chapter 6 - ordinary differential equation

7/28/2019 Chapter 6 - Ordinary Differential Equation

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MDC10103

Advanced Engineering

Mathematics


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Lectures Information

Saifulnizan Jamian

E5-001-06

012-3247301 07-4537334

[email protected]


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Contents

Ordinary Differential Equation

Partial Differential Equation Numerical Optimization

Statistical Methods


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Chapter 6

Ordinary Differential Equation


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Contents

Fourth-order Runge-Kutta method

Fourth-order Adams predictor-corrector method

System of differential equation by Fourth-order Runge-Kutta method

Boundary-value problem: linear shooting method and

finite-difference method


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Introduction

Ordinary differential equations arise in many different contextsthroughout mathematics and science (social and natural) one way

or another, because when describing changes mathematically, the

most accurate way uses differentials and derivatives (related,

though not quite the same). Since various differentials,derivatives, and functions become inevitably related to each other

via equations, a differential equation is the result, governing

dynamical phenomena, evolution and variation. Often, quantities

are defined as the rate of change of other quantities (time

derivatives), or gradients of quantities, which is how they enter

differential equations.

(Wikipedia)


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Example

Newton's second law of motion

t


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n t a - a ue ro em oFirst-Order Differential

EquationConsider an IVP of the first-order differential equation:

)1.6.()(),,(' 00 Eqyxyyxfy ==

Can be solved by:

a)Fourth-order Runge-Kutta Method (RK4)

b)Fourth-order Adam Predictor-Corrector Method


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Fourth-order Runge-KuttaMethod (RK4)

The solution of Eq. (6.1) by RK4 method is:

( )43211 226

1

kkkkyy ii ++++=+

( )

( )342

3

1

21

,;2

,2

2,2;,

kyhxhfkk

yh

xhfk

k

y

h

xhfkyxhfk

iiii

iiii

++=

++=

++==

where:


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Example 6.1

Solve the IVP y= x/y, y(0)=1 at x=0(0.2)1 using

RK4 method. If the exact solution is y=(x2+1)0.5,

find the absolute errors.


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Solution

( )43211

226

1kkkkyy

ii ++++=+

3

42

3

121

2.02.0;

2

1.02.0

2

1.02.0;2.0

ky

xk

ky

xk

k

y

xk

y

xk

i

i

i

i

i

i

i

i

++

=+

+=

+

+==

where:

2.0,),(' === hyxyxfy


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Solution

i xi yi k1 k2 k3 k4 Exact y error

0 0 1.0000 0.0000 0.0200 0.0198 0.0392 1.0000 0.E+00

1 0.2 1.0198 0.0392 0.0577 0.0572 0.0743 1.0198 6.E-07

2 0.4 1.0770 0.0743 0.0898 0.0891 0.1029 1.0770 2.E-06

3 0.6 1.1662 0.1029 0.1150 0.1144 0.1249 1.1662 3.E-06

4 0.8 1.2806 0.1249 0.1340 0.1336 0.1414 1.2806 4.E-06

5 1 1.4142 0.1414 0.1482 0.1478 0.1536 1.4142 4.E-06


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Example 6.2

Solve the IVP y= x2(1-3y), y(0)=1 at x=0(0.2)1

using RK4 method. Find the absolute errors if the

exact solution is

3

1

3

2 3

+=

x

ey


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Solution

( )43211

226

1kkkkyy

ii

++++=+

( ) ( )

( )

( ) ( )( )32

4

22

3

12

2

2

1

312.02.0

;2

311.02.0

2311.02.0;312.0

kyxk

kyxk

kyxkyxk

ii

ii

iiii

++=

++=

++==

where:

( ) 2.0,31),(' 2 === hyxyxfy


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Solution

i xi yi k1 k2 k3 k4 Exact y error0 0 1.0000 0.0000 -0.0040 -0.0040 -0.0159 1.0000 0.E+00

1 0.2 0.9947 -0.0159 -0.0353 -0.0348 -0.0602 0.9947 1.E-06

2 0.4 0.9587 -0.0600 -0.0893 -0.0871 -0.1163 0.9587 3.E-06

3 0.6 0.8705 -0.1160 -0.1409 -0.1372 -0.1536 0.8705 5.E-06

4 0.8 0.7329 -0.1534 -0.1569 -0.1560 -0.1461 0.7329 6.E-06

5 1 0.5786 -0.1472 -0.1247 -0.1328 -0.0972 0.5786 4.E-05


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Fourth-order AdamsPredictor-Corrector Method

The solution of Eq. (6.1) by this method is:

( )

( )2111

3211

519924

9375955

24

++

+

+++=

++=

iiiiP

iiC

iiiiiiP

ffffh

yy

ffffh

yy

Where superscript P and C are stands for predictor andcorrector equation, respectively.

