This document discusses differential equations and methods for solving them numerically. It begins by defining first and second order differential equations. It then introduces Taylor series expansions for solving differential equations and describes Euler's method and the modified Euler's method. Euler's method uses the slope at the start of each interval while modified Euler's method uses the average slope across each interval for improved accuracy. The document provides examples of applying these numerical methods and analyzes the associated errors.

1. By:
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2. Chapter 8 Differential Equations
An equation that defines a relationship between an
unknown function and one or more of its derivatives
is referred to as a differential equation.
A first order differential equation:
dy =
Example:
8-2
f (x, y)
dx
x y x
= = =
5 , with boundary condition 2 at 1.
dy
dx
Solving it, we get 5
y = x +
c
2
2
2
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y x y x
= = = -
Substituting 2 and 1, we obtain 2.5 0.5

3. Example:
dy = -
A second-order differential equation:
d y =
Example:
8-3
f x y dy
( , , ) 2
2
dx
dx
c( y x)
dx
y''= 2x + xy + y'
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4. Taylor Series Expansion
Fundamental case, the first-order ordinary differential
equation:
dy = = =
0 0 f (x) subject to y y at x x
dx
Integrate both sides
ò = òx
dy f x dx
0 x
0
y
y
( )
y g x y f x dx
or ( ) ( ) 0
The solution based on Taylor series expansion:
8-4
= = + òx
x
0
y g x g x x x g x x x g x
= = + - + - +
( ) ( ) ( ) '( ) ( )
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where ( ) and '( ) ( )
''( ) ...
2!
0 0 0 0
0
2
0
0 0
y = g x g x =
f x

5. Example : First-order Differential
Equation
Given the following differential equation:
dy
= 3x2 such that y =1 at x =1
dx
The higher-order derivatives:
8-5
6
6
0 for n 4
d y
d y
3
3
2
2
=
=
= ³
n
dx
n
dx
d y
x
dx
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6. The final solution:
8-6
x d y
x d y
g x x dy
= + - + - + -
( 1)
x x x x x
= + - + - + -
(6 ) ( 1)
( ) 1 ( 1) ( 1)
1 ( 1)(3 ) ( 1)
x x x
= + - + - + -
1 3( 1) 3( 1) ( 1)
where 1
(6)
3!
2!
3!
2!
0
2 3
3
0
2
2
0
3
3 3
2
2 2
=
x
dx
dx
dx
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7. 8-7
Table: Taylor Series Solution
x One Term Two Terms Three Terms Four Terms
1 1 1 1 1
1.1 1 1.3 1.33 1.331
1.2 1 1.6 0.72 1.728
1.3 1 1.9 2.17 2.197
1.4 1 2.2 2.68 2.744
1.5 1 2.5 3.25 3.375
1.6 1 2.8 3.88 4.096
1.7 1 3.1 4.57 4.913
1.8 1 3.4 5.32 5.832
1.9 1 3.7 6.13 6.859
2 1 4 7 8
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8. 8-8
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9. General Case
The general form of the first-order ordinary
differential equation:
dy = = =
0 0 f (x, y) subject to y y at x x
dx
The solution based on Taylor series expansion:
y = g ( x ) = g ( x , y ) + ( x - x ) g '( x , y ) + ( x - x ) g x y +
0 0 0 0 0 0 0
''( , ) ...
2!
0
8-9
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10. Euler’s Method
Only the term with the first derivative is used:
e
g(x) = g(x ) + (x - x ) dy + 0 0
dx
This method is sometimes referred to as the one-step
Euler’s method, since it is performed one step at a
time.
8-10
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11. Example: One-step Euler’s Method
Consider the differential equation:
dy
= 4x2 such that y =1 at x =1
dx
For x =1.1
dy 4x dx y
1 4 1.1
y - = x3 =
Therefore, at x=1.1, y=1.44133 (true value).
8-11
ò = ò 1.1
1
2
1
0.44133
3
1
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12. D = - =
With a step size of ( ) 0.1, we get
(1.1) 1 0.1[4(1) ] 1.4
g
The error 0.04133 (in absolute value).
