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8-2

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:

• Example:

),( yxfdx

dy

5.05.2 obtain we,1 and 2 ngSubstituti

2

5get weit, Solving

.1at 2 conditionboundary with,5

2

2

xyx y

cxy

xyxdx

dy

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8-3

• Example:

• A second-order differential equation:

• Example:

),,(2

2

dx

dyyxf

dx

yd

)( xycdx

dy

'2'' yxyxy

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8-4

Taylor Series Expansion

• Fundamental case, the first-order ordinary differential equation:

Integrate both sides

• The solution based on Taylor series expansion:

00 at subject to )( xxyyxfdx

dy

x

x

y

ydxxfdy

00

)( x

xdxxfyxgy

0

)()(or 0

)()(' and )( where

...)(''!2

)()(')()()(

0000

0

20

00

xfxgxgy

xgxx

xgxxxgxgy

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8-5

Example : First-order Differential Equation

Given the following differential equation:

The higher-order derivatives:

1at 1such that 3 2 xyxdx

dy

4nfor 0

6

6

3

3

2

2

n

n

dx

yd

dx

yd

xdx

yd

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8-6

The final solution:

1 where

)1()1(3)1(31

)6(!3

)1()6(

!2

)1()3)(1(1

!3

)1(

!2

)1()1(1)(

0

32

3

0

22

0

3

33

2

22

x

xxx

xx

xxx

dx

ydx

dx

ydx

dx

dyxxg

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8-7

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

Table: Taylor Series Solution

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8-8school.edhole.com

8-9

General Case

• The general form of the first-order ordinary differential equation:

• The solution based on Taylor series expansion:

00 at subject to ),( xxyyyxfdx

dy

...),(''!2

)(),(')(),()( 00

000000

yxg

xxyxgxxyxgxgy

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8-10

Euler’s Method

• Only the term with the first derivative is used:

• This method is sometimes referred to as the one-step Euler’s method, since it is performed one step at a time.

edx

dyxxxgxg )()()( 00

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8-11

Example: One-step Euler’s Method

• Consider the differential equation:

• For x =1.1

Therefore, at x=1.1, y=1.44133 (true value).

1at 1such that 4 2 xyxdx

dy

1.1

1

2

14 dxxdy

y

44133.03

41

1.1

1

3 xy

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8-12

0.008373. iserror The

1.43296g(1.10)

valueestimated thesteps, fiveafter 0.02, of size stepa For

0.020833. toreduced iserror The

4205.1])05.1(4)[05.110.1()05.1()10.1(

2.12.01])1(4)[00.105.1()1()05.1(

:)05.1 and 1(at

twiceequation sr'apply Eule and 0.05 of size stepa Use

value).absolute (in 0.04133error The

4.1])1(4[1.01)1.1(

get we,1.0)( of size stepa With

2

2

2

0

gg

gg

xx

g

xxx

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8-13

Errors with Euler`s Method

• Local error: over one step size.

Global error: cumulative over the range of the solution.

• The error using Euler`s method can be approximated using the second term of the Taylor series expansion as

• If the range is divided into n increments, then the error at the end of range for x would be n.

].,[ in maximum theis where

!2

)(

02

2

2

220

xxdx

yd

dx

ydxx

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8-14

Example: Analysis of Errors

0088.0)02.0)(1.1)(4(5

022.0)05.0)(1.1)(4(2

044.0)1.0)(1.1(4

:1.1at

error theon limitsupper the0.02. and 0.05, 0.1, of sizes stepFor

)(4)8(!2

)(by bounded iserror theThus,

8

1at 1 that such 4

202.0

205.0

21.0

20

20

2

2

2

x

xxxxxx

xdx

yd

xyxdx

dy

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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 -5.7419355school.edhole.com

8-16

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

Table: Local and Global Errors with a Step Size of 0.05.

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8-17

x Exact solution

Numerical 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

Table: Local and Global Errors with a Step Size of 0.05 (continued).

