# The Ordinary Differential Equations Biology Essay

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Ordinary differential equations have very huge range of applications in a variety of fields, as they express almost systems undergoing change such as exchanges of matter, energy, information or any other quantities, often as they vary in time and/or space their thorough out analytical treatment forms the basis of fundamental theories in Mathematics, Physics and other areas. Mathematicians have studied the nature of differential equations and make many well-developed solution techniques. Frequently, systems expressed by differential equations are so complex, or the systems that they illustrate are so large, that a purely analytical solution to the equations is not tractable. The numerical solution of these types of systems has been the subject of intense research which leads methods to solution of problems. But numerical methods don't offer the exact answer to a given problem and can only tend to a solution, getting closer to it with each iteration. Numerical techniques provide discontinuous points of a curve and thus it is usually time consuming to get a complete curve of results and so in these methods particularly for non-linear equations, stability and convergence should be considered so as to keep away from divergence or unsuitable results.

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Besides these two techniques the series expansion method is also a procedure to solve the differential equations which is in-between to both. Taylor series method is one for the solution of functions in the form of polynomials which provides highly accurate results of differential equations. But this is time-consuming especially for high order equation, and become extremely difficult to find the general nth term derivative for some functions, series doesn't converge quickly a new approach which formulize Taylor series in slight different manner and reduces symbolic computation of the necessary derivatives is Differential transformation method(DTM).

The thought of differential transformation method was first proposed in 1986 by a Chinese "Zhou jiakui" and its main applications are to solve both linear and non-linear initial value problems in electric circuit analysis. This method constructs a semi-analytical numerical technique that uses Taylor series for the solution of differential equations in the form of a polynomial approximation. This is an iterative procedure provides results nearer to exact solution after each iteration. DTM is widespread and is able to solve various kinds of functions both linear ,non-linear, higher order ODEs involving a large number of variables where direct methods would be prohibitively expensive (and in some cases impossible) even with the best available computing power.

In this thesis the DTM is applied to solve higher order differential equations such as greater then second to eighth order differential equations linear and non- linear and method is compared with analytical and numerical methods by taking the suitable points in the given interval of domain. This partially numerical method eliminates cumbersome calculations needed to obtain in both pure numerical and analytical approach. When comparisons are made in studies of problems a very good agreement is observed. This single method can be applied to solve different ODEs subject to specified initial conditions both for linear and non-linear differential equations.

