Matakuliah
Tahun
: S0262-Analisis Numerik
: 2010
Solving Ordinary Differential Equation
Numerically
Pertemuan 8
Material Outline
• Ordinary differential equation
– Euler method
– Runge-Kutta method
DIFFERENTIAL EQUATIONS (ODE)
• Numerical Solution for ODE is absolutely needed for
many mathematical or engineering cases because
many cases in engineering fields that the analytical
solutions do not exist or really hard to be solved
analytically.
• In this session the ODE’s that will be discussed are
first degree ODE that has the general form as
follows:
dy
F ( x, y, y ' )  0, or y ' 
 f ( x, y )
dx
4
 EULER METHOD
 Euler Method in general is the method that is
intended to solve 1st degree ODE .
 This method is a method of step-by- step
with an initial value.
 The initial value usually is given as a part
of the problem. Example:
y'  f ( x, y), and y( x0 )  y0
5
 EULER METHOD
 The method will be explained by first reviewed the following
Taylor series:
h2
hn n
y ( x  h)  y ( x)  hy ' ( x) 
y ' ' ( x)   
y ' ( x)
2!
n!
 In the above eq., if the absolute value of h is very small
(h<1h2<<0), then the 3rd term and the following can be
neglected and yield the following eq:
y(x+h)  y(x) + h y’(x)
 Therefore, the error that is associated with the approximation
will depend on the magnitude of h and the type of the
function.
6
 EULER METHOD
 The Euler Method can also be derived from limit
definition as follow:
y ( x  h)  y ( x )
y ' ( x)  lim
h 0
h
 Therefore, the approximation value of the above
limit can be written as:
y ( x  h)  y ( x )
y ' ( x) 
 y ( x  h)  y ( x)  hy ' ( x)
h
 The accuracy of the approximation formula given
above depends on h, the smaller the h the more
accurate the approximation.
7
 EULER METHOD
If y(x), y(x+h), y(x+2h), … are written as y0, y1, y2, … , then the
above approximation formula can be written as:
,
1
0
0
,
2
1
1
y  y  hy
y  y  hy



,
yn  yn 1  hyn 1
Example: Use the Euler method to solve the following ODE :
y’(x,y) = x + y, y(0) = y0=0;
you may use h=0.2 and compare your result with
analytical solution  y(x)= ex - x -1
8
Answer: y’(x,y) = x + y,
y(0) = y0=0; h=0,2
Euler method:
yn+1  yn + 0,2 y’n
y1 y0 + 0,2 y’0
=0 + 0,2 y’(0,0)=0+0,2x0=0
y2 y1 + 0.2 y’1
= 0 + 0.2 (0.2) = 0.4
In which:
y1 = y(h)
y’1= y’(h,y1)=h+ y1= 0,2; etc
Note: Analytical solution

y(x)= ex - x -1 is found by the
theory of Calculus
9
 EULER METHOD
Solution:
n
xn
Euler
yn=yn-1+hy’n-1
Y’n=xn+yn
Exact
y(x)= ex - x -1
Error
O
0
0 (Initial)
0,000
0,000
0.000
1
0,2
0.000
0.200
0,021
0.021
2
0,4
0.040
0.440
0,092
0.052
3
0,6
0.128
0.688
0,222
0.094
4
0,8
0.274
0.946
0,426
0.152
5
1,0
0.489
1.215
0,718
0.229
6
1,2
7
1,4
10
RUNGE-KUTTA METHOD
 Aside from its simplicity, the numerical error
associated with Euler method is relatively high as
shown in the previous examples.
 Runge-Kutta Method is a good method that can
improve the weakness of Euler method.
 As in the previous discussion, the error in the Euler
method is due to the omission of the 3rd, 4th ,….
terms from Taylor Series in the approximation
formula.
11
 The formula in the Runge-Kutta Method is also
consists of 2 terms, however, in the 2nd term some
additional coefficients will be introduced as given in
the following formula:
 Fourth Order Runge-Kutta Method:
1
yn 1  yn  (k1n  2k1n  2k3n  k 4 n )
6
k1n  hy ' ( xn , yn )
k 2 n  hy ' ( xn  1 2 h, yn  1 2 k1n )
k3n  hy ' ( xn  1 2 h, yn  1 2 k 2 n )
k 4 n  hy ' ( xn  h, yn  k3n )
in which , xn 1  xn  h  x0  nh
12
RUNGE-KUTTA METHOD
 Example: Use the method of Runge-Kutta to solve the following
ODE :
y’(x,y) = x + y; y(0) = y0=0 ; take h=0.2
And compare your result to exact solution: y(x)= ex - x –1
Solution:
yn 1  yn  (k1n  2k 2 n  2k3n  k 4 n ) / 6
y1  y0  (k10  2k 20  2k30  k 40 ) / 6
k10  hy ' ( x0 , y0 )  0.2  0  0
k 20  hy ' ( x0  1 2 h, y0  1 2 k10 )  0.2 y ' (0.1, 0)  0.02
k30  hy ' ( x0  1 2 h, y0  1 2 k 20 )  0.2 y ' (0.1,0.01)  0.022
k 40  hy ' ( x0  h, y0  k30 )  0.2 y ' (0.2,0.022)  0.0444
y1  0  (0  0.04  0.044  0.0444) / 6  0.0214
13
RUNGE-KUTTA METHOD
 Solution :
y’(x,y) = x + y,
y(0) = y0=0; h=0.2
Note: Analytical solution
 y(x)= ex - x -1 is found by
the theory of Calculus
yn 1  yn  (k1n  2k 2 n  2k3n  k4 n ) / 6
y2  y1  (k11  2k 21  2k31  k 41 ) / 6
k11  hy ' ( x1 , y1 )  0.2 y ' (0.2,0.0214)  0.04428
k 21  hy ' ( x1  1 2 h, y1  1 2 k11 )  0.2 y ' (0.3, 0.04354)  0.068708
k31  hy ' ( x1  1 2 h, y1  1 2 k 21 )  0.2 y ' (0.3,0.055754)  0.0711508
k 41  hy ' ( x1  h, y1  k31 )  0.2 y ' (0.4,0.0925508)  0.09851016
y2  0.0214  (0.04428  0.137416  0.1423016  0.09851016) / 6
 0.091818
14
RUNGE-KUTTA METHOD
 Solution :
y’(x,y) = x + y,
y(0) = y0=0; h=0.2
yn 1  yn  (k1n  2k 2 n  2k3n  k 4 n ) / 6
y3  y2  (k12  2k 22  2k32  k 42 ) / 6
k12  hy ' ( x2 , y2 )  
k 22  hy ' ( x2  1 2 h, y2  1 2 k12 )  
k32  hy ' ( x2  1 2 h, y2  1 2 k 22 )  
k 42  hy ' ( x2  h, y2  k32 )  
y2  0.091818  (      ) / 6
15
 Solution :
y’(x,y) = x + y,
y(0) = y0=0; h=0.2
n
xn
R.K:yn=yn-1
+(k1+2k2+2k3+k4)/6
Exact: y(x)=
Error
ex - x -1
0
0
0.0000
0.0000
0.000
1
0.2
0.0214
0.0214
0.000
2
0.4
0.0918
0.0918
0.000
3
0.6
0.2221
0.2221
0.000
4
0.8
0.4255
0.4255
0.000
5
1.0
6
1.2
16
 Exercise:
Use the method of Runge-Kutta to solve the following ODE:
y ' ( x, y )  2 xy  1
y (0)  y0  0
 h  0.1
17