Note: RK4 method is a single point method, where the new approximation solution is just

based on previous point, while Fourth-order Adams Predictor-corrector method is multiple

point method, where the new approximation solution is based on previous four points. To get

last three starting point we may approximate them by RK4 method.


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Example 6.3

Solve the IVP y= x/y, y(0)=1 at x=0(0.2)1 using

Fourth-order Adams Predictor-Corrector method.

If the exact solution is y=(x2+1)0.5, find the

absolute errors.


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Solution

( )43211 226

1kkkkyy ii ++++=+

3

42

3

121

2.02.0;

2

1.02.0

2

1.02.0;2.0

ky

xk

ky

xk

k

y

xk

y

xk

i

i

i

i

i

i

i

i

++

=+

+=

+

+==

where:

2.0,),(' === hy

xyxfy

Obtain y1, y2 and y3 from RK4:


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Solution

i xi yi k1 k2 k3 k4 Exact y error

0 0 1.0000 0.0000 0.0200 0.0198 0.0392 1.0000 0.E+00

1 0.2 1.0198 0.0392 0.0577 0.0572 0.0743 1.0198 6.E-07

2 0.4 1.0770 0.0743 0.0898 0.0891 0.1029 1.0770 2.E-06

3 0.6 1.1662 0.1029 0.1150 0.1144 0.1249 1.1662 3.E-06

Obtain y4 and y5 using Fourth-order Adams Predictor-

Corrector Method

+++=

++=

1

1

2

2

3

3

4

434

0

0

1

1

2

2

3

3

34

519924

2.0

937595524

2.0

y

x

y

x

y

x

y

xyy

y

x

y

x

y

x

y

x

yy

P

C

P


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Solution

i xi yp yi=yc Exact y error

0 0 1.0000 1.0000 0.E+00

1 0.2 1.0198 1.0198 4.E-06

2 0.4 1.0770 1.0770 3.E-05

3 0.6 1.1662 1.1662 1.E-05

4 0.8 1.2799 1.2807 1.2806 8.E-05

5 1 1.4139 1.4143 1.4142 1.E-04

+++=

++=

2

2

3

3

4

4

5

5

45

1

1

2

2

3

3

4

445

519924

2.0

9375955242.0

y

x

y

x

y

x

y

xyy

yx

yx

yx

yxyy

P

C

P


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Example 6.4

Solve the IVP y= x2(1-3y), y(0)=1 at x=0(0.2)1

using Fourth-order Adams Predictor-Corrector

method. Find the absolute errors if the exact

solution is

3

1

3

2

'

3

+=

x

ey


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Solution

( ) ( ) ( ) ( )( )

( ) ( ) ( ) ( )( )12

2

2

3

24

2

434

001

2

2

2

3

2

34

31315311931924

2.0

31931373159315524

2.0

123

123

yxyxyxyxyy

yxyxyxyxyy

PC

P

+++=

++=

Obtain y1, y2 and y3 from RK4.

Obtain y4 and y5 using Fourth-order Adams Predictor-

Corrector Method

( ) ( ) ( ) ( )( )

( ) ( ) ( ) ( )( )22

3

2

4

25

2

545

112

2

3

2

4

2

45

31315311931924

2.0

31931373159315524

2.0

234

234

yxyxyxyxyy

yxyxyxyxyy

PC

P

+++=

++=


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Solution

i xi yp yi=yc Exact y error

0 0 1.0000 1.0000 0.E+00

1 0.2 0.9947 0.9947 1.E-05

2 0.4 0.9587 0.9587 3.E-05

3 0.6 0.8705 0.8705 1.E-05

4 0.8 0.7277 0.7337 0.7329 8.E-04

5 1 0.5800 0.5782 0.5786 4.E-04


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System of First-OrderDifferential Equations

