 # The Ordinary Differential Equations Biology Essay

Ordinary differential equations have really immense scope of applications in a assortment of Fieldss, as they express about systems undergoing alteration such as exchanges of affair, energy, information or any other measures, frequently as they vary in clip and/or infinite their thorough out analytical intervention forms the footing of cardinal theories in Mathematics, Physics and other countries. Mathematicians have studied the nature of differential equations and do many well-developed solution techniques. Frequently, systems expressed by differential equations are so complex, or the systems that they illustrate are so big, that a strictly analytical solution to the equations is non manipulable. The numerical solution of these types of systems has been the topic of intense research which leads methods to solution of jobs. But numerical methods do n’t offer the exact reply to a given job and can merely be given to a solution, acquiring closer to it with each loop. Numerical techniques provide discontinuous points of a curve and therefore it is normally clip devouring to acquire a complete curve of consequences and so in these methods peculiarly for non-linear equations, stableness and convergence should be considered so as to maintain away from divergency or unsuitable consequences.

Besides these two techniques the series enlargement method is besides a process to work out the differential equations which is mediate to both. Taylor series method is one for the solution of maps in the signifier of multinomials which provides extremely accurate consequences of differential equations. But this is time-consuming particularly for high order equation, and go highly hard to happen the general n-th term derived function for some maps, series does n’t meet rapidly a new attack which formulize Taylor series in little different mode and reduces symbolic calculation of the necessary derived functions is Differential transmutation method ( DTM ) .

The idea of differential transmutation method was foremost proposed in 1986 by a Chinese “ Zhou jiakui ” and its chief applications are to work out both additive and non-linear initial value jobs in electric circuit analysis. This method constructs a semi-analytical numerical technique that uses Taylor series for the solution of differential equations in the signifier of a multinomial estimate. This is an iterative process provides consequences nearer to demand solution after each loop. DTM is widespread and is able to work out assorted sorts of maps both additive, non-linear, higher order ODEs affecting a big figure of variables where direct methods would be prohibitively expensive ( and in some instances impossible ) even with the best available calculating power.

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In this thesis the DTM is applied to work out higher order differential equations such as greater so 2nd to eighth order differential equations additive and non- linear and method is compared with analytical and numerical methods by taking the suited points in the given interval of sphere. This partly numerical method eliminates cumbrous computations needed to obtain in both pure numerical and analytical attack. When comparings are made in surveies of jobs a really good understanding is observed. This individual method can be applied to work out different ODEs topic to specified initial conditions both for additive and non-linear differential equations.

## 2.2 Differential Transformation Method

Al-Salawalha and Noorani ( 2007 ) developed the differential transmutation method ( DTM ) to work out Lorenz system in which DTM solution is compared with fourth-order Runge-Kutta method which is sufficiently accurate for comparing intents. Erturk ( 2007 ) applied these techniques to sixth-order boundary value job and equations of Lane Emden type and concluded that DTM is really fast convergent, precise and cost efficient tool for work outing Lane-Emden equations.Patrico and Rosa ( 2007 ) solved out Adection-diffusion jobs by utilizing DTM in their paper the attack for work outing PDEs is presented and Eigen values which depends on parametric quantities that appear in PDEs are studied. Nejatbakhsh ( 2007 ) proposed the DTM to work out fuzzed built-in equations. El-Shahed ( 2008 ) used DTM to work out non-linear oscillating systems and demonstrated DTM does non necessitate little parametric quantities in the equations so restriction of traditional disturbance methods can be eliminated.Attarnejad and Shahba ( 2008 ) applied DTM in free quiver analysis of revolving non-prismatic beams. Kai-Tai and Bin ( 2009 ) used DTM for non-linear differential equations in which this method is demonstrated by managing ( 1+3 ) dimensional Burgers equations and two dimensional heat and diffusion equations their work shows dependability and efficiency of proposed method.Demir and Sungu ( 2009 ) used DTM for numerical solution for category of non-linear Emeden-flower equations, as a category of non remarkable initial value job they introduced and implemented their method and illustrated the efficiency and simpleness of method. Abazari, N and Abazari, R ( 2009 ) has obtained the solution of non-linear second-order pantograph equations by differential transmutation method. Attarnejad ( 2009 ) applied DTM in free quiver analysis of Timoshenko beams resting on two-parametric elastic foundation. In order to happen out solution of Cauchy reaction-diffusion jobs DTM and fluctuation loop method was applied by Othman and Mahdy ( 2010 ) and the concluded that both methods provide same consequences. Rashidi ( 2010 ) presented the DTM attack to the job of micro-polar flow in porous channel with mass injection and first-class understanding was noted between DTM and fourth-order Range Kutta method. Biazar and Mohammadi ( 2010 ) applied DTM to generalise Burgers Huxley equation and concluded that DTM is more effectual and convenient every bit compared to spheres decomposition method and fluctuation loop method.

## 3.1.1 LINEAR DIFFERENTIAL EQUATIONS

A differential equation is additive if there does non be generations along with dependent variables and their derived functions or all coefficients are maps of dependent variables.

## 3.1.2 NON-LINEAR DIFFERENTIAL EQUATIONS

Differential equations that do non fulfill the definition of additive differential equations are non-linear.

## 3.1.3 TAYLOR SERIES

Taylor series method is one of the most crude algorithms for approximative solution ordinary differential equations.The enterprise of the rehabilitation of these algorithms is based on the approximative computation of higher derived functions. The Taylor series of a existent or complex map degree Fahrenheit ( ten ) that is boundlessly differentiable in a vicinity of a existent or complex figure a is the power series

## 3.2 DIFFERENTIAL TRANSFORMATION METHOD ( DTM )

The differential transmutation method is utilised to work out higher order initial value jobs. The application of present method gives rapid convergence and unbelievable truth. This attack can easy be applied additive and non-linear jobs and reduces the needed computational attempt with this method exact solution may be obtained without any demand of onerous calculations. To clear up and represent capableness of proposed technique assorted jobs have been solved. When comparing is made with the surveies a really good understanding is observed. Consequences are compared with analytical solution by taking suited points to given interval of sphere.