Use a step size of 0.05 and apply Euler's equation twice
x x
= =
(at 1 and 1.05) :
g g
= + - = + =
(1.05) (1) (1.05 1.00)[4(1) ] 1 0.2 1.2
= + - =
(1.10) (1.05) (1.10 1.05)[4(1.05) ] 1.4205
The error is reduced to 0.020833.
For a step size of 0.02, after five steps, the estimated value
8-12
g(1.10) 1.43296
The error is 0.008373.
2
2
2
0
=
=
= + =
g g
x x x
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13. Errors with Euler`s Method
Local error: over one step size.
Global error: cumulative over the range of the solution.
The error e using Euler`s method can be approximated using the
second term of the Taylor series expansion as
2 2
0
e = ( x - x )
d y
2!
2
2
x x
d y
dx
where is the maximum in [ , ].
2 0
dx
If the range is divided into n increments, then the error at the end
of range for x would be ne.
8-13
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14. Example: Analysis of Errors
8-14
= = =
4 such that 1 at 1
dy
d y
Thus, the error is bounded by ( )
For step sizes of 0.1, 0.05, and 0.02. the upper limits on the error
= =
4(1.1)(0.1) 0.044
= =
2(4)(1.1)(0.05) 0.022
5(4)(1.1)(0.02) 0.0088
at 1.1:
(8 ) 4 ( )
2!
8
2
e
e
0.02
2
0.05
2
0.1
2
0
2
0
2
2
2
= =
=
= - = -
=
e
e
x
x x x x x x
x
dx
x y x
dx
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15. 8-15
Table: Local and Global Errors with a Step Size of 0.1.
x Exact
solution
Numerical
Solution
Local
Error(%)
Global
Error(%)
1 1 1 0 0
1.1 1.4413333 1.4 -2.8677151 -2.8677151
1.2 1.9706667 1.884 -2.300406 -4.3978349
1.3 2.596 2.46 -1.9003595 -5.238829
1.4 3.3253333 3.136 -1.6038492 -5.6936648
1.5 4.1666667 3.92 -1.396 -5-92
1.6 5.128 4.82 -1.1960478 -6.0062402
1.7 6.2173333 5.844 -1.0508256 -6.004718
1.8 7.4426667 7 -0.9315657 -5.947689
1.9 8.812 8.296 -0.8321985 -5.8556514
2 10.333333 9.74 -0.7483871 -5.7419355 school.edhole.com

16. 8-16
Table: Local and Global Errors with a Step Size of 0.05.
x Exact
solution
Numerical
Solution
Local
Error(%)
Global
Error(%)
1 1 1 0 0
1.05 1.2101667 1.2 -0.8401047 -0.8401047
1.1 1.4413333 1.4205 -0.7400555 -1.4454209
1.15 1.6945 1.6625 -0.6589948 -1.8884627
1.2 1.9706667 1.927 -0.5920162 -2.2158322
1.25 2.2708333 2.215 -0.5357798 -2.4587156
1.3 2.596 2.5275 -0.4879301 -2.6386749
1.35 2.9471667 2.8655 -0.4467568 -2.771023
1.4 3.3253333 3.23 -0.4109864 -2.8668805
1.45 3.7315 3.622 -0.3796507 -2.9344768
1.5 4.1666667 4.4025 -0.352 -2.98
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17. 8-17
Table: Local and Global Errors with a Step Size of 0.05
(continued).
x Exact
solution
Numerica
l Solution
Local
Error(%)
Global
Error(%)
1.55 4.6318333 4.4925 -0.3274441 -3.0081681
1.6 5.128 4.973 -0.3055122 -3.0226209
1.65 5.6561667 5.485 -0.2858237 -3.0261956
1.7 6.2173333 6.0295 -0.2680678 -3.0211237
1.75 6.8125 6.6075 -0.2519878 -3.0091743
1.8 7.4426667 7.22 -0.2373701 -2.9917592
1.85 8.1088333 7.868 -0.2240355 -2.9700121
1.9 8.812 8.5525 -0.2118323 -2.9448479
1.95 9.5531667 9.2745 -0.2006316 -2.9170083
2 10.333333 10.035 -0.1903226 -2.8870968
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18. 8-18
Table: Local and Global Errors with a Step Size of 0.02.