## 2.1 High Order Initial and Boundary Value Problems

Differential transformation method (DTM) is very valuable implement for analytical and numerical solution of problems of differential equations both linear and non-linear, Partial and ordinary differential equations subject to the specified initial and boundary conditions .This method is applied to initial value problems of higher order ordinary differential equations. Ordinary differential equations have very huge scope of applications in various fields of Life.Sakai and Usmani (1983) has obtainedSpline Solutions for non-linear fourth-Order two-point boundary value problems. Wu and Liu (2000) has solved initial value problems by generalized differential quadrature rule in which structural dynamics problems which include four different kinds of vibration equations, third and fourth-order ODEs.Mai-Duy (2004) solved eighth-order initial value problem of ODEs using indirect radial basis functions collocation method. Kayode (2007) applied an efficient zero-stable method for fourth order differential equations; the derived method is capable of handling problems where it is either linear or non-linear.Mestrovic (2007) has solved eighth-order initial value and boundary value problems of ODEs by modified decomposition method. Mohyod-Din and Noor (2007) has applied variational iteration decomposition method to solve eighth-order ODEs.Mohyod-Din (2008) has solved twelfth order boundary value problems by using variation iteration method.Taiwo and Ogunlaran (2008) has applied cubic spline collocation tau method to solve out fourth-order ODEs.Mohyod-Din and Noor (2008) proposed variation iteration method and variation iteration decomposition method for fifth-order higher dimensional initial and boundary value problems. Jwamer and problem by using lacunary-spline method and concluded that required method approximates the solution of problem and accuracy can be improved by reducing step size. Karmer (2010) has obtained the solution of fourth-order initial value) has obtained spline solution of linear fourth order boundary value problems. Olabode (2009) presented the six-step scheme for solution of fourth-order ordinary differential equations in which approach of collocation is adoptive in the derivation scheme and then applied to special fourth-order initial value problems of ODEs. The solution of fourth-order initial value problems of linear and non-linear ODEs are also obtained by six zero-stable methods which reduced computational burden involved in the method of reducing such equations in to system of first-order equations, and concluded that this method satisfied the basic requirements for convergence of linear multi-step methods. A multi-derivative collocation method is applied to non-linear fifth-order ODEs with initial conditions by Kayode and Awoyemi (2010).Farajeyan and Jalilian (2010) has solved out fifth-order boundary value problems in off-step points and compared their methods with non-polynomial spline solved by Sidiqiet-alMohyid-Din et-al (2009) presented variational parameters method to solve out sixth-order boundary value problems of ODEs to both linear and non-linear and compared their method with variation iteration method,Homotopy perturbation method, and Adomian,s decomposition method and commented that proposed method provides more realistic solution then these methods.Abdel-Halim (2009) has found out the solution of twelfth-order boundary value problems by applying variation iteration method and concluded that adopted method is highly accurate to analytical method.Mirmoradiet-al (2009) and Ul-Islam (2009) also found out the solution of twelfth-order boundary value problems by homotopy perturbation method and differential transformation method. Siddiqiet al (2009) has solved eighth-order boundary value problems using variational.Viswanadhamand Krishna (2010) also obtained the solution of sixth-order boundary value problems by using Septic B-spline collocation method for sixth-order boundary value problems and demonstrated the applicability of proposed method iteration method. Mohyod-Din and Yildirim (2010) applied homotopy perturbation method to solve ninth and tenth order boundary value problems the proposed iterative scheme provides the solutions without any restrictive assumptions. The solution of twelfth-order boundary value problems has also been seek out by Ali et-al (2010) by using optimal homotopy asymptotic method and compares their method with differential transformation method. Nawaz et-al(2010) has also used the optimal homotopy asymptotic method to solve twelfth-order boundary value problems.

## 2.2 Differential Transformation Method

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Al-Salawalha and Noorani(2007) developed the differential transformation method (DTM) to solve Lorenz system in which DTM solution is compared with fourth-order Runge-Kutta method which is sufficiently accurate for comparison purposes. Erturk (2007) applied these techniques to sixth-order boundary value problem and equations of Lane Emden type and concluded that DTM is very fast convergent, precise and cost efficient tool for solving Lane-Emden equations.Patrico and Rosa (2007) solved out Adection-diffusion problems by using DTM in their paper the approach for solving PDEs is presented and eigen values which depends on parameters that appear in PDEs are studied. Nejatbakhsh (2007) proposed the DTM to solve fuzzy integral equations. El-Shahed (2008) used DTM to solve non-linear oscillatory systems and demonstrated DTM does not require small parameters in the equations so limitation of traditional perturbation methods can be eliminated.Attarnejad and Shahba(2008) applied DTM in free vibration analysis of rotating non-prismatic beams. Kai-Tai and Bin (2009) used DTM for non-linear differential equations in which this method is demonstrated by handling (1+3) dimensional Burgers equations and two dimensional heat and diffusion equations their work shows reliability and efficiency of proposed method.Demir and Sungu (2009) used DTM for numerical solution for class of non-linear Emeden-flower equations, as a class of non singular initial value problem they introduced and implemented their method and illustrated the efficiency and simplicity of method. Abazari,N and Abazari, R (2009) has obtained the solution of non-linear second-order pantograph equations by differential transformation method. Attarnejad (2009) applied DTM in free vibration analysis of Timoshenko beams resting on two-parametric elastic foundation. In order to find out solution of Cauchy reaction-diffusion problems DTM and variation iteration method was applied by Othman and Mahdy (2010) and the concluded that both methods provide same results. Rashidi (2010) presented the DTM approach to the problem of micro-polar flow in porous channel with mass injection and excellent agreement was noted between DTM and fourth-order Range Kutta method. Biazar and Mohammadi(2010) applied DTM to generalized burgers Huxley equation and concluded that DTM is more effective and convenient as compared to domains decomposition method and variation iteration method.

