Numerical Solution of Ordinary Differential Equations

Presentation on Numerical Solution of Ordinary Differential Equations

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

2. NUMERICAL SOLUTION OF ORDINARY DIFFERENTIAL EQUATIONS
CONTENTS
PowerSeriesapproximations
SolutionbyTaylorseries(Type1)
Euler’s Method
Runge-KuttaMethod

6. Substitutingthesederivativesandtruncatingtheseriesin(1)givetheapproximatesolutionatx.

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-KuttaMethod
Theuseofthepreviousmethodstosolvethedifferentialequationnumericallyisrestrictedduetoeitherslow
convergenceorduetolabourinvolved,especiallyinTaylorseriesmethod.But,inRunge-Kuttamethods,thederivatives
ofhigherorderarenotrequiredandwerequireonlythegivenfunctionvaluesatdifferentpoints.Sincethederivationof
fourthorderRunge-Kuttamethodistedious,wewillderiveRunge-kuttamethodofsecondorder.
Second order Runge-Kuttamethod (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=∆??????.
Second order R.K. algorithm

10. Since the derivation of third and fourth order Runge-Kuttaalgorithms are tedious, we state them below for use.
The third order Runge-Kuttamethod algorithm is given below:
Third order
R.K. algorithm
The fourth order Runge-Kuttamethod 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-Kuttamethod,

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-Kuttamethod 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