# Euler Method

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```					                          Euler Method

Major: All Engineering Majors

Authors: Autar Kaw, Charlie Barker

http://numericalmethods.eng.usf.edu
Transforming Numerical Methods Education for STEM

12/10/2011                http://numericalmethods.eng.usf.edu    1
Euler Method

http://numericalmethods.eng.usf.edu
Euler’s Method
y

 f x, y , y0  y0
dy
True value
dx
y1, Predicted
Rise                                                                             value
Slope                                      x0,y0                              Φ
Run
y1  y 0
                                                    Step size, h
x1  x 0
 f  x0 , y 0                                                                                       x

y1  y 0  f x0 , y 0 x1  x0 
Figure 1 Graphical interpretation of the first step of Euler’s method
 y0  f  x0 , y0 h

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Euler’s Method
y

yi 1  yi  f  xi , yi h                                                    True Value

h  xi 1  xi                                                            yi+1, Predicted value

Φ
yi
h
Step size
x
xi                 xi+1

Figure 2. General graphical interpretation of Euler’s method

4                                                               http://numericalmethods.eng.usf.edu
How to write Ordinary Differential
Equation
How does one write a first order differential equation in the form of

 f  x, y 
dy
dx
Example

 2 y  1.3e  x , y0  5
dy
dx
is rewritten as

 1.3e  x  2 y, y0  5
dy
dx
In this case

f x, y   1.3e  x  2 y
5                                                          http://numericalmethods.eng.usf.edu
Example
A ball at 1200K is allowed to cool down in air at an ambient temperature
of 300K. Assuming heat is lost only due to radiation, the differential
equation for the temperature of the ball is given by
d
dt
           
 2.2067 1012  4  81108 , 0  1200K

Find the temperature at   t  480 seconds using Euler’s method. Assume a step size of
h  240 seconds.

6                                                              http://numericalmethods.eng.usf.edu
Solution
Step 1:
d
dt

 2.2067 10 12  4  81108           

f t ,   2.2067  10 12  4  81  10 8   
 i 1   i  f ti , i h
1   0  f t0 , 0 h
 1200  f 0,1200240
 1200   2.2067 1012 12004  81108 240
 1200   4.5579240
 106.09K
1   is the approximate temperature at t  t1  t0  h  0  240  240
 240   1  106 .09 K
7                                                             http://numericalmethods.eng.usf.edu
Solution Cont
Step 2:     For i  1,     t1  240 , 1  106 .09

 2  1  f t1 , 1 h
 106.09  f 240,106.09240
                 
 106.09   2.2067  1012 106.094  81 108 240  
 106.09  0.017595240
 110.32 K

 2 is the approximate temperature at t  t2  t1  h  240  240  480
 480    2  110 .32 K

8                                                           http://numericalmethods.eng.usf.edu
Solution Cont

The exact solution of the ordinary differential equation is given by the
solution of a non-linear equation as
  300
0.92593ln            1.8519 tan 1 0.00333   0.22067 103 t  2.9282
  300

The solution to this nonlinear equation at t=480 seconds is

 (480)  647.57K

9                                                            http://numericalmethods.eng.usf.edu
Comparison of Exact and
Numerical Solutions
1400

1200
Temperature, θ(K)

1000
Exact Solution
800

600

400
h=240
200

0
0   100   200        300         400    500
Time, t(sec)

Figure 3. Comparing exact and Euler’s method

10                                                                   http://numericalmethods.eng.usf.edu
Effect of step size
Table 1. Temperature at 480 seconds as a function of step size, h

Step, h    (480)          Et        |єt|%
480      −987.81     1635.4         252.54
240       110.32     537.26         82.964
120      546.77      100.80         15.566
60      614.97      32.607         5.0352
30      632.77      14.806         2.2864

 (480)  647.57K   (exact)

11                                                  http://numericalmethods.eng.usf.edu
Comparison with exact results
1500

1000              Exact solution
Temperature, θ(K)

500
h=120
h=240
0
0   100      200         300       400       500
-500
Tim e, t (sec)         h=480
-1000

-1500

Figure 4. Comparison of Euler’s method with exact solution for different step sizes

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Effects of step size on Euler’s
Method
800

400
Temperature,θ(K)

0
0   100   200       300        400      500
-400
Step size, h (s)
-800

-1200

Figure 5. Effect of step size in Euler’s method.

13                                                             http://numericalmethods.eng.usf.edu
Errors in Euler’s Method
It can be seen that Euler’s method has large errors. This can be illustrated using
Taylor series.
1 d2y                         1 d3y
y i 1  y i 
dy
xi 1  xi                 xi 1  xi  
2
xi 1  xi 3  ...
dx xi , yi                 2! dx 2 x , y                 3! dx 3 x , y
i   i                   i   i

yi 1  yi  f ( xi , yi )xi 1  xi        f ' ( xi , yi )xi 1  xi   f ' ' ( xi , yi )xi 1  xi   ...
1                              2 1                              3

2!                               3!
As you can see the first two terms of the Taylor series
yi 1  yi  f  xi , yi h are the Euler’s method.

The true error in the approximation is given by
f xi , yi  2 f xi , yi  3                Et  h 2
Et                 h               h  ...
2!               3!
14                                                                      http://numericalmethods.eng.usf.edu
For all resources on this topic such as digital audiovisual
lectures, primers, textbook chapters, multiple-choice
tests, worksheets in MATLAB, MATHEMATICA, MathCad
and MAPLE, blogs, related physical problems, please
visit

http://numericalmethods.eng.usf.edu/topics/euler_meth
od.html
THE END

http://numericalmethods.eng.usf.edu

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