## 3.1.1 LINEAR DIFFERENTIAL EQUATIONS

A differential equation is linear if there does not exist multiplications along with dependent variables and their derivatives or all coefficients are functions of dependent variables.

## 3.1.2 NON-LINEAR DIFFERENTIAL EQUATIONS

Differential equations that do not satisfy the definition of linear differential equations are non-linear.

## 3.1.3 TAYLOR SERIES

Taylor series method is one of the most primitive algorithms for approximate solution ordinary differential equations .The initiative of the rehabilitation of these algorithms is based on the approximate calculation of higher derivatives. The Taylor series of a real or complex function f(x) that is infinitely differentiable in a neighborhood of a real or complex number a is the power series

## 3.2 DIFFERENTIAL TRANSFORMATION METHOD (DTM)

The differential transformation method is utilized to solve higher order initial value problems. The application of present method gives rapid convergence and incredible accuracy. This approach can easily be applied linear and non-linear problems and reduces the required computational effort with this method exact solution may be obtained without any need of burdensome computations. To clarify and exemplify capability of proposed technique various problems have been solved. When comparison is made with the studies a very good agreement is observed. Results are compared with analytical solution by taking suitable points to given interval of domain.

The idea of the differential transform was first recommended in 1986 by a Chinese "Zhou jiakui" and its major applications therein are to solve both linear and non-linear initial value problems of electric circuit analysis.DTM is a transformed technique based on a Taylor series formula and is a helpful implement to obtain analytic solution of differential equations. In this method certain transformation rules are applied to governing differential equations and boundary conditions of the system are transformed into recurrence equations that finally lead to a solution of system of algebraic equations, the solution of thesealgebraic equations is desired solution of the problem.

## 3.2.1 DEFINITION

Consider a function y(x) which is analytic in a domain D and suppose x=x0is a point in D. The function y(x) is then given by a power series whose center is located at x0. The differential transform of the nth derivative of a function y(x)in one variable is defined as below:

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(3.1)

In Eq. (3.1) y(x) is the original function and Y(k) is the transformed function and differential inverse transform of Y(k) is defined as follows

(3.2)

In real applications, function y (x) is expressed by a finite series and Eq. (3.2) can be written as.

(3.3)

Combining Eq.s (3.1) and (3.2) the new expression for y(x) reduces to

(3.4)

Eq. (3.4) implies the idea of differential transform is resultant from Taylor series method. Though differential transformation method is not capable to calculate the derivatives, relative derivatives can be computed by an iterative manner which can be described by the transformed equations of original functions. Eq. (3.4) shows the following part

(3.5)

Is negligibly small. Actually, n is decided by the convergence of natural frequency. The following 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 proof

## PROOF

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

This completes proof

## PROOF

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

This completes proof

## 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 proof

In general

Where

## PROOF

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

This completes proof

Some important functions and there differential transformations are given in table 3.1

## Differential Transformation

Cos(x)

Sin(x)

exp(x)

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 form of a polynomial. It is different from the high-order Taylor series method which requires symbolic computation of the necessary derivatives of the data functions. The Taylor series method is computationally time-consuming especially for high order equations. The differential transform is an iterative procedure for obtaining analytic Taylor series solutions of differential equations. It can be said that differential transform method is a universal one, and is able to solve various kinds of differential equations.

## Structural dynamics problems:

Structural dynamics is the subset of structural analysis which covers the behavior of structures as subjected to the dynamics. A simple second-order dynamics equation is a vibrating equation of single degree of freedom.

(4.1)

Where =, m is the mass c is the damping value ,k is elastic coefficient p and qare existing forces, amplitude and frequency respectively ,and are acceleration, velocity and displacement respectively,The structural dynamics equations has two initial conditions related velocity band displacement all the initial conditions are as follows.