The thought of the differential transform was foremost recommended in 1986 by a Chinese “ Zhou jiakui ” and its major applications therein are to work out both additive and non-linear initial value jobs of electric circuit analysis.DTM is a transformed technique based on a Taylor series expression and is a helpful implement to obtain analytic solution of differential equations. In this method certain transmutation regulations are applied to regulating differential equations and boundary conditions of the system are transformed into return equations that eventually lead to a solution of system of algebraic equations, the solution of thesealgebraic equations is coveted solution of the job.

## 3.2.1 Definition

See a map Y ( x ) which is analytic in a sphere D and say x=x0is a point in D. The map Y ( x ) is so given by a power series whose centre is located at x0. The differential transform of the n-th derived function of a map Y ( ten ) in one variable is defined as below:

( 3.1 )

In Eq. ( 3.1 ) Y ( x ) is the original map and Y ( K ) is the transformed map and differential opposite transform of Y ( K ) is defined as follows

( 3.2 )

In existent applications, map Y ( x ) is expressed by a finite series and Eq. ( 3.2 ) can be written as.

( 3.3 )

Uniting Eq.s ( 3.1 ) and ( 3.2 ) the new look for Y ( ten ) reduces to

( 3.4 )

Eq. ( 3.4 ) implies the thought of differential transform is attendant from Taylor series method. Though differential transmutation method is non capable to cipher the derived functions, comparative derived functions can be computed by an iterative mode which can be described by the transformed equations of original maps. Eq. ( 3.4 ) shows the undermentioned portion

( 3.5 )

Is negligibly little. Actually, n is decided by the convergence of natural frequence. The undermentioned theorems that can be deduced from Eq. ( 3.1 ) and ( 3.2 ) are given as:

## Proof

Substitute the above Eq. in to Eq. ( 3.1 )

This completes cogent evidence

## Proof

Substitute the above Eq. in to Eq. ( 3.1 )

This completes cogent evidence

## Proof

Substitute the above Eq. in to Eq. ( 3.1 )

This completes cogent evidence

## Proof

Substitute the above Eq. in to Eq. ( 3.1 )

( 3.6 )

By Labniz theorem

( 3.7 )

Substitute Eq. ( 3.7 ) to Eq. ( 3.6 )

This completes cogent evidence

In general

Where

## Proof

Substitute the above Eq. in to Eq. ( 3.1 )

This completes cogent evidence

Some of import maps and there differential transmutations are given in table 3.1

## Differential Transformation

Cos ( x )

Sin ( x )

exp ( ten )

1/k

exp ( -x )

K

This method constructs a semi-analytical numerical technique that uses Taylor series for the solution of differential equations in the signifier of a multinomial. It is different from the high-order Taylor series method which requires symbolic calculation of the necessary derived functions of the information maps. The Taylor series method is computationally time-consuming particularly for high order equations. The differential transform is an iterative process for obtaining analytic Taylor series solutions of differential equations. It can be said that differential transform method is a cosmopolitan one, and is able to work out assorted sorts of differential equations.

## Structural kineticss jobs:

Structural kineticss is the subset of structural analysis which covers the behaviour of constructions as subjected to the kineticss. A simple second-order kineticss equation is a vibrating equation of individual grade of freedom.

( 4.1 )

Where = , m is the mass degree Celsius is the muffling value, K is elastic coefficient P and qare bing forces, amplitude and frequence severally, and are acceleration, speed and supplanting severally, The structural kineticss equations has two initial conditions related speed set supplanting all the initial conditions are as follows.

( 4.2 ) The differential transmutation method is applied to four different sorts of vibrating equations. All has the following informations with units omitted

( 4.3 )

As discussed above, the differential transform and reverse differential transform by a finite series is given as below.

( 4.4 )

( 4.5 )

Where Eq.s ( 4.4 ) and ( 4.5 ) are differential transforms of present equation and initial conditions severally. The transformed initial conditions can be obtain by puting t=0, and t0 =0 in Eq. ( 4.5 ) which are as below.

( 4.6 )

( 4.7 )

( 1 )

## Analytic Solution:

0a‰¤ta‰¤1 ( 2 )

Using DTM to eq. ( 1 )

( 3 )

At k=0 in Eq. ( 3 )

Y ( 2 ) =

( 4 )

Similarly

( 5 )

( 6 )

## Y ( K )

0

1.000000000000000

1

2.000000000000000

2

-0.500000000000000

3

-0.333333333333333

4

0.041666666666667

5

0.016666666666667

6

-0.001388888888889

7

-0.000396825396825

8

0.000024801587302

9

0.000005511463845

10

-0.000000275573192

11

-0.000000050104217

12

0.000000002087676

13

0.000000000321181

14

-0.000000000011471

15

-0.000000000001529

16

0.000000000000048

17

0.000000000000006

18

-0.000000000000000

19

-0.000000000000000

In the same manner seting values into return equation ( 3 ) continually we can obtain tabular array of estimates utilizing MATLAB

Table 1Transformed loops of return Eq. ( 3 ) )

Substituting values from above tabular array into equation ( 4.5 ) at t0=0 the needed approximative solution of Eq. ( 1 ) is as follows in the signifier of multinomials.

