225 
 

 American Scientific Research Journal for Engineering, Technology,  and Sciences  (ASRJETS) 
ISSN (Print) 2313-4410, ISSN (Online) 2313-4402 

© Global Society of Scientific Research and Researchers  
http://asrjetsjournal.org/  

 

Solution of Second Order Linear and Nonlinear Two Point 

Boundary Value Problems Using Legendre Operational 

Matrix of Differentiation 

Abdurkadir Edeo* 

Bule Hora University, College of Natural and Computational Sciences, Department of Mathematics, Bule Hora 

144, Ethiopia 

Email: edeoabdurkadir@gmail.com 

 

 

Abstract 

In this paper, an approach using Tau method based on Legendre operational matrix of differentiation [2&5] has 

been addressed to find the solutions of second order linear and nonlinear two point boundary value problems of 

ordinary differential equations. In the implementation of this approach, the given second order two point 

boundary problems is converted into a system of algebraic equations, whose solutions are the Legendre 

coefficients. The validity and efficiency of the method has also been illustrated with numerical examples 

supported by graphs.  

Keywords: Boundary Value Problems (BVPs); Legendre Operational Matrix of Differentiation; Linear and 

nonlinear ordinary differential equations. 

1. Introduction  

According to [1], second order two point boundary value problems of ordinary differential equations are 

equations of the form 

𝑦𝑦′′ = 𝑓𝑓(𝑥𝑥,𝑦𝑦,𝑦𝑦′)   , 𝑎𝑎 ≤  𝑥𝑥 ≤  𝑏𝑏, With the boundary conditions on the solution prescribed by 

𝑦𝑦(𝑎𝑎)  =  𝛼𝛼 ,  𝑦𝑦(𝑏𝑏)  =  𝛽𝛽, for some constants 𝛼𝛼 and 𝛽𝛽.  

------------------------------------------------------------------------ 

* Corresponding author.  

http://asrjetsjournal.org/


American Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2019) Volume 51, No  1, pp 225-234 

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Second order differential equations with various types of boundary conditions are among many of linear and 

nonlinear problems occurring in science and engineering which can be solved either analytically or numerically 

[3]. Second order two point Boundary value problems are encountered in many engineering fields including 

optimal control, beam deflections, heat flow, and various dynamical systems [4]. In the literature of numerical 

analysis for solving a two point second order boundary value problem (BVP) of differential equations, many 

authors have attempted to obtain higher accuracy rapidly by using a numerous methods. Among various 

numerical techniques, finite difference method has been widely used but it takes more computational costs to get 

high accuracy [3]. In [4] the author has applied a cubic B-spline method to find the solutions of both linear and 

nonlinear second order two point BVPs of ordinary differential equations. The authors in [6] have used an 

extended cubic B-spline method for solving linear two point BVPs. In [7] the authors found the solution of two 

point boundary value problems using quartic B-spline method. But these B-spline methods require dividing the 

interval [𝑎𝑎, 𝑏𝑏] into 𝑛𝑛 subintervals and the construction of cubic or extended cubic B-splines in each subinterval. 

In addition with these methods we need to solve 𝑛𝑛 + 1 systems of linear or nonlinear equations to arrive at 

better accuracy. Therefore, like the finite difference method these methods are also computational too cost. In 

[8] the authors have developed Galerkin method to approximate the solution of second order Neumann and 

Cauchy linear boundary value problems. The authors in [9] derived a new difference scheme for solving the 

linear and nonlinear second order two-point boundary value problem by using the quartic spline interpolation 

and Taylor expansion.  

Finding the solution of ordinary differential equations numerically is not only concerned with getting better 

accuracy. It is also concerned with saving computational speed and effort. This paper is therefore aimed at 

finding the solutions of general second order linear and nonlinear two point boundary value problems of 

ordinary differential equations of the form: 

𝑦𝑦′′(𝑥𝑥) + 𝑝𝑝(𝑥𝑥)𝑦𝑦′(𝑥𝑥) + 𝑓𝑓(𝑥𝑥,𝑦𝑦) = 𝑔𝑔(𝑥𝑥) , 0 ≤ 𝑥𝑥 ≤ 1                                                               (1.1) 

Subject to the boundary conditions: 

𝑦𝑦(0) = 𝛼𝛼  ,   𝑦𝑦(1) = 𝛽𝛽                                                                                                          (1.2) 

 by using Legendre operational matrix of differentiation.  