(4.2) The differential transformation method is applied to four different kinds of vibrating equations. All has the following data with units omitted

(4.3)

As discussed above, the differential transform and inverse 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 respectively. The transformed initial conditions can be obtain by setting t=0, and t0 =0 in Eq. (4.5) which are as below.

(4.6)

(4.7)

(1)

## Analytical Solution:

0â‰¤tâ‰¤1 (2)

Applying 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 way putting values into recurrence equation (3) continually we can obtain table of approximations using MATLAB

Table 1Transformed iterations of recurrence Eq. (3))

Substituting values from above table into equation (4.5) at t0=0 the required approximate solution of Eq. (1) is as follows in the form of polynomials.

## Table 2

Comparison between Analytical and DTM solution as 0â‰¤tâ‰¤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

Error approximations =Analytical Solution -DTM Solution

## N=20

0

0

0

0

0.1

-0.000000002220446Ã-1.0e-7

-0.000000002220446Ã-1.0e-7

-0.000000002220446Ã-1.0e-7

0.2

-0.000000011102230Ã-1.0e-7

-0.000000011102230Ã-1.0e-7

0

0.3

-0.000000879296636Ã-1.0e-7

-0.000000879296636Ã-1.0e-7

-0.000000002220446Ã-1.0e-7

0.4

-0.000020641266474Ã-1.0e-7

-0.000020641266474Ã-1.0e-7

0

0.5

-0.000239170905303Ã-1.0e-7

-0.000239170905303Ã-1.0e-7

-0.000000004440892Ã-1.0e-7

0.6

-0.001768225565968Ã-1.0e-7

-0.001768225565968Ã-1.0e-7

-0.000000002220446Ã-1.0e-7

0.7

-0.009588001503857Ã-1.0e-7

-0.009588001503857Ã-1.0e-7

0.000000008881784Ã-1.0e-7

0.8

-0.041433563247040Ã-1.0e-7

-0.041433563247040Ã-1.0e-7

0.000000004440892Ã-1.0e-7

0.9

-0.150549026578517Ã-1.0e-7

-0.150549026578517Ã-1.0e-7

0.000000093258734Ã-1.0e-7

1.0

-0.477083093031183Ã-1.0e-7

-0.477083093031183Ã-1.0e-7

0.000000510702591Ã-1.0e-7

## 4.1.2 Second kind of vibrating equation:

The second kind of vibrating equation and its analytical solution are given as below

(1)

## Analytical Solution:

0â‰¤tâ‰¤1 (2)

Applying DTM to Eq. (1)

(3)

At k=0 in eq. (3)

(4)

Similarly

(6)

] (7)

In the same manner putting values into recurrence equation (3) continually we can obtain table of approximations using MATLAB

## Table 4

Transformed iterations of recurrence 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 table into Eq. (4.5) at t0=0 the required approximate solution of Eq. (1) is as follows in the form of polynomials.

## 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

Error approximations =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 kind of vibrating equation:

The third kind of homogeneous vibrating equation and its analytical solution are as below

(1)

## Analytical Solution:

(2)

Applying DTM to Eq. (1)

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

(4) Similarly

(5)

In the same manner putting values into recurrence Eq. (3) continually we can obtain table of approximations using 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 table into Eq. (4.5) at t0=0 the required approximate solution of Eq. (1) is as follows in the form of polynomials

(7)

## Table 8

Comparison between analytical and differential transformation 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