## Table 2

Comparison between Analytical and DTM solution as 0a‰¤ta‰¤1

## N=20

0

1.000000000000000

1.000000000000000

1.000000000000000

1.000000000000000

0.1

1.194670998571682

1.194670998571682

1.194670998571682

1.194670998571682

0.2

1.377405239431364

1.377405239431365

1.377405239431365

1.377405239431364

0.3

1.546376902448285

1.546376902448373

1.546376902448373

1.546376902448285

0.4

1.699897678620186

1.699897678622250

1.699897678622250

1.699897678620186

0.5

1.836433639098779

1.836433639122696

1.836433639122696

1.836433639098779

0.6

1.954620561699749

1.954620561876572

1.954620561876572

1.954620561699749

0.7

2.053277561759871

2.053277562718671

2.053277562718671

2.053277561759870

0.8

2.131418891146211

2.131418895289567

2.131418895289567

2.131418891146211

0.9

2.188263787525631

2.188263802580534

2.188263802580534

2.188263787525622

1.0

2.223244323192242

2.223244323192242

2.223244275483880

2.223244275483933

## Table 3

Mistake estimates =Analytical Solution -DTM Solution

## N=20

0

0

0

0

0.1

-0.000000002220446A-1.0e-7

-0.000000002220446A-1.0e-7

-0.000000002220446A-1.0e-7

0.2

-0.000000011102230A-1.0e-7

-0.000000011102230A-1.0e-7

0

0.3

-0.000000879296636A-1.0e-7

-0.000000879296636A-1.0e-7

-0.000000002220446A-1.0e-7

0.4

-0.000020641266474A-1.0e-7

-0.000020641266474A-1.0e-7

0

0.5

-0.000239170905303A-1.0e-7

-0.000239170905303A-1.0e-7

-0.000000004440892A-1.0e-7

0.6

-0.001768225565968A-1.0e-7

-0.001768225565968A-1.0e-7

-0.000000002220446A-1.0e-7

0.7

-0.009588001503857A-1.0e-7

-0.009588001503857A-1.0e-7

0.000000008881784A-1.0e-7

0.8

-0.041433563247040A-1.0e-7

-0.041433563247040A-1.0e-7

0.000000004440892A-1.0e-7

0.9

-0.150549026578517A-1.0e-7

-0.150549026578517A-1.0e-7

0.000000093258734A-1.0e-7

1.0

-0.477083093031183A-1.0e-7

-0.477083093031183A-1.0e-7

0.000000510702591A-1.0e-7

## 4.1.2 Second sort of vibrating equation:

The 2nd sort of vibrating equation and its analytical solution are given as below

( 1 )

## Analytic Solution:

0a‰¤ta‰¤1 ( 2 )

Using DTM to Eq. ( 1 )

( 3 )

At k=0 in combining weight. ( 3 )

( 4 )

Similarly

( 6 )

] ( 7 )

In the same mode seting values into return equation ( 3 ) continually we can obtain tabular array of estimates utilizing MATLAB

## Table 4

Transformed loops of return Eq. ( 3 )

## Y ( K )

0

1.000000000000000

1

2.000000000000000

2

0

3

-0.333333333333333

4

-0.083333333333333

5

0.016666666666667

6

0.008333333333333

7

-0.000396825396825

8

-0.000347222222222

9

0.000005511463845

10

0.000008267195767

11

-0.000000050104217

12

-0.000000129435893

13

0.000000000321181

14

0.000000001445314

15

-0.000000000001529

16

-0.000000000012140

17

0.000000000000006

18

0.000000000000080

19

-0.000000000000000

HerePutting values from above tabular array into Eq. ( 4.5 ) at t0=0 the needed approximative solution of Eq. ( 1 ) is as follows in the signifier of multinomials.

## Table 5

Comparison between analytical and differential transform method

## N=20

0

1.000000000000000

1.000000000000000

1.000000000000000

1.000000000000000

0.1

1.199650194383944

1.199658508290185

1.199658508290185

1.199658508290185

0.2

1.397073767377483

1.397205860702081

1.397205860702080

1.397205860702080

0.3

1.589710527186928

1.590371465590335

1.590371465590177

1.590371465590177

0.4

1.774682440172093

1.776737257930723

1.776737257926456

1.776737257926456

0.5

1.948860391106190

1.953771603969460

1.953771603913522

1.953771603913522

0.6

2.108946515177418

2.118867964960000

2.118867964498021

2.118867964498017

0.7

2.251569243221287

2.269387667320681

2.269387664551305

2.269387664551264

0.8

2.373387635029029

2.402706019546526

2.402706006429009

2.402706006428669

0.9

2.471201141846882

2.516260924708550

2.516260872840763

2.516260872838526

1.0

2.542060656289027

2.607603064373899

2.607602886598754

2.607602886586701

## Table 6

Mistake estimates =Analytical Solution -DTM Solution

## N=20

0

0

0

0

0.1

-0.000008313906242

-0.000008313906242

-0.000008313906242

0.2

-0.000132093324598

-0.000132093324597

-0.000132093324597

0.3

-0.000660938403407

-0.000660938403249

-0.000660938403249

0.4

-0.002054817758630

-0.002054817754363

-0.002054817754363

0.5

-0.004911212863270

-0.004911212807332

-0.004911212807332

0.6

-0.009921449782583

-0.009921449320603

-0.009921449320600

0.7

-0.017818424099394

-0.017818421330018

-0.017818421329978

0.8

-0.029318384517497

-0.029318371399980

-0.029318371399640

0.9

-0.045059782861668

-0.045059730993881

-0.045059730991644

1.0

-0.065542408084872

-0.065542230309728

-0.065542230297674

## 4.1.3 Third sort of vibrating equation:

The 3rd sort of homogenous vibrating equation and its analytical solution are as below

( 1 )

## Analytic Solution:

( 2 )

Using DTM to Eq. ( 1 )

( 3 ) At k=0 in Eq. ( 3 )

( 4 ) Similarly

( 5 )

In the same mode seting values into return Eq. ( 3 ) continually we can obtain tabular array of estimates utilizing MATLAB