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8-18

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

Table: Local and Global Errors with a Step Size of 0.02.

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8-19

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 intervalb. 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.

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8-20

Example : Modified Euler’s Method

1at 1 that such xyyxdx

dy

10768.1)07684.1(1.01)0.11.1()0.1()1.1( 1e.

07684.1)15369.11(2

1 1d.

15369.11.11.1 c.1

1.1)1(1.01)0.11.1()0.1()1.1( 1b.

111 a.1

:1.0for iterationfirst theof steps five The

1.1

1

1

a

a

dx

dygg

dx

dy

dx

dy

dx

dygg

dx

dy

x

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8-21

23193.1)1.12.1()1.1()2.1( 2e.

24251.1)(2

1 2d.

32732.122345.12.1 c.2

22345.1)15771.1(1.010768.1)1.12.1()1.1()2.1( 2b.

15771.110768.11.1 a.2

:interval second for the steps The

2.11.1

2.1

1.1

1.1

a

a

dx

dygg

dy

dx

dy

dx

dy

dx

dx

dy

dx

dygg

yxdx

dy

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8-22

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

xhxxx

xxgyxgy

hyxhfyhxfyxfyy

ii

iiii

iiiiiiii

,

),( ),( where

))],(,(),([5.0

1

1

1

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8-23

• 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 :

),(1 ii yxfS

),( 12 hSyhxfS ii

hSSyy ii )(5.0 211

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8-24

Third-order Runge-Kutta Methods

• The following is an example of the third-order Runge-Kutta methods :

hyxhfyhxhfyxhfyhxf

yxhfyhxfyxfyy

iiiiiiii

iiiiiiii

)))],(5.0,5.0(2),(,(

)),(5.0,5.0(4),([6

11

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8-25

• The computational steps for the third-order method:

1. Evaluate the slope at (xi,yi).

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

)5.0,5.0( 12 hSyhxfS ii

)2,( 213 hShSyhxfS ii

hSSSyy ii ]4[6

13211

),(1 ii yxfS

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8-26

Fourth-order Runge-Kutta Methods

1. Compute the slope S1 at (xi,yi).

2. Estimate y at the mid-point of the interval.

3. Estimate the slope S2 at mid-interval.

4. Revise the estimate of y at mid-interval

. at that such ),( 00 hxxxyyyxfdx

dy

),(1 ii yxfS

),(22/1 iiii yxfh

yy

)5.0,5.0( 12 hSyhxfS ii

22/1 2S

hyy ii

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8-27

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.

)5.0,5.0( 23 hSyhxfS ii

31 hSyy ii

),( 34 hSyhxfS ii

)22(6 43211 SSSSh

yy ii

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8-28

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.

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8-29

Euler-trapezoidal Method

• Euler’s method is the predictor algorithm.

• The trapezoidal rule is the corrector equation.

• Eluer formula (predictor):

• Trapezoidal rule (corrector):

The corrector equation can be applied as many times as necessary to get convergence.