Error approximations =analytical Solution -differential transformation method

## N=20

0

0

0

0

0.1

0.036497624074094Ã-1.0e-4

0.036497624074094Ã-1.0e-4

0.036497624074094Ã-1.0e-4

0.2

0.083926532012946Ã-1.0e-4

0.083926532019607Ã-1.0e-4

0.083926532019607Ã-1.0e-4

0.3

0.143463299375757Ã-1.0e-4

0.143463299897562Ã-1.0e-4

0.143463299897562Ã-1.0e-4

0.4

0.216001579451675Ã-1.0e-4

0.216001591635262Ã-1.0e-4

0.216001591635262Ã-1.0e-4

0.5

0.302131484788415Ã-1.0e-4

0.302131624547730Ã-1.0e-4

0.302131624547730Ã-1.0e-4

0.6

0.402123620386607Ã-1.0e-4

0.402124643192892Ã-1.0e-4

0.402124643192892Ã-1.0e-4

0.7

0.515917125221144Ã-1.0e-4

0.515922614185982Ã-1.0e-4

0.515922614181541Ã-1.0e-4

0.8

0.643109809339570Ã-1.0e-4

0.643133280986419Ã-1.0e-4

0.643133280955333Ã-1.0e-4

0.9

0.782946266015827Ã-1.0e-4

0.783030642015348Ã-1.0e-4

0.783030641819948Ã-1.0e-4

1.0

0.934296354886399Ã-1.0e-4

0.934560840568643Ã-1.0e-4

0.934560839507270Ã-1.0e-4

## 4.1.4 Fourth-Kind of Vibrating Equation

The fourth kind of vibrating equation and its analytical solution are as below

0â‰¤tâ‰¤1 (1)

Analytical Solution:) (2) where C=

D=

B=C

A=

Applying DTM to q. (1)

(3)

By substituting k=0,1,2,â€¦We can obtain table of approximations

Table 10 Transformed iterations of recurrence 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 table into Eq. (4.5) at t0=0 the required approximate solution of eq. (1) is as follows in the form of polynomials

## Table 11

Comparison between analytical and differential transformation 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 approximation =Analytical method -Differential transformation method

## N=20

0

0

0

0

0.1

-0.000000000370814Ã-1.0e-4

0

0

0.2

-0.000000189104288Ã-1.0e-4

0

0

0.3

-0.000007258742496Ã-1.0e-4

0

0

0.4

-0.000096488570467Ã-1.0e-4

0.000000000002220Ã-1.0e-4

0

0.5

-0.000717233703540Ã-1.0e-4

0.000000000097700Ã-1.0e-4

0

0.6

-0.003690762182540Ã-1.0e-4

0.000000001318945Ã-1.0e-4

0

0.7

-0.014733007858148Ã-1.0e-4

0.000000012034818Ã-1.0e-4

0

0.8

-0.048831943986194Ã-1.0e-4

0.000000082103213Ã-1.0e-4

0

0.9

-0.140408500812761Ã-1.0e-4

0.000000448014958Ã-1.0e-4

0.000000000013323Ã-1.0e-4

1.0

-0.360877653049485Ã-1.0e-4

0.000002049369563Ã-1.0e-4

0.000000000111022Ã-1.0e-4

## The given equation is:

(1)

Subject to specified initial conditions

Analytical Solution (2)

## DTM Solution

The DTM of kth order derivative of a function y(x) is defined as.

(3)

And inverse differential transform is defined as.

(4)

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

(7)- (9) We get

(5)

(6)

(7)

Applying DTM to Eq. (1)

(8) Putting values to Eq. (13) recurrently we can obtain the table of approximations.

## 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 table to inverse differential transform we can obtain the solution of differential equation in form of polynomials