## Y ( K )

0

2.000000000000000

1

-0.600000000000000

2

-0.313333333333333

3

0.057833333333333

4

0.014510000000000

5

-0.002169611111111

6

-0.000314481746032

7

0.000042674077381

8

0.000003893645613

9

-0.000000513092871

10

-0.000000030732298

11

0.000000004143170

12

0.000000000165131

13

-0.000000000023944

14

-0.000000000000627

15

0.000000000000104

16

0.000000000000002

Heresubstituting values from above tabular array into Eq. ( 4.5 ) at t0=0 the needed approximative solution of Eq. ( 1 ) is as follows in the signifier of multinomials

( 7 )

## Table 8

Comparison between analytical and differential transmutation method

## N=20

0

1.000000000000000

1.000000000000000

1.000000000000000

1.000000000000000

0.1

1.193696242661779

1.193692592899371

1.193692592899371

1.193692592899371

0.2

1.373598759750577

1.373590367097376

1.373590367097375

1.373590367097375

0.3

1.538056408079735

1.538042061749797

1.538042061749745

1.538042061749745

0.4

1.685588009517719

1.685566409359774

1.685566409358555

1.685566409358555

0.5

1.814895384052218

1.814865170903739

1.814865170889763

1.814865170889763

0.6

1.924874434006965

1.924834221644926

1.924834221542646

1.924834221542646

0.7

2.014624188615535

2.014572596903013

2.014572596354116

2.014572596354117

0.8

2.083453739335702

2.083389428354768

2.083389426007603

2.083389426007607

0.9

2.130887017948563

2.130808723321962

2.130808714884362

2.130808714884381

1.0

2.156665391408469

2.156571961772980

2.156571935324412

2.156571935324518

## Table 9

Mistake estimates =analytical Solution -differential transmutation method

## N=20

0

0

0

0

0.1

0.036497624074094A-1.0e-4

0.036497624074094A-1.0e-4

0.036497624074094A-1.0e-4

0.2

0.083926532012946A-1.0e-4

0.083926532019607A-1.0e-4

0.083926532019607A-1.0e-4

0.3

0.143463299375757A-1.0e-4

0.143463299897562A-1.0e-4

0.143463299897562A-1.0e-4

0.4

0.216001579451675A-1.0e-4

0.216001591635262A-1.0e-4

0.216001591635262A-1.0e-4

0.5

0.302131484788415A-1.0e-4

0.302131624547730A-1.0e-4

0.302131624547730A-1.0e-4

0.6

0.402123620386607A-1.0e-4

0.402124643192892A-1.0e-4

0.402124643192892A-1.0e-4

0.7

0.515917125221144A-1.0e-4

0.515922614185982A-1.0e-4

0.515922614181541A-1.0e-4

0.8

0.643109809339570A-1.0e-4

0.643133280986419A-1.0e-4

0.643133280955333A-1.0e-4

0.9

0.782946266015827A-1.0e-4

0.783030642015348A-1.0e-4

0.783030641819948A-1.0e-4

1.0

0.934296354886399A-1.0e-4

0.934560840568643A-1.0e-4

0.934560839507270A-1.0e-4

## 4.1.4 Fourth-Kind of Vibrating Equation

The 4th sort of vibrating equation and its analytical solution are as below

0a‰¤ta‰¤1 ( 1 )

Analytic Solution: ) ( 2 ) where C=

D=

B=C

A=

Using DTM to q. ( 1 )

( 3 )

By replacing k=0,1,2, aˆ¦We can obtain tabular array of estimates

Table 10 Transformed loops of return Eq. ( 3 )

## Y ( K )

0

1.000000000000000

1

2.000000000000000

2

-0.600000000000000

3

-0.146666666666667

4

0.053666666666667

5

-0.010406666666667

6

-0.001615444444444

7

0.001064506349206

8

0.000015540892857

9

-0.000037003342372

10

0.000000197356836

11

0.000000735433603

12

-0.000000007623741

13

-0.000000009794568

14

0.000000000111850

15

0.000000000094837

16

-0.000000000001059

17

-0.000000000000702

18

0.000000000000007

19

0.000000000000004

20

-0.000000000000000

Heresubstituting values from above tabular array into Eq. ( 4.5 ) at t0=0 the needed approximative solution of combining weight. ( 1 ) is as follows in the signifier of multinomials

## Table 11

Comparison between analytical and differential transmutation method

## N=20

0

1.000000000000000

1.000000000000000

1.000000000000000

1.000000000000000

0.1

1.193858594424458

1.193858594424458

1.193858594424458

1.193858594424495

0.2

1.374909113458111

1.374909113458111

1.374909113458111

1.374909113477022

0.3

1.542448467242302

1.542448467242302

1.542448467242302

1.542448467968177

0.4

1.695875763496114

1.695875763496114

1.695875763496114

1.695875773144971

0.5

1.834678689119651

1.834678689119651

1.834678689119641

1.834678760843022

0.6

1.958420298740042

1.958420298740042

1.958420298739910

1.958420667816260

0.7

2.066726685385884

2.066726685385884

2.066726685384680

2.066728158686670

0.8

2.159275965028245

2.159275965028245

2.159275965020034

2.159280848222643

0.9

2.235788953070564

2.235788953070562

2.235788953025762

2.235802993920645

1.0

2.296021848365647

2.296021848365636

2.296021848160710

2.296057936130952

## Table 12

Error estimate =Analytical method -Differential transmutation method

## N=20

0

0

0

0

0.1

-0.000000000370814A-1.0e-4

0

0

0.2

-0.000000189104288A-1.0e-4

0

0

0.3

-0.000007258742496A-1.0e-4

0

0

0.4

-0.000096488570467A-1.0e-4

0.000000000002220A-1.0e-4

0

0.5

-0.000717233703540A-1.0e-4

0.000000000097700A-1.0e-4

0

0.6

-0.003690762182540A-1.0e-4

0.000000001318945A-1.0e-4

0

0.7

-0.014733007858148A-1.0e-4

0.000000012034818A-1.0e-4

0

0.8

-0.048831943986194A-1.0e-4

0.000000082103213A-1.0e-4

0

0.9

-0.140408500812761A-1.0e-4

0.000000448014958A-1.0e-4

0.000000000013323A-1.0e-4

1.0

-0.360877653049485A-1.0e-4

0.000002049369563A-1.0e-4

0.000000000111022A-1.0e-4

## The given equation is:

( 1 )

Capable to specified initial conditions

Analytic Solution ( 2 )

## DTM Solution

The DTM of kth order derived function of a map Y ( x ) is defined as.