,*,*,1

iiji dx

dyhyy

][2 1,1,*

,*,1

jii

iji dx

dy

dx

dyhyy

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8-30

Example 8-6: Euler-trapezoidal Mehtod

1at 1 that such :Problem xyyxdx

dy

1.1)1(1.01

1.0

111

is 1.1at for estimate )(predictor initial The

0,1

0,0,*00,1

0,0

y

dx

dyyy

dx

dy

xy

15369.11.11.1

:estimate theimprove toused is equationcorrector The

0,1

dx

dy

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8-31

10789.115782.112

1.01

2

15782.110789.11.1

10789.115771.112

1.01

2

15771.110768.11.1

10768.115369.112

1.01

2

2,10,0,*03,1

2,1

1,10,0,*02,1

1,1

0,10,0,*01,1

dx

dy

dx

dyhyy

dx

dy

dx

dy

dx

dyhyy

dx

dy

dx

dy

dx

dyhyy

. have weAnd

.1.1at 1.10789 toconverges , Since

3,1,*1

2,13,1

yy

xyyy

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8-32

22367.1)15782.1(1.010789.1

:equationpredictor the,2.1at of estimate For the

,*1,*10,2

dx

dyhyy

xy

33203.123215.12.1

23215.132744.115782.12

1.010789.1

2

32744.122367.12.1

:equationcorrector The

2,2

1,2,*1,*11,2

1,2

dx

dy

dx

dy

dx

dyhyy

dx

dy

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8-33

1.23239. is 2.1at of estimate The

.iterations threein converges algorithmcorrector theAgain,

23239.133215.115782.12

1.010789.1

2

33215.123238.12.1

23238.133203.115782.12

1.010789.1

2

3,2,*1,*13,2

3,2

2,2,*1,*12,2

xy

dx

dy

dx

dyhyy

dx

dy

dx

dy

dx

dyhyy

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8-34

Milne-Simpson Method

• Milne’s equation is the predictor euqation.• The Simpson’s rule is the corrector formula. • Milne’s equation (predictor):

For the two initial sampling points, a one-step method such as Euler’s equation can be used.

• Simpsos’s rule (corrector):

]22[3

4

,*2,*1,*,*30,1

iiiii dx

dy

dx

dy

dx

dyhyy

]4[3 ,*1,*,1

,*1,1

iiji

iji dx

dy

dx

dy

dx

dyhyy

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8-35

Example 8-7: Milne-Simpson Mehtod

.4.1 and 3.1at estimate want toWe

1at 1 that such :Problem

xxy

xyyxdx

dy

1 1 1

1.1 1.10789 1.15782

1.2 1.23239 1.33215

Assume that we have the following values, obtained from the Euler-trapezoidal method in Example 8-6.

x y dx

dy

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8-36

51917.136560.13.1

36560.1)33215.1(1.023239.1

:used be can method sEuler'

, 3.1at for estimate )(predictor initial thecompute To

0,3

,*2,*20,3

dx

dy

dx

dyhyy

xy

37474.1

15782.1)33215.1(451917.13

1.010789.1

43

1.0

:formularcorrector The

,*1,*20,3,*11,3

dx

dy

dx

dy

dx

dyyy

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8-37

37492.1

15782.1)33215.1(452434.13

1.010789.1

52434.137491.13.1

37491.1

15782.1)33215.1(452424.13

1.010789.1

52424.137474.13.1

3,3

2,3

2,3

1,3

y

dx

dy

y

dx

dy

The computations for x=1.3 are complete.

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8-38

The Milne predictor equation for estimating y at x=1.4:

53762.1

15782.1233215.152434.123

1.041

223

4

,*1,*2,*3,*00,4

dx

dy

dx

dy

dx

dyhyy

73610.153762.14.11,4

dx

dy

The corrector formular:

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8-39

complete. isit Then

53791.133215.152434.1473617.13

1.023239.1

73617.153791.14.1

53791.1

33215.152434.1473601.13

1.023239.1

43

2,4

2,4

,*2,*30,4,*21,4

y

dx

dy

dx

dy

dx

dy

dx

dyhyy

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8-40

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:

nnx xbxbbby 2

210ˆ

dxeFx 2

dx

dy

dx

yde

ˆ where

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8-41

4. Find the minimum of F with respect to the unknowns (b1,b2, b3,…,bn) , that is

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).

xallii

dxb

ee

b

F02

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8-42

Example 8-8: Least-squares Method

2/2

:solution Analytical

1.x0 interval for theit Solve

0at 1 that such :Problem

xey

xyxydx

dy

1

1

0

10

10

ˆ

1ˆ

is modellinear theThus .1 yields

)0(1ˆ

conditionboundary theUsing

ˆ

bdx

yd

xby

b

bby

xbby

• First, assume a linear model is used:

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8-43

0)543

2

2(

0)1)](1([22

1

)1(

:functionerror The

0

51

431

2

1

0 0

211

1

2

1

111

x

x x

xbxxbxxb

dxxxbxbdxdb

dee

xdb

de

xbxbxybe

xy

x

x

3215

32

151

1ˆ

Thus, .bget we,1 withintegral above

thesolve ,10 range thein interested are weSince

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8-44

x True y Value Numerical y Value Error (%)

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

Table: A linear model for the least-squares method

2/2xey xy 32151ˆ

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8-45

• Next, to improve the accuracy of estimates, a quadratic model is used:

xxxbxb

xbxbxxbbxyxbbe

xbbdx

yd

b

bbby

xbxbby

)2()1(

)1(22

is functionerror The

2ˆ

.1 yields

)0()0(1ˆ

conditionboundary theUsing

ˆ

32

21

2212121

21

0

2210

2210

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8-46

4

1

12

5

15

8

:limitupper theas 1 Using

046524

3

3

2

0121

1

21

0

462

51

2422

2

31

1

0

232

21

2

1

bb

x

xxbxbxxbxb

xbxb

dxxxxxbxb

xb

e

x

x

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8-47

2

21

21

0

572

51

252

42

412

1

0

332

21

3

2

78776.014669.01ˆ

.78776.0 and 14669.0get We15

7

105

71

20

9

:limitupper theas 1 Using

05753

2

5

4

3

4

4

2

02212

2

xxy

bb

bb

x

xxbxbxxbxbxbxb

dxxxxxxbxb

xxb

e

x

x

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8-48

x True y Value Numerical y Value Error (%)

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

Table: A quadratic model for the least-squares method

2/2xey 278776.014669.01ˆ xxy

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8-49

Galerkin Method

• Example: Galerkin Method

The same problem as Example 8-8.

Use the quadratic approximating equation.

ii

i

x i

b

ew

w

niedxw

method, squaresleast For the

factor. nga weighti is where

...2,1 0

. and Let 221 xwxw

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8-50

2

21

21

1

0

232

21

1

0

32

21

85526.026316.01ˆ

:result final The4

1

3

1

15

2

3

1

15

7

12

1

:equations normal following get the We

0])2()1([

0])2()1([

xxy

bb

bb

dxxxxxbxb

xdxxxxbxb

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8-51

Table: Example for the Galerkin method

x True y value Numerical y value Error (%)

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

2/2xey 285526.026316.01ˆ xxy

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8-52

Higher-Order Differential Equations

• Second order differential equation:

Transform it into a system of first-order differential equations.

dx

dy

dx

dyyyy

ydx

dy

yyxfdx

dy

121

21

212

and where

),,(

dx

dyyxf

dx

yd,,

2

2

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8-53

• In general, any system of n equations of the following type can be solved using any of the previously discussed methods:

),...,,(

),...,,(

),...,,(

),...,,(

21

2133

2122

2111

nnn

n

n

n

yyyxfdx

dy

yyyxfdx

dy

yyyxfdx

dy

yyyxfdx

dy

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8-54

Example: Second-order Differential Equation

EI

XX

EI

M

dX

Yd 2

2

2 10 :Problem

ZdX

dYEI

XX

EI

M

dX

dZ

10

:into med transforbe canIt 2

hZYXfYY

hZYXfZZ

ZYXEI

iiiii

iiiii

),,(

),,(

:equations following thesolve tomethod s Euler'Use

02314.0 and 0,0at 3600 Assume

11

21

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8-55

Table: Second-order Differential Equation

Using a Step Size of 0.1 FtX

(ft)

Y

(ft)

Exact Z Exact Y

(ft)

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

dX

dZ

dX

dYZ

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8-56

Table: Second-order Differential Equation

Using a Step Size of 0.1 Ft (continued)

dX

dZ

dX

dYZ

X

(ft)

Y

(ft)

Exact Z Exact Y

(ft)

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

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