## 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 approximate solution of Eq. (1)

## Table 15

Error approximation =Analytic method - Differential transformation method

## N=20

0

0

0

0

0.1

0.000000000008882 Ã-1.0e-4

0.000000000017764 Ã-1.0e-4

0.000000000017764 Ã-1.0e-4

0.2

-0.000000010871304 Ã-1.0e-4

0

0

0.3

-0.000000956932311 Ã-1.0e-4

-0.000000000004441 Ã-1.0e-4

-0.000000000004441 Ã-1.0e-4

0.4

-0.000023057316056 Ã-1.0e-4

-0.000000000004441 Ã-1.0e-4

0.000000000008882 Ã-1.0e-4

0.5

-0.000273233817794 Ã-1.0e-4

-0.000000000506262 Ã-1.0e-4

0

0.6

-0.002067088424873 Ã-1.0e-4

-0.000000009501289 Ã-1.0e-4

0.000000000002220 Ã-1.0e-4

0.7

-0.011474211925044 Ã-1.0e-4

-0.000000113382637 Ã-1.0e-4

0

0.8

-0.050781388632182 Ã-1.0e-4

-0.000000972643632 Ã-1.0e-4

0.000000000004996 Ã-1.0e-4

0.9

-0.189050050639628 Ã-1.0e-4

-0.000006486323700 Ã-1.0e-4

0

1.0

-0.614088518242628 Ã-1.0e-4

-0.000035463700865 Ã-1.0e-4

0.000000000008882 Ã-1.0e-4

## Problem 2

(1)Subject to specified initial conditions.

Analytical Solution (2)

Applying DTM to Eq. (1) same as in problem (1)

(3)

(4)

Similarly

(5)

Applying DTM to Eq. (1)

(6) Substituting values to equation (13) recurrently we can obtain the table of approximations

## Table 16

Solution of recurrence 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:

## Error

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)

Subject to the initial conditions

(2)

## DTM Solution:

The DTM of kth order derivative of a function y(x) is defined as.

(3)

And inverse differential transform is defined as.

(4)

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

(5)

(6)

(7)

(8)

Applying DTM to eq. (1)

(9)

Substituting values to recurrence Eq. (16) continuously we can obtain the table of approximations.

## 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 using values from above table

(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 problem is

5(1)

Subject to the initial conditions:

(2)

The Analytical solution:

(3)

## DTM solution

Applying DTM to Eq. (1)

(4)

Where c (k) is the transformation of cos(x)

(5)

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

## âŸ¹

âŸ¹ (6)

Substitute k= I Eq. (9)

## âŸ¹

âŸ¹ (7)

Substitute k=2 Eq. (9)

## âŸ¹

âŸ¹ (8)

Continuing the same process we can obtain table of approximation using 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 table in to Eq. (5) The DTM solution of Eq.(1) 1s obtain as:

As 0â‰¤xâ‰¤1 (9)

Which is the desired solution of problem 2 taking the points in the interval comparison to exact solution the table is given below

## Table 22

Solving for n=25:

## Relative Error

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

0â‰¤xâ‰¤1 (1) with the initial conditions:

(2)

Analytical Solution:(3)

## Solution:

DTM of kth order derivative of a function y(x) is defined as.

(4)

And inverse differential transform is defined as.

(5)

(6)

(7)

(8)

(9)

(10) (11)

(12)

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

(13)

(14)

(14)

(15)

(16)

(17)

(18)

(19)

Applying DTM to Eq. (1)

(20)

Substituting k=0 in recurrence Eq. (20)

(21)

Substitute k=1in Eq. (20)

(22)

Substitute k=2 in Eq. (20)

(23)