x Exact solution Numerical Solution Local Error(%) Global Error(%)
1 1 1 0 0
1.02 1.0816107 1.08 -0.1489137 -0.1489137
1.04 1.1664853 1.163232 -0.1408219 -0.2789005
1.06 1.254688 1.24976 -0.1334728 -0.392767
1.08 1.3462827 1.339648 -0.1267688 -0.4928138
1.1 1.4413333 1.43296 -0.120629 -0.5809436
1.2 1.9706667 1.95312 -0.0963464 -0.8903924
1.3 2.596 2.56848 -0.0793015 -1.0600924
1.4 3.3253333 3.28704 -0.0667201 -1.1515638
1.5 4.1666667 4.1168 -0.057088 -1.1968
1.6 5.128 5.06876 -0.049506 -1.2137285
1.7 6.2173333 6.14192 -0.0434055 -1.212953
1.8 7.4426667 7.35328 -0.0384092 -1.2010032
1.9 8.812 8.70784 -0.0342563 -1.1820245
2 10.333333 10.2136 -0.0307613 -1.1587097
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19. Modified Euler’s Method
 Use an average slope, rather than the slope at the start
of the interval :
a. Evaluate the slope at the start of the interval
b. Estimate the value of the dependent variable y at the
end of the interval using the Euler’s metod.
c. Evaluate the slope at the end of the interval.
d. Find the average slope using the slopes in a and c.
e. Compute a revised value of the dependent variable y
at the end of the interval using the average slope of
step d with Euler’s method.
8-19
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20. Example : Modified Euler’s Method
dy
= x y such that y = 1 at x = 1
dx
D =
The five steps of the first iteration for 0.1:
= =
dy
1a. 1 1 1
1
g g dy
= + - = + =
1b. (1.1) (1.0) (1.1 1.0) 1 0.1(1) 1.1
= =
dy
1c. 1.1 1.1 1.15369
1.1
dx
dy
1d. 1
1
= + =
(1 1.15369) 1.07684
2
g g dy
= + - = + =
1e. (1.1) (1.0) (1.1 1.0) 1 0.1(1.07684) 1.10768
a
a
dx
dx
dx
dx
x
8-20
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21. The steps for the second interval :
= = =
dy
2a. 1.1 1.10768 1.15771
1.1
g g dy
= + - = + =
2b. (1.2) (1.1) (1.2 1.1) 1.10768 0.1(1.15771) 1.22345
8-21
1.1
= =
dy
2c. 1.2 1.22345 1.32732
1.2
dx
dx
2d. 1
dx
dx
= + =
( ) 1.24251
2
1.1 1.2
g g dy
= + - =
2e. (1.2) (1.1) (1.2 1.1) 1.23193
a
a
dx
dy
dy
dy
dx
x y
dx
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22. Second-order Runge-Kutta Methods
The modified Euler’s method is a case of the second-order
Runge-Kutta methods. It can be expressed as
y = y + 0.5[ f ( x , y ) + f ( x + h , y +
hf ( x , y ))]
h
i i i i i i i i
y = g x y = g x + D
x
where ( ), ( ),
i i i i
x = x + D x h = D
x
i +
i
+
+
,
1
1
1
8-22
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23.  The computations according to Euler’s method:
1. Evaluate the slope at the start of an interval, that is,
at (xi,yi) .
2. Evaluate the slope at the end of the interval (xi+1,yi+1) :
3. Evaluate yi+1 using the average slope S1 of and S2 :
8-23
( , ) 1 i i S = f x y
( , ) 2 1 S f x h y hS i i = + +
y y S S h i i 0.5( ) 1 1 2 = + + +
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24. Third-order Runge-Kutta Methods
The following is an example of the third-order Runge-
Kutta methods :
1
y y f x y f x h y hf x y
= + + + + + +
[ ( , ) 4 ( 0.5 , 0.5 ( , ))
6
i i i i i i i i
f ( x h , y hf ( x , y ) 2 hf ( x 0.5 h , y 0.5 hf ( x , y )))]
h
i i i i i i i i
1
+ - + + +
8-24
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25.  The computational steps for the third-order method:
1. Evaluate the slope at (xi,yi).
( , ) 1 i i S = f x y
2. Evaluate a second slope S2 estimate at the mid-point
in of the step as
3. Evaluate a third slope S3 as
4. Estimate the quantity of interest yi+1 as
8-25
( 0.5 , 0.5 ) 2 1 S f x h y hS i i = + +
( , 2 ) 3 1 2 S f x h y hS hS i i = + - +
1
y = y + [ S + 4 S + S ]
h i +
1 i 1 2 3 6
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26. Fourth-order Runge-Kutta Methods
dy = = = D =
( , ) such that at . 0 0 f x y y y x x x h
dx
1. Compute the slope S1 at (xi,yi).
2. Estimate y at the mid-point of the interval.
i 1/ 2 i 2 i i y = y + h f x y +
3. Estimate the slope S2 at mid-interval.
4. Revise the estimate of y at mid-interval
8-26
( , ) 1 i i S = f x y
( , )
( 0.5 , 0.5 ) 2 1 S f x h y hS i i = + +
y y h S i i = + + school.edhole.com
1/ 2 2 2

27. 5. Compute a revised estimate of the slope S3 at mid-interval.
6. Estimate y at the end of the interval.
7. Estimate the slope S4 at the end of the interval
8. Estimate yi+1 again.
8-27
( 0.5 , 0.5 ) 3 2 S f x h y hS i i = + +
1 3 y y hS i i = + +
( , ) 4 3 S f x h y hS i i = + +
1 6 1 2 3 4 y y h S S S S i i = + + + + +
( 2 2 )
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28. Predictor-Corrector Methods
Unless the step sizes are small, Euler’s method and
Runge-Kutta may not yield precise solutions.
The Predictor-Corrector Methods iterate several times
over the same interval until the solution converges to
within an acceptable tolerance.
Two parts: predictor part and corrector part.
8-28
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29. Euler-trapezoidal Method
 Euler’s method is the predictor algorithm.
 The trapezoidal rule is the corrector equation.
 Eluer formula (predictor):
,*
y = y + h dy +
i 1, j i ,*
dx
i
 Trapezoidal rule (corrector):
y y h dy
+ = + +
dy
[ ]
2 ,* 1, 1
i 1, j i ,*
dx
i i j
+ -
dx
The corrector equation can be applied as many times as
necessary to get convergence.
8-29
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30. Example 8-6: Euler-trapezoidal Mehtod
dy
Problem: = x y such that y = 1 at x = 1
dx
The initial (predictor) estimate for at 1.1 is
= =
1 1 1
é
dy
y y 0.1
dy
1 0.1(1) 1.1
1,0
0,0
0,0
1,0 0,*
= + =
ù
úû
êë
= +
=
y
dx
dx
y x
8-30
The corrector equation is used to improve the estimate :
1.1 1.1 1.15369
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1,0
= =
dy
dx

31. 8-31
[ ]
[ ]
1 0.1
é
y y h dy
= + +
= =
ù
dy
1.1 1.10768 1.15771
1 0.1
é
dy
y y h dy
= + +
dy
= =
ù
1.1 1.10789 1.15782
1 0.1
[1 1.15782] 1.10789
2
dy
y y h dy
2
1 1.15771 1.10789
2
2
1 1.15369 1.10768
2
2
dy
0,0 1,2
1,1
1,2
1,3 0,*
0,0 1,1
1,2 0,*
0,0 1,0
1,1 0,*
ù
= + + = úû
êë é
= + +
= + + = úû
êë
= + + = úû
êë
dx
dx
dx
dx
dx
dx
dx
dx
y y y x
= =
Since , converges to 1.10789 at 1.1.
1,3 1,2
y =
y
And we have .
1,* 1,3
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32. 8-32
=
For the estimate of at 1.2, the predictor equation :
y y h dy
2,0 1,* 1,* = + = + =
1.10789 0.1(1.15782) 1.22367
dx
y x
The corrector equation :
= =
1.2 1.22367 1.32744
ù
dy
y y h dy
= + +
dy
[ ]
2
é
1.10789 0.1
= + + =
1.15782 1.32744 1.23215
2
1.2 1.23215 1.33203
2,1
2,2
1,* 2,1
2,1 1,*
= =
úû
êë
dy
dx
dx
dx
dx
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33. 8-33
y y h dy
= + +
dy
dx
1,* 2,2
ù
úû
[ ]
2
2,2 1,*
é
êë
dx
1.10789 0.1
= + + =
1.15782 1.33203 1.23238
2
= =
1.2 1.23238 1.33215
dy
dx
y y h dy
= + +
dy
dx
1,* 2,3
ù
úû
[ ]
2
2,3
2,3 1,*
é
êë
dx
1.10789 0.1
= + + =
1.15782 1.33215 1.23239
2
Again, the corrector algorithm converges in three iterations.