Consider the initial-value problem of first-order differential

equations

( )

( )yxtgdt

dy

yxtfdt

dx

,,

,,

=

=

with ( ) ( ) 0000 , ytyxtx ==

The solution of both ODEs using RK4 method will

be given by:( )

( )43211

43211

226

1

226

1

ggggyy

ffffxx

ii

ii

++++=

++++=

+

+


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System of First-OrderDifferential Equations

7 - 25

( )

( )

( )

( )33422

3

121

33422

3

121

,,;2

,2

,2

2

,

2

,

2

;,,

,,;2

,2

,2

2,

2,

2;,,

gyfxhthggg

yf

xh

thgg

gy

fx

hthggyxthgg

gyfxhthffgyfxhthff

gy

fx

hthffyxthff

iiiiii

ii

iiiii

iiiiii

ii

iiiii

+++=

+++=

+++==

+++=

+++=

+++==

where:


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Example 5

A simple RL-electrical circuit consists of electrical current i (in

ampheres), resistanceR (in ohms), inductanceL (in hengrys), and

electromotive forceE(t) (in volts), as shown in Figure below.

Given that q (in coulombs) is the charge, and R, and L are assumed

constants. Given that i = dq/dtand di/dt= d2q/dt2. According toKirchoffs second law, the current I satisfies the differential equation

Given thatR = 15,L = 3,E(t) = 120 and the initial condition q = 0 and i

= 0 when t= 0. Find i and q for 0t 2 with t= 0.25 using RK4

method. If the exact solution are q = 8(e-5t 1)/5 +8tand i = 8(1-e-5t),

find the absolute errors fori and q.

( )tEdt

dqR

dt

xdL =+

2

2


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SolutionGiven

hencedt

qd

dt

diand

dt

dqi

2

2

== ( )tEdt

dqR

dt

qdL =+

2

2

becomes

( )tERidt

diL =+

The second order differential becomes system of first ODE below:

( )

( ) iiiqtgL

RitEdtdi

iqtfidt

dq

540315120,,)(

,,

====

==


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SolutionUsing RK4 method,

( ) ( )4321143211 226

1;22

6

1ggggiiffffqq kkkk ++++=++++= ++

( )

( )

( )( )342

3

1

21

342

31

21

54025.0;2

54025.0

254025.0;54025.0

25.0;2

25.0;2

25.0;25.0

gigg

ig

g

igig

gifg

ifg

ifif

+=

+=

+==

+=

+=

+==

where:


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Solutionk t q i f1 g1 f2 g2 f3

0 0 0.0000 0.0000 0.0000 10.0000 1.2500 3.7500 0.4688

1 0.25 0.8919 5.5404 1.3851 3.0745 1.7694 1.1530 1.5292

2 0.5 2.5512 7.2438 1.8109 0.9453 1.9291 0.3545 1.8553

3 0.75 4.4465 7.7675 1.9419 0.2906 1.9782 0.1090 1.9555

4 1 6.4143 7.9285 1.9821 0.0894 1.9933 0.0335 1.9863

5 1.25 8.4044 7.9780 1.9945 0.0275 1.9979 0.0103 1.9958

6 1.5 10.4014 7.9932 1.9983 0.0084 1.9994 0.0032 1.9987

7 1.75 12.4004 7.9979 1.9995 0.0026 1.9998 0.0010 1.9996

8 2 14.4001 7.9994 1.9998 0.0008 1.9999 0.0003 1.9999

g3 f4 g4 exact q exact i error q error i

7.6563 1.9141 0.4297 0.0000 0.0000 0.0000 0.0000

2.3539 1.9736 0.1321 0.8584 5.7080 0.0335 0.1676

0.7237 1.9919 0.0406 2.5313 7.3433 0.0199 0.0995

0.2225 1.9975 0.0125 4.4376 7.8119 0.0089 0.04440.0684 1.9992 0.0038 6.4108 7.9461 0.0035 0.0176