( 3 )

And reverse differential transform is defined as.

( 4 )

As x0 =0 in order to acquire transform initial conditions put x=0 in Eq.

( 7 ) – ( 9 ) We get

( 5 )

( 6 )

( 7 )

Using DTM to Eq. ( 1 )

( 8 ) Puting values to Eq. ( 13 ) recurrently we can obtain the tabular array of estimates.

## Y ( K )

0

5.000000000000000

1

-5.000000000000000

2

0.500000000000000

3

-1.500000000000000

4

-0.541666666666667

5

-0.275000000000000

6

-0.084722222222222

7

-0.025595238095238

8

-0.006274801587302

9

-0.001413690476190

10

-0.000281360229277

11

-0.000051331770082

12

-0.000008544856635

13

-0.000001315717462

14

-0.000000187902284

15

-0.000000025058991

16

-0.000000003132135

17

-0.000000000368506

18

-0.000000000040944

19

-0.000000000004310

20

-0.000000000000431

21

-0.000000000000041

22

-0.000000000000004

Substituting values from tabular array to inverse differential transform we can obtain the solution of differential equation in signifier of multinomials

## N=20

0

5.000000000000000

5.000000000000000

5.000000000000000

5.000000000000000

0.1

4.503442995987397

4.503442995987396

4.503442995987395

4.503442995987395

0.2

4.007039566674863

4.007039566675950

4.007039566674863

4.007039566674863

0.3

3.499376448548929

3.499376448644623

3.499376448548930

3.499376448548930

0.4

2.966923861220081

2.966923863525813

2.966923861220081

2.966923861220080

0.5

2.393500761666350

2.393500788989731

2.393500761666400

2.393500761666350

0.6

1.759625149842015

1.759625356550858

1.759625149842965

1.759625149842015

0.7

1.041723348208621

1.041724495629814

1.041723348219960

1.041723348208621

0.8

0.211166432331796

0.211171510470659

0.211166432429060

0.211166432331796

0.9

-0.766905033774798

-0.766886128769735

-0.766905033126166

-0.766905033774798

1.0

-1.935015388128720

-1.934953979276896

-1.935015384582350

-1.935015388128721

( 9 )

## Table 14

Comparison between exact and approximative solution of Eq. ( 1 )

## Table 15

Error estimate =Analytic method – Differential transmutation method

## N=20

0

0

0

0

0.1

0.000000000008882 A-1.0e-4

0.000000000017764 A-1.0e-4

0.000000000017764 A-1.0e-4

0.2

-0.000000010871304 A-1.0e-4

0

0

0.3

-0.000000956932311 A-1.0e-4

-0.000000000004441 A-1.0e-4

-0.000000000004441 A-1.0e-4

0.4

-0.000023057316056 A-1.0e-4

-0.000000000004441 A-1.0e-4

0.000000000008882 A-1.0e-4

0.5

-0.000273233817794 A-1.0e-4

-0.000000000506262 A-1.0e-4

0

0.6

-0.002067088424873 A-1.0e-4

-0.000000009501289 A-1.0e-4

0.000000000002220 A-1.0e-4

0.7

-0.011474211925044 A-1.0e-4

-0.000000113382637 A-1.0e-4

0

0.8

-0.050781388632182 A-1.0e-4

-0.000000972643632 A-1.0e-4

0.000000000004996 A-1.0e-4

0.9

-0.189050050639628 A-1.0e-4

-0.000006486323700 A-1.0e-4

0

1.0

-0.614088518242628 A-1.0e-4

-0.000035463700865 A-1.0e-4

0.000000000008882 A-1.0e-4

## Problem 2

( 1 ) Subject to specified initial conditions.

Analytic Solution ( 2 )

Using DTM to Eq. ( 1 ) same as in job ( 1 )

( 3 )

( 4 )

Similarly

( 5 )

Using DTM to Eq. ( 1 )

( 6 ) Substituting values to equation ( 13 ) recurrently we can obtain the tabular array of estimates

## Table 16

Solution of return Eq. ( 6 )

## Y ( K )

0

0.500000000000000

1

-1.000000000000000

2

1.000000000000000

3

1.000000000000000

4

0.333333333333333

5

0.033333333333333

6

-0.011111111111111

7

-0.004761904761905

8

-0.000793650793651

9

-0.000044091710758

10

0.000008818342152

11

0.000002405002405

12

0.000000267222489

13

0.000000010277788

14

-0.000000001468255

15

-0.000000000293651

16

-0.000000000024471

17

-0.000000000000720

18

0.000000000000080

19

0.000000000000013

20

0.000000000000001

21

0.000000000000000

Substituting values into Eq. ( 7 ) at x0=0

+0

( 7 )

## Table 17

Solving for n=20:

## Mistake

0

0.500000000000000

0.500000000000000

0

0.1

0.411033655071385

0.411033655071474

-0.000000000000088

0.2

0.348543225883140

0.348543225928290

-0.000000000045150

0.3

0.319771805688618

0.319771807424332

-0.000000001735714

0.4

0.332820822995523

0.332820846112278

-0.000000023116755

0.5

0.396661010045277

0.396661182278522

-0.000000172233245

0.6

0.521126585154520

0.521127473840234

-0.000000888685714

0.7

0.716889061011709

0.716892619530846

-0.000003558519137

0.8

0.995406746985406

0.995418582763902

-0.000011835778496

0.9

1.368845680203319

1.368879844267690

-0.000034164064371

1.0

1.849967407347072

1.850055590769046

-0.000088183421974

## Problem 1

The given equation is

( 1 )

Capable to the initial conditions

( 2 )

## DTM Solution:

The DTM of kth order derived function of a map Y ( x ) is defined as.