=
y x
The estimate of at 1.2 is 1.23239.
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34. Milne-Simpson Method
 Milne’s equation is the predictor euqation.
 The Simpson’s rule is the corrector formula.
 Milne’s equation (predictor):
y y 4
h dy
dy
+ - = + - +
dy
i i dx
i i i
For the two initial sampling points, a one-step
method such as Euler’s equation can be used.
 Simpsos’s rule (corrector):
8-34
[2 2 ]
3
,* 1,* 2,*
1,0 3,*
- -
dx
dx
dy
y y h dy
+ - = + + +
dy
[ 4 ]
3 1, ,* 1,*
i 1, j i 1,*
dx
dx
i j i i
+ -
dx
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35. Example 8-7: Milne-Simpson Mehtod
x y y x
= = =
dy
Problem: such that 1 at 1
y x x
= =
dx
We want to estimate at 1.3 and 1.4.
Assume that we have the following values,
obtained from the Euler-trapezoidal method
in Example 8-6.
dy
x y dx
8-35
1 1 1
1.1 1.10789 1.15782
1.2 1.23239 1.33215
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36. To compute the initial (predictor) estimate for at 1.3 ,
Euler's method can be used :
y y hdy
= + = + =
1.23239 0.1(1.33215) 1.36560
1.3 1.36560 1.51917
3,0
2,*
3,0 2,*
= =
=
dy
dx
dx
y x
8-36
The corrector formular :
4
ù
dy
dy
[ ]
y y 0.1
dy
1.10789 0.1
= + + +
1.37474
1.51917 4(1.33215) 1.15782
3
3
3,0 2,* 1,*
3,1 1,*
=
úû
êë é
= + + +
dx
dx
dx
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37. 8-37
= =
1.3 1.37474 1.52424
[ ]
1.10789 0.1
= + + +
=
= =
1.3 1.37491 1.52434
[ ]
1.10789 0.1
= + + +
1.37492
1.52434 4(1.33215) 1.15782
3
1.37491
1.52424 4(1.33215) 1.15782
3
3,1
3,2
3,2
3,3
=
dy
dx
y
dy
dx
y
The computations for x=1.3 are complete.
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38. The Milne predictor equation for estimating y at x=1.4:
8-38
y y 4
h dy
ù
é
dy
= + - +
dy
( ) [ ( ) ( )]
1 4 0.1
= + - +
1.53762
2 1.52434 1.33215 2 1.15782
3
2 2
3
3,* 2,* 1,*
4,0 0,*
=
úû
êë
dx
dx
dx
The corrector formular:
1.4 1.53762 1.73610
4,1
= =
dy
dx
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39. 8-39
dy
y y h dy
= + + +
4
ù
úû
dy
[ ( ) ]
3
é
êë
1.23239 0.1
= + + +
1.53791
1.73601 4 1.52434 1.33215
3
=
= =
1.4 1.53791 1.73617
[ ( ) ]
1.23239 0.1
1.73617 4 1.52434 1.33215 1.53791
3
Then it is complete.
4,2
4,2
4,0 3,* 2,*
4,1 2,*
= + + + =
dy
dx
y
dx
dx
dx
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40. Least-Squares Method
 The procedure for deriving the least-squares
function:
1. Assume the solution is an nth-order polynomial:
2. Use the boundary condition of the ordinary
differential equation to evaluate one of (bo,b1,b2,
…,bn).
3. Define the objective function:
8-40
n
x ny = b + b + b x2 ++ b x
0 1 2 ˆ
x = ò 2
F e dx
dy
dx
e = dy - ˆ
where
dx
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41. 4. Find the minimum of F with respect to the unknowns
(b1,b2, b3,…,bn) , that is
F = 2 e ¶
e
0
¶
dx
b
b
5. The integrals in Step 4 are called the normal
equations; the solution of the normal equations yields
value of the unknowns (b1,b2, b3,…,bn).