0.0210 1.9998 0.0012 8.4031 7.9846 0.0013 0.0065

0.0065 1.9999 0.0004 10.4009 7.9956 0.0005 0.0023

0.0020 2.0000 0.0001 12.4003 7.9987 0.0002 0.0008

0.0006 2.0000 0.0000 14.4001 7.9996 0.0001 0.0003

oun ary a ue ro em


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oun ary- a ue ro em(BVP) of Second-Order

Differential EquationConsider a linear second-order differential equation)2.6.()()()()(')()(")( Eqxdxyxcxyxbxyxa =++

Can be solved by:

a)Shooting method

b)Finite-Difference Method (FDM)

With the boundary conditions (BCs)nn yxyyxy == )(;)( 00


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Shooting Method

Let

?)(;)()()()(')(

)(;)('

0

00

==++==

xzxdyxczxbxzxa

yxyzxy

Since so we can use it as our initial

guess of z(x0). Then, the system of IVP can be solved usingRK4 method. After the first trial, we will notice that the BC

y(xn) is not equal to yn, so we have to adjust (shoot) the initial

guess of z(x0) until it matches the BC y(xn)=yn. Therefore this

method is called shooting method.

Here Eq. (6.2) can be transformed into a system of IVP

as below:

)(')(")(' xzxythenzxy n ==

,'0

0

xx

yyyz

n

n

==


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Example 6

A simple RL-electrical circuit consists of electrical current i (in

ampheres), resistanceR (in ohms), inductanceL (in hengrys), and

electromotive forceE(t) (in volts), as shown in Figure below.

Given that q (in coulombs) is the charge, and R, and L are assumed

constants. Given that i = dq/dtand di/dt= d2q/dt2. According toKirchoffs second law, the current I satisfies the differential equation

Given thatR = 15,L = 3,E(t) = 120 and the BCs as follow:

q = 0 at t= 0 and q = 14.4 at t= 2

Find i and q for 0t 2 with h = 0.25 using Shooting method.

( )tEdt

dqR

dt

xdL =+

2

2


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Solution

7 - 33

STEP 1

Since the second initial condition i is unknown at t= 0, we can

guess this equal to 2.7

02

04.14)0( =

==dt

dqi

Solving by RK4 as in Example 5 gives the results in following

table.


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0 0 0.0000 7.2000 1.8000 1.0000 1.9250 0.3750 1.8469

1 0.25 1.8892 7.7540 1.9385 0.3075 1.9769 0.1153 1.9529

2 0.5 3.8551 7.9244 1.9811 0.0945 1.9929 0.0354 1.9855

3 0.75 5.8447 7.9767 1.9942 0.0291 1.9978 0.0109 1.9955

4 1 7.8414 7.9929 1.9982 0.0089 1.9993 0.0034 1.9986

5 1.25 9.8404 7.9978 1.9995 0.0027 1.9998 0.0010 1.9996

6 1.5 11.8401 7.9993 1.9998 0.0008 1.9999 0.0003 1.9999

7 1.75 13.8400 7.9998 1.9999 0.0003 2.0000 0.0001 2.00008 2 15.8400 7.9999 2.0000 0.0001 2.0000 0.0000 2.0000

g3 f4 g4

0.7656 1.9914 0.0430

0.2354 1.9974 0.0132

0.0724 1.9992 0.0041

0.0223 1.9998 0.00120.0068 1.9999 0.0004

0.0021 2.0000 0.0001

0.0006 2.0000 0.0000

0.0002 2.0000 0.0000

0.0001 2.0000 0.0000


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Solution

7 - 35

STEP 2

Since q(2) = 15.84 is higher than the value of BC given, so we

have to reduce our initial guess ofi(0). Let try i(0) = 1 and resolve

again using RK4.