( 3 )

And reverse differential transform is defined as.

( 4 )

As x0 =0 in order to acquire transform initial conditions put x=0 in Eq. ( 4 )

( 5 )

( 6 )

( 7 )

( 8 )

Using DTM to eq. ( 1 )

( 9 )

Substituting values to recurrence Eq. ( 16 ) continuously we can obtain the tabular array of estimates.

## Table 18

Approximate solutions of Eq. ( 9 )

## Y ( K )

0

0

1

4.1

2

0

3

-4.65

4

0

5

2.0175

6

-0.0041666666666667

7

-0.437480158730159

8

-0.0135168650793651

9

0.0150768849206349

10

-0.141347552910053

11

-0.637237100669392

12

-3.38322100385468

13

-21.1060125220555

14

-150.863006832612

15

-1217.45631518106

16

-10959.8671904436

17

-108971.707839356

18

-1186710.05121155

19

-14054200.5718693

20

-179903323.500211

Substituting the above values into Eq. ( 4 ) at x0=0 and utilizing values from above tabular array

( 10 )

## Table 19

Comparison between exact and DTM solution

## N=20

0

0

0

0

0

0.1

0.405370298310152

0.405370126951091

0.405370126936499

0.405370126929922

0.2

0.783445404444206

0.783439692229503

0.783439583420165

0.783436434405318

0.3

1.10929914563354

1.10925238586293

1.10922140915576

1.10066744189367

0.4

1.36253266085616

1.36230563837833

1.36036830607035

1.1089870959927

0.5

1.52903005575763

1.52815256916026

1.47807725359092

-202.422924091622

0.6

1.60215546248805

1.59910998966171

0.871215446346099

-7575.14512156101

0.7

1.58328542407052

1.57344911639245

– 5.4936098062074

-161797.0475774

0.8

1.48162986966944

1.45203464800982

– 49.465973292177

– 2300876.13130477

0.9

1.31335846243213

1.23074151643712

-290.46569295093

-23969171.892451

1.0

1.10011111223214

0.886065641534391

-1392.55972699872

-195265558.258049

## N=20

0

0

0

0.1

1.71359061218457e-007

1.71373653490292e-007

1.7138023045149e-007

0.2

5.71221470346828e-006

5.82102404156348e-006

8.97003888800096e-006

0.3

4.67597706088618e-005

7.77364777817535e-005

0.00863170373986888

0.4

0.000227022477830241

0.00216435478581234

2.4715197568489

0.5

0.000877486597372856

0.0509528021667105

203.951954147379

0.6

0.00304547282634116

0.730940016141955

7576.7472770235

0.7

0.00983630767806898

7.07689523027787

161798.630862824

0.8

0.0295952216596163

50.9476031618466

2300877.61293464

0.9

0.0826169459950155

291.779051413367

23969173.2058096

1.0

0.214045470697753

1393.65983811095

195265559.35816

## Problem 2

Theprojected initial value job is

5 ( 1 )

Capable to the initial conditions:

( 2 )

The Analytic solution:

( 3 )

## DTM solution

Using DTM to Eq. ( 1 )

( 4 )

Where degree Celsius ( K ) is the transmutation of cos ( ten )

( 5 )

Substituting k=0 in to recurrence Eq. ( 9 )

## aY?

aY? ( 6 )

Substitute k= I Eq. ( 9 )

## aY?

aY? ( 7 )

Substitute k=2 Eq. ( 9 )

## aY?

aY? ( 8 )

Continuing the same procedure we can obtain tabular array of estimate utilizing MATLAB:

## Table 21

Approximate solution of Eq. ( 4 )

## Y ( K )

0

2.000000000000000

1

0

2

0

3

0

4

-0.291666666666667

5

0

6

-0.051388888888889

7

0

8

-0.003298611111111

9

0

10

-0.000340884038801

11

0

12

0.000072373453450

13

0

14

-0.000039580002688

15

0

16

0.000022062561899

17

0

18

-0.000013253902583

19

0

20

0.000008424778799

21

0

22

-0.000005604579088

23

0

24

0.000003870301317

Substituting values from above tabular array in to Eq. ( 5 ) The DTM solution of Eq. ( 1 ) 1s obtain as:

As 0a‰¤xa‰¤1 ( 9 )

Which is the coveted solution of job 2 taking the points in the interval comparing to demand solution the tabular array is given below

## Table 22

Solving for n=25:

## Relative Mistake

0

2.000000000000000

2.000000000000000

0

0.1

2.000008333333829

1.999970725865633

0.000037607468196

0.2

2.000133333460318

1.999529811782220

0.000603521678097

0.3

2.000675003254466

1.997599314689780

0.003075688564686

0.4

2.002133365841340

1.992319751306367

0.009813614534973

0.5

2.005208527096754

1.980953278253333

0.024255248843421

0.6

2.010800833151946

1.959743048885204

0.051057784266742

0.7

2.020011192915431

1.923723240706365

0.096287952209067

0.8

2.034141655652010

1.866472170629576

0.167669485022435

0.9

2.054696353719439

1.779798812594181

0.274897541125258

1.0

2.083382940683384

1.653353241905638

0.430029698777746

## Problem 1

0a‰¤xa‰¤1 ( 1 ) with the initial conditions:

( 2 )

Analytic Solution: ( 3 )

## Solution:

DTM of kth order derived function of a map Y ( x ) is defined as.