This study is a contribution towards finding the solutions of second order linear and nonlinear two point 

boundary value problems of ordinary differential equations, by using Tau method based on Legendre 

operational matrix of differentiation. Even though many authors have attempted to obtain higher accuracy 

rapidly by using a numerous methods like finite difference method and spline methods such as cubic B-spline 

method, an extended cubic B-spline method and quartic B-spline method, each of these methods are 

computational too cost and slow when compared to the present method. Thus, the purpose of this study was to 

find the solutions of second order linear and nonlinear two point boundary value problems of ordinary 

differential equations, by using Tau method based on Legendre operational matrix of differentiation.  

The advantage of this method over the other methods is therefore 



American Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2019) Volume 51, No  1, pp 225-234 

227 
 

• It is computational  less cost,  

•  It needs less computational time and effort and  

• It has better accuracy.  

The remainder of this paper has been organized in the following order and procedures.  

• In section 2 Legendre polynomial and its operational matrix of differentiation has been discussed.  

• In section 3 the application of this method has been discussed in brief.  

• In section 4 this method has been used to solve some numerical examples supported by graphs.  

• The paper has been concluded and recommended in section 5. 

2. Legendre Polynomials and its Operational Matrix of Differentiation  

Legendre Polynomials are defined on the interval [−1,1] and can be determined with the aids of the following 

recurrence formulae [5]. 

𝐿𝐿0(𝑧𝑧) = 1 , 𝐿𝐿1(𝑧𝑧) = 𝑧𝑧 

𝐿𝐿𝑟𝑟+1(𝑧𝑧) = 2𝑟𝑟+1
𝑟𝑟+1

(𝑧𝑧)𝐿𝐿𝑟𝑟(𝑧𝑧) − 𝑟𝑟
𝑟𝑟+1

𝐿𝐿𝑟𝑟−1(𝑧𝑧) ;    𝑟𝑟 = 1,2,3, . . .                                                                  ( 2.1) 

In order to use this polynomials on the interval [0, 1], the so-called shifted Legendre Polynomials is defined by 

introducing 𝑧𝑧 = 2𝑥𝑥 − 1. Let the shifted Legendre polynomial  𝐿𝐿𝑟𝑟(2𝑥𝑥 − 1) be denoted by 𝑃𝑃𝑟𝑟(𝑥𝑥). Then  𝑃𝑃𝑟𝑟(𝑥𝑥) can 

be obtained as follows: 

𝑃𝑃0(𝑥𝑥) = 1 , 𝑃𝑃1(𝑥𝑥) = 2𝑥𝑥 − 1 

𝑃𝑃𝑟𝑟+1(𝑥𝑥) = 2𝑟𝑟+1
𝑟𝑟+1

(2𝑥𝑥 − 1)𝑃𝑃𝑟𝑟(𝑥𝑥) − 𝑟𝑟
𝑟𝑟+1

𝑃𝑃𝑟𝑟−1(𝑥𝑥) ;    𝑟𝑟 = 1,2,3, . . .                                                         (2.2) 

The analytical form of the shifted Legendre polynomial 𝑃𝑃𝑟𝑟(𝑥𝑥) of degree r is given by; 

𝑃𝑃𝑟𝑟(𝑥𝑥) = ∑ (−1)𝑟𝑟+𝑘𝑘𝑟𝑟
𝑘𝑘=0  (𝑟𝑟+𝑘𝑘)!

(𝑟𝑟−𝑘𝑘)!
 𝑥𝑥𝑘𝑘

(𝑘𝑘!)2
                                                                                                      (2.3) 

NB:  𝑃𝑃𝑟𝑟(0) = (−1)𝑟𝑟  and  𝑃𝑃𝑟𝑟(1) = 1 

The orthogonality condition for these shifted Legendre polynomials is: 

∫ 𝑃𝑃𝑟𝑟(𝑥𝑥)𝑃𝑃𝑠𝑠(𝑥𝑥)1
0 𝑑𝑑𝑥𝑥 = �

1
2𝑟𝑟+1

      𝑓𝑓𝑓𝑓𝑟𝑟  𝑟𝑟 = 𝑠𝑠 
0             𝑓𝑓𝑓𝑓𝑟𝑟  𝑟𝑟 ≠ 𝑠𝑠

                                                                                           (2.4) 

Any function 𝑦𝑦(𝑥𝑥) ∈ 𝐿𝐿2[0,1] can be approximated in terms of  𝑃𝑃𝑟𝑟(𝑥𝑥) by: 