Calculating values with MATLAB we get the following table

## 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

## Error

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

## Error

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

Error approximations =Analytical solution -DTM solution

## N=20

0

0

0

0

0.1

-0.076161299489286Ã-1.0e-12

0.000444089209850Ã-1.0e-12

0.000444089209850Ã-1.0e-12

0.2

-0.152766688188422Ã-1.0e-12

0.000222044604925Ã-1.0e-12

0.000222044604925Ã-1.0e-12

0.3

-0.229372076887557Ã-1.0e-12

0.000222044604925Ã-1.0e-12

0.000222044604925Ã-1.0e-12

0.4

-0.305755420981768Ã-1.0e-12

0.000222044604925Ã-1.0e-12

0.000222044604925Ã-1.0e-12

0.5

-0.382360809680904Ã-1.0e-12

0.000222044604925Ã-1.0e-12

0.000222044604925Ã-1.0e-12

0.6

-0.458966198380040Ã-1.0e-12

-0.000444089209850Ã-1.0e-12

-0.000444089209850Ã-1.0e-12

0.7

-0.532018873400375Ã-1.0e-12

0

0

0.8

-0.583533221742982Ã-1.0e-12

0.001332267629550Ã-1.0e-12

0

0.9

-0.520916643154123Ã-1.0e-12

0.009769962616701Ã-1.0e-12

0.000444089209850Ã-1.0e-12

1.0

0.050626169922907Ã-1.0e-12

0.050626169922907Ã-1.0e-12

-0.000444089209850Ã-1.0e-12

Here we can note that as for as n increases error decreases, so the method is very fatly convergent then other methods such as fourth order Range Kutta method and IRBF (Irregular basis functions networks) Collocation method. Good accuracy and high rate of convergence are obtained For example, at the centre spacing of 1/5, error-norms are 8.70eâˆ’6and 1.76e âˆ’ 10 for the RK4 method and IRBF collocation method.

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

The equation is

(1)

0â‰¤xâ‰¤1

With the initial conditions:

(2)

Analytical solution: (3)

## DTM Solution

Applying DTM to Eq.(1) same as in problem (1)

(4)

Substitute k=0 in recurrence Eq. (4)

(5)

Substitute k=1 in recurrence Eq. (21)

(6)

Substitute k=2 in recurrence Eq. (21)

(7)

Continuing the same process we can obtain the table of approximations by using MATLAB

## Table 26

FirstNineteenth iterations of recurrence 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)

## Error

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:

## Error

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

0.000000000111022Ã-10-006

0.000000000111022Ã-10-006

0.000000000111022Ã-10-006

0.2

-0.000000004884981Ã-10-006

0.000000000333067Ã-10-006

0.000000000333067Ã-10-006

0.3

-0.000000456412685Ã-10-006

-0.000000000111022Ã-10-006

-0.000000000111022Ã-10-006

0.4

-0.000010906275882Ã-10-006

-0.000000000111022Ã-10-006

-0.000000000111022Ã-10-006

0.5

-0.000128176136371Ã-10-006

-0.000000000222045Ã-10-006

-0.000000000222045Ã-10-006

0.6

-0.000961510870923Ã-10-006

-0.000000000444089Ã-10-006

-0.000000000222045Ã-10-006

0.7

-0.005291213911462Ã-10-006

-0.000000002664535Ã-10-006

-0.000000000222045Ã-10-006

0.8

-0.023210600330614Ã-10-006

-0.000000021538327Ã-10-006

-0.000000000277556Ã-10-006

0.9

-0.085628303697627Ã-10-006

-0.000000141525680Ã-10-006

-0.000000000721645Ã-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

1.0Ã-10-7

0.50

0.8243606

0.8243606

0.8343575

0

3.1Ã-10-6

0.75

0.5292500

0.5292500

0.5291946

0

5.5Ã-10-5

## Transformation Method

Most of the problems in our daily life are usually non-linear and are illustrated by non-linear differential equations. A number of them are solved via numerical procedures and numerous are calculated by analytic techniques of perturbation. In analytical methods of perturbation one should use the small parameter in differential equations. For that reason, locating the small parameters and applying them to the equation are insufficiency of the Perturbation methods. In recent times, much awareness has been given over the newly developed techniques to make approximate analytic solutions of non-linear differential equations without declaring the deficiencies of problems. Therefore, this method can be applied to both non-linear differential equations as well as linear differential equations without linearization, discretization or perturbation.

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

Differential Transformation method for direct solution of fourth order differential equations reduces the computational burden and computer time wastage involved in the method of reducing such equations to a system of first order equations the problem calculated below shows the reliability of proposed approach

## Problem

Fourth order non-linear initial value problems:

(1) (2)

Analytical Solution (3)

Applying transformation rules to Eq. (2) we can obtain transform initial conditions.

(A)

Asputting x=0 in eq (A) and using initial conditions as follows.

(4)

(5)

1.500000000000000 (6)

(7)

The differential transformation of Eq. (1) is

(8)

At k=0 in recurrence Eq. (8)

(9)

At k=1

(10)

Substituting values again and again to recurrence Eq. (8) by using 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 required approximate solution of Eq. (1) is as below in the form of polynomials:

## Table 32

Comparison between exact and DTM Solution at N=20

## Error

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