8-41
ò =
¶
¶
all x
i i
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42. Example 8-8: Least-squares Method
xy y x
dy
= = =
Problem: such that 1 at 0
Solve it for the interval 0 x 1.
Analytical solution :
y ex
2 / 2 dx
=
£ £
8-42
• First, assume a linear model is used:
y = b +
b x
0 1
Using the boundary condition
ˆ 1 (0)
y = = b +
b
yields 1. Thus the linear model is
y = +
b x
1
1
0
0 1
ˆ 1
ˆ
ˆ
b
dy
dx
b
=
=
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43. 8-43
e = b - xy = b - x +
b x
e de
x
de
x x
ò ò
dx b x b x x dx
= - + - =
x
) 0
(1 )
The error function :
= -
db
0 0
b x x b x x b x
3 4 5
2
1
2
(
2 2[ (1 )](1 ) 0
0
5
1
3 4
1
2
1
2
1 1
1
2
1
1 1 1
- - + + =
db
£ £
Since we are interested in the range 0 1, solve the
above integral with 1, we get b . Thus,
y x
x
x
15
32
15
32
1
ˆ 1
= +
= =
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44. 8-44
x True y Value Numerical y Value Error (%)
y = ex2 / 2 y ˆ = 1+ 15
x 32
0 1. 1. -
0.2 1.0202 1.0938 7.2
0.4 1.0833 1.1875 9.6
0.6 1.1972 1.2812 7.0
0.8 1.3771 1.3750 0.0
1.0 1.6487 1.46688 -10.9
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45. Next, to improve the accuracy of estimates, a
quadratic model is used:
8-45
ˆ
y = b + b x +
b x
2
0 1 2
Using the boundary condition
ˆ 1 (0) (0 )
y = = b + b +
b
=
0 1 2
yields 1.
0
b b x
dy
= +
2
1 2
The error function is
2
e = b + b x - xy = b + b x - x + b x +
b x
2 2 (1 )
b x b x x x
dx
b
= - + - -
(1 ) (2 )
ˆ
3
2
2
1
2
1 2 1 2 1 2
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46. 8-46
= -
1
e
[ ( ) ( ) ]( )
b x b x x x x dx
- + - - - =
1 2 1 0
b x b x b x b x x b x b x x
=
¶
ò
é
x
Using 1 as the upper limit :
1
4
5
12
8
15
0
4 2 5 6 4
3
3
2
1 2
0
6 4
2
5
1
4 2
2 2
2
3
1
1
0
3 2
2
2
1
2
1
+ =
ù
= úû
êë
- + - - + + +
¶
b b
x
b
x
x
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47. 8-47
x x
= -
[ ( ) ( ) ]( )
b x b x x x x x dx
e
¶
ò
b
- + - - - =
2 1 2 2 0
b x b x b x b x x b x b x x
x
=
é
Using 1 as the upper limit :
7
15
b 71
b
+ =
105
9
20
b b
= - =
We get 0.14669 and 0.78776.
2
0
1 2
1 2
0
7 5
2
5
1
5 2
2
4
2
4
2 1
1
3 3
2
2
1
3
2
ˆ 1 0.14669 0.78776
0
3 5 7 5
2
5
4
3
4
4
2
2
y x x
x
x
= - +
ù
= úû
êë
- + - - + + +
¶
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48. 8-48
x True y Value Numerical y Value Error (%)
y = ex2 / 2 yˆ = 1- 0.14669x + 0.78776x2
0 1. 1. -
0.2 1.0202 1.0022 -1.8
0.4 1.0833 1.0674 0.0
0.6 1.1972 1.1956 0.0
0.8 1.3771 1.38668 0.0
1.0 1.6487 1.6411 0.0
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49. Galerkin Method
ò = =
w edx i n
w
0 1,2...
where is a weighting factor.
i
x i
For the least squares method,
 Example: Galerkin Method
w = ¶
e
b
i
i
¶
The same problem as Example 8-8.
Use the quadratic approximating equation.
8-49
Let and 2.