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0 0 0.0000 1.0000 0.2500 8.7500 1.3438 3.2813 0.6602

1 0.25 1.0304 5.8478 1.4620 2.6902 1.7982 1.0088 1.5881

2 0.5 2.7323 7.3383 1.8346 0.8271 1.9380 0.3102 1.8733

3 0.75 4.6407 7.7966 1.9491 0.2543 1.9809 0.0954 1.9611

4 1 6.6125 7.9375 1.9844 0.0782 1.9941 0.0293 1.9880

5 1.25 8.6038 7.9808 1.9952 0.0240 1.9982 0.0090 1.9963

6 1.5 10.6012 7.9941 1.9985 0.0074 1.9994 0.0028 1.9989

7 1.75 12.6004 7.9982 1.9995 0.0023 1.9998 0.0009 1.99978 2 14.6001 7.9994 1.9999 0.0007 1.9999 0.0003 1.9999

g3 f4 g4

6.6992 1.9248 0.3760

2.0597 1.9769 0.1156

0.6333 1.9929 0.0355

0.1947 1.9978 0.01090.0599 1.9993 0.0034

0.0184 1.9998 0.0010

0.0057 1.9999 0.0003

0.0017 2.0000 0.0001

0.0005 2.0000 0.0000


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Solution

STEP 3Again this time, q(2) = 14.6 is still higher than the value of BC

given. If we try to reduce the initial guess ofi(0) slowly, we will

achieve at one stage that the result ofq(2) match the BC given.

However, a faster way rather than slowly adjusting the initial

guess is using the linear interpolation based on the data obtainedin step 1 and step 2 ( or after two trials).

j 0 1 2

q(2) 15.84 14.6 14.4

i(0) 7.2 1?

0006.0

)1(84.156.14

84.154.14)2.7(

6.1484.15

6.144.14)4.14(

)( 101

00

10

1

=

+

=

+

=

f

iqq

qqi

qq

qqqf


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Solution

STEP 4So in the third trial we use the initial guess ofi(0) = -0.0006 and

resolve again using RK4. This time we will see the BCs is

fulfilled.


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0 0 0.0000 -0.0006 -0.0002 10.0008 1.2499 3.7503 0.4686

1 0.25 0.8918 5.5402 1.3850 3.0748 1.7694 1.1530 1.5292

2 0.5 2.5511 7.2437 1.8109 0.9454 1.9291 0.3545 1.8552

3 0.75 4.4464 7.7675 1.9419 0.2907 1.9782 0.1090 1.9555

4 1 6.4142 7.9285 1.9821 0.0894 1.9933 0.0335 1.9863

5 1.25 8.4043 7.9780 1.9945 0.0275 1.9979 0.0103 1.9958

6 1.5 10.4012 7.9932 1.9983 0.0084 1.9994 0.0032 1.9987

7 1.75 12.4003 7.9979 1.9995 0.0026 1.9998 0.0010 1.99968 2 14.4000 7.9994 1.9998 0.0008 1.9999 0.0003 1.9999

g3 f4 g4

7.6568 1.9141 0.4297

2.3541 1.9736 0.1321

0.7238 1.9919 0.0406

0.2225 1.9975 0.01250.0684 1.9992 0.0038

0.0210 1.9998 0.0012

0.0065 1.9999 0.0004

0.0020 2.0000 0.0001

0.0006 2.0000 0.0000


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Example 7

The radial temperature distribution in a cylinder is governed by

Using shooting method to find the radial temperature distribution if the

inner radius is 5 units and outer radius is 10 units. The inner and outersurface are maintained at 120C and 60 C respectively with h = 1.

01

2

2

=+dt

dT

rdr

Td


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Solution

7 - 41

STEP 1

T(5) =120, T(10) =60

( )

?)5(;),,(

,,

2

2

====

==

ssTrgr

s

dr

ds

dr

Td

sTrfsdr

dT

Since we do not have the information of s(5), we can guess that

12510

12060)5( =

==dr

dTs

Solving by RK4.


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SolutionUsing RK4 method,

( ) ( )4321143211 226

1;22

6

1ggggssffffTT iiii ++++=++++= ++

hr

gsg

hr

gs

g

hr

gs

g

r

sg

gsfg

sfg

sfsf

+

+=

+

+=

+

+==

+=+=+==

14

2

3

1

21

3423121

;

2

2;

2

2;

;2;2;

where:


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Solutioni r T s f1 g1 f2 g2 f3

0 5 120.0000 -12.0000 -12.0000 2.4000 -10.8000 1.9636 -11.01821 6 109.0612 -10.0000 -10.0000 1.6667 -9.1667 1.4103 -9.2949

