Euler Method
Major: All Engineering Majors
Authors: Autar Kaw, Charlie Barker
http://numericalmethods.eng.usf.edu
Transforming Numerical Methods Education for STEM
Undergraduates
3/22/2016
http://numericalmethods.eng.usf.edu
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Euler Method
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Euler’s Method
y
dy
 f x, y , y 0  y 0
dx
Slope

Rise
Run

y1  y0
x1  x0
 f  x0 , y 0 
y1  y0  f x0 , y0 x1  x0 
 y0  f x0 , y0 h
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True value
Φ
x0,y0
y1, Predicted
value
Step size, h
x
Figure 1 Graphical interpretation of the first step of Euler’s method
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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
xi
xi+1
x
Figure 2. General graphical interpretation of Euler’s method
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How to write Ordinary Differential
Equation
How does one write a first order differential equation in the form of
dy
 f  x, y 
dx
Example
dy
 2 y  1.3e  x , y 0   5
dx
is rewritten as
dy
 1.3e  x  2 y, y 0   5
dx
In this case
f x, y   1.3e  x  2 y
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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
 2.2067 10 12  4  81108 , 0  1200 K
dt

Find the temperature at

t  480 seconds using Euler’s method. Assume a step size of
h  240 seconds.
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Solution
Step 1:
d
 2.2067  10 12  4  81 10 8
dt



f t ,   2.2067 10 12  4  81108

 i 1   i  f ti ,  i h
1   0  f t0 ,  0 h
 1200  f 0,1200 240



 1200   2.2067 10 12 1200 4  81108 240
 1200   4.5579 240
1
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 106.09 K
is the approximate temperature at t  t1  t0  h  0  240  240
 240  1  106.09K
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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 12 106.09 4  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  110.32K
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Solution Cont
The exact solution of the ordinary differential equation is given by the
solution of a non-linear equation as
0.92593 ln
  300
 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)  647.57 K
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Comparison of Exact and
Numerical Solutions
Temperature, θ(K)
1400
1200
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
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Effect of step size
Table 1. Temperature at 480 seconds as a function of step size, h
Step, h
(480)
Et
|єt|%
480
240
120
60
30
−987.81
110.32
546.77
614.97
632.77
1635.4
537.26
100.80
32.607
14.806
252.54
82.964
15.566
5.0352
2.2864
 (480)  647.57 K
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(exact)
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Comparison with exact results
Temperature, θ(K)
1500
1000
Exact solution
500
h=120
h=240
0
0
-500
100
200
Tim e, t (sec)
300
400
500
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
Temperature,θ(K)
800
400
0
0
100
200
300
400
500
-400
Step size, h (s)
-800
-1200
Figure 5. Effect of step size in Euler’s method.
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Errors in Euler’s Method
It can be seen that Euler’s method has large errors. This can be illustrated using
Taylor series.
dy
1 d2y
1 d3y
2
3
xi 1  xi  




y i 1  y i 
x

x

x

x
 ...
i

1
i
i

1
i
2
3
dx xi , yi
2! dx x , y
3! dx x , y
i
yi 1  yi  f ( xi , yi )xi 1  xi  
i
i
i
1
1
2
3
f ' ( xi , yi )xi 1  xi   f ' ' ( xi , yi )xi 1  xi   ...
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
Et 
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f xi , yi  2 f xi , yi  3
h 
h  ...
2!
3!
Et  h 2
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Additional Resources
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
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