1 2 w = x w = x
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50. 8-50
b x b x x x xdx
- + - - =
[ (1 ) (2 ) ] 0
b x b x x x x dx
- + - - =
[ (1 ) (2 ) ] 0
ò
We get the following normal equations :
2
7
b b
+ =
1 2
15
1
b b
+ =
1 2
1
0
3 2
2
2
1
1
0
3
2
2
1
1
3
1
4
3
1
12
2
15
The final result :
ˆ 1 0.26316 0.85526
y = - x +
x
ò
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51. Table: Example for the Galerkin method
x True y value Numerical y value Error
(%)
y = ex2 / 2 yˆ = 1- 0.26316x + 0.85526x2
0 1. 1. --
0.2 1.0202 0.9816 0.0
0.4 1.0833 1.0316 0.0
0.6 1.1972 1.1500 0.0
0.8 1.3771 1.3368 0.0
1.0 1.6487 1.5921 0.0
8-51
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52. Higher-Order Differential Equations
Second order differential equation:
2
d y , , 2
f x y dy
= æ
ö çè
÷ø
dx
dx
Transform it into a system of first-order differential
equations.
dy
dx
( , , )
y y y dy
dx
y
dy
dy
dx
f x y y
dx
= = =
=
=
1
1 2
2
1
1 2
2
where and
8-52
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53. In general, any system of n equations of the following
type can be solved using any of the previously
discussed methods:
8-53
( , , ,... )
n
( , , ,... )
n
( , , ,... )
3 1 2
( , , ,... )
1 2
3
2 1 2
2
1 1 2
dy
1
n n
n
n
f x y y y
dy
dy
dy
dx
f x y y y
dx
f x y y y
dx
f x y y y
dx
=
=
=
=

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54. Example: Second-order Differential Equation
8-54
2 10 Problem: = = -
d Y X X
2
EI
M
EI
dX
2
It can be transformed into :
M
= = -
Z
dZ
dY
dX
X X
EI
EI
dX
=
10
2
EI X Y Z
= = = = -
Assume 3600 at 0, 0 and 0.02314
Use Euler's method to solve the following equations :
Z = Z +
f ( X , Y , Z )
h
i +
i i i i
1 2
= +
Y Y f ( X , Y , Z )
h
i +
1 i 1
i i i
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55. Table: Second-order Differential Equation
Using a Step Size of 0.1 Ft
X
(ft)
Y
(ft)
Exact Z Exact Y
(ft)
dZ
Z = dY
0 0 -0.0231481 0 -0.0231481 0
0.1 0.000275 -0.0231481 -0.0023148 -0.0231344 -0.0023144
0.2 0.0005444 -0.0231206 -0.0046296 -0.0230933 -0.004626
0.3 0.0008083 -0.0230662 -0.0069417 -0.0230256 -0.0069321
0.4 0.0010667 -0.0229854 -0.0092483 -0.0229319 -0.0092302
0.5 0.0013194 -0.0228787 -0.0115469 -0.0228125 -0.0115177
0.6 0.0015667 -0.0227468 -0.0138347 -0.0226681 -0.0137919
0.7 0.0018083 -0.0225901 -0.0161094 -0.0224994 -0.0160505
0.8 0.0020444 -0.0224093 -0.0183684 -0.0223067 -0.018291
0.9 0.002275 -0.0222048 -0.0206093 -0.0220906 -0.020511
8-55
dX
dX
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56. Table: Second-order Differential Equation
Using a Step Size of 0.1 Ft (continued)
X
(ft)
Y
(ft)
Exact Z Exact Y
(ft)
dZ
Z = dY
1 0.0025 -0.0219773 -0.0228298 -0.0218519 -0.0227083
2 0.0044444 -0.0185565 -0.0434305 -0.0183333 -0.04296663
3 0.0058333 -0.0134412 -0.0298019 -0.0131481 -0.0588194
4 0.0066667 -0.007187 -0.0704998 -0.0068519 -0.0688889
5 0.0069444 -0.0003495 -0.0746352 0.00000000 -0.071228
6 0.0066667 0.0065157 -0.0718747 0.0068519 -0.0688889
7 0.0058333 0.0128532 -0.06244066 0.0131481 -0.0588194
8 0.0044444 0.0181074 -0.0471107 0.0183333 -0.042963
9 0.0025 0.0217227 -0.0272183 0.0278519 -0.0227083
10 0.000000 0.0231435 -0.00466523 0.0231481 0.000000
8-56
dX
dX
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