2 7 99.8123 -8.5714 -8.5714 1.2245 -7.9592 1.0612 -8.0408

3 8 91.8005 -7.5000 -7.5000 0.9375 -7.0313 0.8272 -7.0864

4 9 84.7336 -6.6667 -6.6667 0.7407 -6.2963 0.6628 -6.3353

5 10 78.4120 -6.0000 -6.0000 0.6000 -5.7000 0.5429 -5.7286

g3 f4 g4

2.0033 -9.9967 1.6661

1.4300 -8.5700 1.2243

1.0721 -7.4993 0.9374

0.8337 -6.6663 0.7407

0.6669 -5.9998 0.6000

0.5456 -5.4544 0.4959


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Solution

7 - 44

STEP 2

Since T(10) = 78.412 is higher than the value of BC given, so we

have to reduce our initial guess of s(5). Let try s(5) = -15 and

resolve again using RK4.


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0 5 120.0000 -15.0000 -15.0000 3.0000 -13.5000 2.4545 -13.77271 6 106.3264 -12.5000 -12.5000 2.0833 -11.4583 1.7628 -11.6186

2 7 94.7654 -10.7143 -10.7143 1.5306 -9.9490 1.3265 -10.0510

3 8 84.7506 -9.3750 -9.3750 1.1719 -8.7891 1.0340 -8.8580

4 9 75.9170 -8.3333 -8.3333 0.9259 -7.8704 0.8285 -7.9191

5 10 68.0150 -7.5000 -7.5000 0.7500 -7.1250 0.6786 -7.1607

g3 f4 g4

2.5041 -12.4959 2.0826

1.7875 -10.7125 1.5304

1.3401 -9.3741 1.1718

1.0421 -8.3329 0.9259

0.8336 -7.4997 0.7500

0.6820 -6.8180 0.6198


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Solution

STEP 3Again this time, T(10) = 68.015 is still higher than the value of

BC given. Use a faster way rather than slowly adjusting the initial

guess, using the linear interpolation based on the data obtained in

step 1 and step 2 ( or after two trials).

j 0 1 2

T 78.412 68.015 60

s -12 -15?

3127.17

)15(412.78015.68

412.7860)12(

015.68412.78

015.6860)60(

)( 101

00

10

1

=

+

=

+

=

f

sTT

TTsTT

TTTf


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SolutionSTEP 4

So in the third trial we use the initial guess ofs(5) = -17.3127 and

resolve again using RK4. This time we will see the BCs is

fulfilled.


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0 5 120.0000 -17.3127 -17.3127 3.4625 -15.5814 2.8330 -15.89621 6 104.2183 -14.4273 -14.4273 2.4045 -13.2250 2.0346 -13.4099

2 7 90.8747 -12.3662 -12.3662 1.7666 -11.4829 1.5311 -11.6007

3 8 79.3159 -10.8204 -10.8204 1.3526 -10.1442 1.1934 -10.2237

4 9 69.1203 -9.6182 -9.6182 1.0687 -9.0838 0.9562 -9.1401

5 10 59.9999 -8.6564 -8.6564 0.8656 -8.2235 0.7832 -8.2648

g3 f4 g4

2.8902 -14.4225 2.4037

2.0631 -12.3642 1.7663

1.5468 -10.8195 1.3524

1.2028 -9.6176 1.0686

0.9621 -8.6561 0.8656

0.7871 -7.8692 0.7154


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Example 8

Solve the BVP ofexy+xy-5(1 +x)y =x3

with the following BCs

y(0) = 2 , y(2) = 5

Atx = 0(0.5)2 using shooting method.


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Solution

BVP Fi it Diff


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BVP: Finite-DifferenceMethod

To solve Eq. (6.2) using FDM, the interval ofx is divided inton subintervals with width = h such thatxi =x0 + ih. Then

substitute

Into the Eq. (6.2) yields a tridiagonal system ofn - 2 equations

to be solved (solve for y from i = 1 to n - 10

h

yy

xyandh

yyy

xy

iiiii

2)('

2

)("

11

2

11 ++

=

+

=


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Example 9

Solve the BVP ofexy+xy-5(1 +x)y =x3

with the following BCs

y(0) = 2 , y(2) = 5

Atx = 0(0.5)2 using FDM.


