,1.8071)
=0.1385
=(0.1)f(0.7,1.8764)
=0.1386

34. By Runge-Kutta method of fourth order,
=(0.1)f(0.75,1.9455)
=0.1383

36. 2. What is Truncation error?
Truncation errors are the errors that result from using an approximation in place of an exact mathematical
procedure.

38. Reference Text Book: Numerical Methods – P.Kandasamy, K.Thilagavathy & K.Gunavathi,
S.Chand & Company Ltd., New Delhi, 2014.

39. Thank you