( 4 )

And reverse differential transform is defined as.

( 5 )

( 6 )

( 7 )

( 8 )

( 9 )

( 10 ) ( 11 )

( 12 )

As x0 =0 in order to acquire transform initial conditions put x=0 in Eq. ( 5 ) – ( 12 ) We get

( 13 )

( 14 )

( 14 )

( 15 )

( 16 )

( 17 )

( 18 )

( 19 )

Using DTM to Eq. ( 1 )

( 20 )

Substituting k=0 in return Eq. ( 20 )

( 21 )

Substitute k=1in Eq. ( 20 )

( 22 )

Substitute k=2 in Eq. ( 20 )

( 23 )

Calculating values with MATLAB we get the followers tabular array

## Y ( K )

0

1.000000000000000

1

1.000000000000000

2

0.500000000000000

3

0.166666666666667

4

0.041666666666667

5

0.008333333333333

6

0.001388888888889

7

0.000198412698413

8

0.000024801587302

9

0.000002755731922

10

0.000000275573192

11

0.000000025052108

12

0.000000002087676

13

0.000000000160590

15

0.000000000011471

16

0.000000000000765

17

0.000000000000048

18

0.000000000000003

19

0.000000000000000

20

0.000000000000000

Substituting these values in to Eq. ( 5 ) The DTM Solution of Eq. ( 1 ) is as below

Solving for n=10

## Mistake

0

1.000000000000000

1.000000000000000

0

0.1

1.105170918075648

1.105170918075724

-0.000000000000076

0.2

1.221402758160170

1.221402758160323

-0.000000000000153

0.3

1.349858807576003

1.349858807576233

-0.000000000000229

0.4

1.491824697641270

1.491824697641576

-0.000000000000306

0.5

1.648721270700128

1.648721270700511

-0.000000000000382

0.6

1.822118800390509

1.822118800390968

-0.000000000000459

0.7

2.013752707470477

2.013752707471009

-0.000000000000532

0.8

2.225540928492468

2.225540928493051

-0.000000000000584

0.9

2.459603111156950

2.459603111157471

-0.000000000000521

1.0

2.718281828459046

2.718281828458995

0.000000000000051

Table 25 Solving for n=15

## Mistake

0

1.000000000000000

1.000000000000000

0

0.1

1.105170918075648

1.105170918075647

0.000000000000000

0.2

1.221402758160170

1.221402758160170

0.000000000000000

0.3

1.349858807576003

1.349858807576003

0.000000000000000

0.4

1.491824697641270

1.491824697641270

0.000000000000000

0.5

1.648721270700128

1.648721270700128

0.000000000000000

0.6

1.822118800390509

1.822118800390509

-0.000000000000000

0.7

2.013752707470477

2.013752707470477

0

0.8

2.225540928492468

2.225540928492467

0.000000000000001

0.9

2.459603111156950

2.459603111156940

0.000000000000010

1.0

2.718281828459046

2.718281828458995

0.000000000000051

## Table 25

Mistake estimates =Analytical solution -DTM solution

## N=20

0

0

0

0

0.1

-0.076161299489286A-1.0e-12

0.000444089209850A-1.0e-12

0.000444089209850A-1.0e-12

0.2

-0.152766688188422A-1.0e-12

0.000222044604925A-1.0e-12

0.000222044604925A-1.0e-12

0.3

-0.229372076887557A-1.0e-12

0.000222044604925A-1.0e-12

0.000222044604925A-1.0e-12

0.4

-0.305755420981768A-1.0e-12

0.000222044604925A-1.0e-12

0.000222044604925A-1.0e-12

0.5

-0.382360809680904A-1.0e-12

0.000222044604925A-1.0e-12

0.000222044604925A-1.0e-12

0.6

-0.458966198380040A-1.0e-12

-0.000444089209850A-1.0e-12

-0.000444089209850A-1.0e-12

0.7

-0.532018873400375A-1.0e-12

0

0

0.8

-0.583533221742982A-1.0e-12

0.001332267629550A-1.0e-12

0

0.9

-0.520916643154123A-1.0e-12

0.009769962616701A-1.0e-12

0.000444089209850A-1.0e-12

1.0

0.050626169922907A-1.0e-12

0.050626169922907A-1.0e-12

-0.000444089209850A-1.0e-12

Here we can observe that every bit for as n additions mistake decreases, so the method is really fatly convergent so other methods such as 4th order Range Kutta method and IRBF ( Irregular footing maps webs ) Collocation method. Good truth and high rate of convergence are obtained For illustration, at the Centre spacing of 1/5, error-norms are 8.70ea?’6and 1.76e a?’ 10 for the RK4 method and IRBF collocation method.

## DTM solution of 8th order initial value job:

The equation is

( 1 )

0a‰¤xa‰¤1

With the initial conditions:

( 2 )

Analytic solution: ( 3 )

## DTM Solution

Using DTM to Eq. ( 1 ) same as in job ( 1 )

( 4 )

Substitute k=0 in return Eq. ( 4 )

( 5 )

Substitute k=1 in return Eq. ( 21 )

( 6 )

Substitute k=2 in return Eq. ( 21 )

( 7 )

Continuing the same procedure we can obtain the tabular array of estimates by utilizing MATLAB

## Table 26

FirstNineteenth loops of return equation 4

## Y ( K )

0

1.000000000000000

1

0

2

-0.500000000000000

3

-0.333333333333333

4

-0.125000000000000

5

-0.033333333333333

6

-0.006944444444444

7

-0.001190476190476

8

-0.000173611111111

9

-0.000022045855379

10

-0.000002480158730

11

-0.000000250521084

12

-0.000000022964433

13

-0.000000001927085

14

-0.000000000149120

15

-0.000000000010706

16

-0.000000000000717

17

-0.000000000000045

18

-0.000000000000003

19

-0.000000000000000

Substitute the above values to Eq. ( 4 )