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Solution

7 - 53

STEP 1

( )

( ) 3112

11

3

15

2

2

5.0,15'"

iiiii

iiiix

x

xyx

h

yyx

h

yyye

hxyxxyye

i =+

+

+==++

++

Multiply by h2 = 0.25 at both hand-side yields

( ) ( ) ( ) 31111 25.0125.125.02 iiiiiiiiix xyxyyxyyye i =+++ ++

Rearranged this equation

( ) ( )( ) ( )3

11 25.025.0125.1225.0 iiix

ii

x

ii

x xyxeyxeyxe iii =++++ +


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SolutionSTEP 2

Put in the BCs in the table below

l xi yi

0 0 2

1 0.5 y1

2 1.0 y2

3 1.5 y3

4 2 5There are 3 value of y to be solved. Therefore 3

simultaneous equations are required.


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SolutionSTEP 3

Obtain the equations by inserting value ofx based on ii (exi

0.25xi)yi-

1

+ (-2exi-1.25(1+xi)

yi + (exi

0.25xi)Yi+1 = 0.25xi

3

1 1.524 y0 + -5.172 y1 + 1.774 y2 = 0.031

2 2.468 y1 + -7.937 y2 + 2.968 y3 = 0.250

3 4.107 y2 + -12.088 y3 + 4.857 y4 = 0.844


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SolutionSTEP 4

Substituting BCs, y(0)=2=y0; y(2)=5=y4i (exi

0.25xi)yi-

1

+ (-2exi-1.25(1+xi)

yi + (exi

0.25xi)yi+1 = 0.25xi

3

1 1.524 2 + -5.172 y1 + 1.774 y2 = 0.031

2 2.468 y1 + -7.937 y2 + 2.968 y3 = 0.250

3 4.107 y2 + -12.088 y3 + 4.857 5 = 0.844

i (exi 0.25xi)

yi-

1

+ (-2exi-1.25(1+xi)

yi + (exi

0.25xi)yi+1 = 0.25xi

3

1 -5.172 y1 + 1.774 y2 = -3.017

2 2.468 y1 + -7.937 y2 + 2.968 y3 = 0.250

3 4.107 y2 + -12.088 y3 = -23.441

yields


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Solution

STEP 5

Solve by Gauss-Seidel Iteration method

=

441.23

250.0

017.3

088.12107.40

968.2937.7468.2

0774.1172.5

3

2

1

y

y

y

088.12

107.4441.23

937.7

968.2468.2250.0

172.5

774.1107.3

)1(

2)1(

3

)(

1

)1(

1)1(

2

)(

2)1(

1

=

=

=

++

++

+

kk

kkk

kk

yy

yy

y

yy


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Solutionk y1 y2 y3

0 0.000 0.000 0.000

1 0.583 0.150 1.990

2 0.635 0.910 2.248

3 0.895 1.088 2.309

4 0.956 1.129 2.323

5 0.971 1.139 2.326

6 0.974 1.141 2.327

Since max{|y(6) y(5)|} = max {0.003,0.002,0.001} = 0.003 < =0.005,

so the solution is y1 = 0.974, y2 = 1.141 and y3=2.327.


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Exercise

7 - 59


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SolutionSTEP 3

Obtain the equations by inserting value ofx based on ii yi-

1

+ yi + Yi+1 =

1 0.976 y0 + -1.7681 y1 + 0.888 y2 = 0.9924

2 1.0814 y1 + -1.8589 y2 + 0.8894 y3 = 0.0135

3 1.1556 y2 + -1.8711 y3 + 0.8436 y4 = 0.0112

4 1.1978 y3 + -1.8037 y4 =

-1.4902


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SolutionSTEP 3

Obtain the equations by inserting value ofx based on ii yi-

1

+ yi + Yi+1 =

1 y0 + -1.7681 y1 + 0.888 y2 = 0.9924

2 1.0814 y1 + -1.8589 y2 + 0.8894 y3 = 0.0135

3 1.1556 y2 + -1.8711 y3 + 0.8436 y4 = 0.0112

4 1.1978 y3 + -1.8037 y4 =

-1.4902

chapter 6 - ordinary differential equation

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