( 10 )

## Mistake

0.1

0.994653826268083

0.994653826268083

0.000000000000000

0.2

0.977122206528136

0.977122206528136

0.000000000000000

0.3

0.944901165303202

0.944901165303202

-0.000000000000000

0.4

0.895094818584762

0.895094818584762

-0.000000000000000

0.5

0.824360635350064

0.824360635350064

-0.000000000000000

0.6

0.728847520156204

0.728847520156204

-0.000000000000000

0.7

0.604125812241143

0.604125812241143

-0.000000000000000

0.8

0.445108185698493

0.445108185698494

-0.000000000000000

0.9

0.245960311115695

0.245960311115696

-0.000000000000001

## Table28

Solving for n=20:

## Mistake

0.1

0.994653826268083

0.994653826268083

0.000000000000000

0.2

0.977122206528136

0.977122206528141

-0.000000000000005

0.3

0.944901165303202

0.944901165303659

-0.000000000000456

0.4

0.895094818584762

0.895094818595668

-0.000000000010906

0.5

0.824360635350064

0.824360635478240

-0.000000000128176

0.6

0.728847520156204

0.728847521117714

-0.000000000961511

0.7

0.604125812241143

0.604125817532357

-0.000000005291214

0.8

0.445108185698493

0.445108208909094

-0.000000023210600

0.9

0.245960311115695

0.245960396743999

-0.000000085628304

## N=20

0.1

0.000000000111022A-10-006

0.000000000111022A-10-006

0.000000000111022A-10-006

0.2

-0.000000004884981A-10-006

0.000000000333067A-10-006

0.000000000333067A-10-006

0.3

-0.000000456412685A-10-006

-0.000000000111022A-10-006

-0.000000000111022A-10-006

0.4

-0.000010906275882A-10-006

-0.000000000111022A-10-006

-0.000000000111022A-10-006

0.5

-0.000128176136371A-10-006

-0.000000000222045A-10-006

-0.000000000222045A-10-006

0.6

-0.000961510870923A-10-006

-0.000000000444089A-10-006

-0.000000000222045A-10-006

0.7

-0.005291213911462A-10-006

-0.000000002664535A-10-006

-0.000000000222045A-10-006

0.8

-0.023210600330614A-10-006

-0.000000021538327A-10-006

-0.000000000277556A-10-006

0.9

-0.085628303697627A-10-006

-0.000000141525680A-10-006

-0.000000000721645A-10-006

## Table 29

Error =Exact Solution -DTM Solution

## Table 30

Comparison between DTM and Modified decomposition method

0.25

0.9630191

0.9630191

0.9630190

0

1.0A-10-7

0.50

0.8243606

0.8243606

0.8343575

0

3.1A-10-6

0.75

0.5292500

0.5292500

0.5291946

0

5.5A-10-5

## Transformation Method

Most of the jobs in our day-to-day life are normally non-linear and are illustrated by non-linear differential equations. A figure of them are solved via numerical processs and legion are calculated by analytic techniques of disturbance. In analytical methods of disturbance one should utilize the little parametric quantity in differential equations. For that ground, turn uping the little parametric quantities and using them to the equation are inadequacy of the Perturbation methods. In recent times, much consciousness has been given over the freshly developed techniques to do approximative analytic solutions of non-linear differential equations without declaring the lacks of jobs. Therefore, this method can be applied to both non-linear differential equations every bit good as additive differential equations without linearization, discretization or disturbance.

## 4.3.1 Application of DTM to Fourth Order Initial -Value Problem

Differential Transformation method for direct solution of 4th order differential equations reduces the computational load and computing machine clip wastage involved in the method of cut downing such equations to a system of first order equations the job calculated below shows the dependability of proposed attack

## Problem

Fourth order non-linear initial value jobs:

( 1 ) ( 2 )

Analytic Solution ( 3 )

Using transmutation regulations to Eq. ( 2 ) we can obtain transform initial conditions.

( A )

Asputting x=0 in combining weight ( A ) and utilizing initial conditions as follows.

( 4 )

( 5 )

1.500000000000000 ( 6 )

( 7 )

The differential transmutation of Eq. ( 1 ) is

( 8 )

At k=0 in return Eq. ( 8 )

( 9 )

At k=1

( 10 )

Substituting values once more and once more to recurrence Eq. ( 8 ) by utilizing MATLAB:

## Table 31

Approximate solutions of Eq. ( 8 )

## Y ( K )

0

1.000000000000000

1

1.000000000000000

2

1.500000000000000

3

0.166666666666667

4

0.041666666666667

5

-0.008333333333333

6

-0.015277777777778

7

0.000198412698413

8

-0.000024801587302

9

-0.000024801587302

10

0.000003582451499

11

-0.000000025052108

12

-0.000000014613730

13

0.000000003693580

14

-0.000000000149120

15

-0.000000000003824

16

0.000000000001482

17

-0.000000000000098

18

0.000000000000001

19

0.000000000000000

20

-0.000000000000000

Substituting the above values to Eq. ( A ) the needed approximative solution of Eq. ( 1 ) is as below in the signifier of multinomials:

## Table 32

Comparison between exact and DTM Solution at N=20

## Mistake

0

1.000000000000000

1.000000000000000

0

0.1

1.115170918075648

1.115170734741791

0.000000183333857

0.2

1.261402758160170

1.261396358019414

0.000006400140756

0.3

1.439858807576003

1.439806153798558

0.000052653777445

0.4

1.651824697641271

1.651585724920408

0.000238972720863

0.5

1.898721270700128

1.897939776316091

0.000781494384038

0.6

2.182118800390509

2.180044109315150

0.002074691075359

0.7

2.503752707470476

2.498986844797966

0.004765862672510

0.8

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