𝑦𝑦�(𝑥𝑥) = ∑ 𝑐𝑐𝑟𝑟𝑃𝑃𝑟𝑟(𝑥𝑥)∞
𝑟𝑟=0                                                                                                                            (2.5) 



American Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2019) Volume 51, No  1, pp 225-234 

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Where the coefficients 𝑐𝑐𝑟𝑟 are given by  

𝑐𝑐𝑟𝑟 = (2𝑟𝑟 + 1)∫ 𝑦𝑦(𝑥𝑥)1
0 𝑃𝑃𝑟𝑟(𝑥𝑥)𝑑𝑑𝑥𝑥 ;   𝑟𝑟 = 1,2,3, . . .                                                                                (2.6) 

By considering only the first 𝑚𝑚 + 1 terms of the series (2.5) we get; 

𝑦𝑦𝑚𝑚(𝑥𝑥) = ∑ 𝑐𝑐𝑟𝑟𝑃𝑃𝑟𝑟(𝑥𝑥)𝑚𝑚
𝑟𝑟=0 = 𝐶𝐶𝑇𝑇𝜑𝜑(𝑥𝑥)                                                                                                      (2.7) 

Where  𝐶𝐶𝑇𝑇 = [𝑐𝑐0, 𝑐𝑐1, .  .  .  , 𝑐𝑐𝑚𝑚] is the shifted Legendre coefficient and 

            𝜑𝜑(𝑥𝑥) = [𝑝𝑝0(𝑥𝑥), 𝑝𝑝1(𝑥𝑥), … , 𝑝𝑝𝑚𝑚(𝑥𝑥)]𝑇𝑇 is the shifted Legendre vector. 

The derivative of the vector 𝜑𝜑(𝑥𝑥) can be expressed as:    

𝑑𝑑𝑑𝑑(𝑥𝑥)
𝑑𝑑𝑥𝑥

= 𝐷𝐷(1)𝜑𝜑(𝑥𝑥)                                                                                                                                (2.8) 

 

Where  𝐷𝐷(1) is (𝑚𝑚 + 1) × (𝑚𝑚 + 1) operational matrix of derivative which is given by 

𝐷𝐷(1) = �𝑑𝑑𝑖𝑖𝑖𝑖�=�
2(2𝑗𝑗 + 1)  𝑓𝑓𝑓𝑓𝑟𝑟 𝑗𝑗 = 𝑖𝑖 − 𝑘𝑘

0  𝑓𝑓𝑜𝑜ℎ𝑒𝑒𝑟𝑟𝑒𝑒𝑖𝑖𝑠𝑠𝑒𝑒
     ;                     � 𝑘𝑘 = 1,3, … ,𝑚𝑚       𝑖𝑖𝑓𝑓 𝑚𝑚 𝑖𝑖𝑠𝑠 𝑓𝑓𝑑𝑑𝑑𝑑

𝑘𝑘 = 1,3, … ,𝑚𝑚− 1   𝑖𝑖𝑓𝑓 𝑚𝑚 𝑖𝑖𝑠𝑠 𝑒𝑒𝑒𝑒𝑒𝑒𝑛𝑛 

For example for 𝑚𝑚 even we have; 

𝐷𝐷(1) =  2

⎝

⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎛

0 0 0 0 ... 0 0 0
1 0 0 0 ... 0 0 0
0 3 0 0 ... 0 0 0
1 0 5 0 ... 0 0 0
. . . . ... . . .
1 0 5 0 ... 0 0
0 3 0 7 ... 0 0

τ
ω ⎠

⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎞

  Where 𝜏𝜏 = 2𝑚𝑚 − 3,𝜔𝜔 = 2𝑚𝑚 − 1          (2.9) 

From equation (2.8) it can be generalized for any 𝑛𝑛 ∈ 𝑁𝑁 as: 

𝑑𝑑𝑛𝑛𝑑𝑑(𝑥𝑥)
𝑑𝑑𝑥𝑥𝑛𝑛

= (𝐷𝐷(1))𝑛𝑛𝜑𝜑(𝑥𝑥) = 𝐷𝐷(𝑛𝑛)𝜑𝜑(𝑥𝑥)  ,𝑛𝑛 = 1, 2, 3, …                                                                         (2.10) 

Where, (𝐷𝐷(1))𝑛𝑛  denotes matrix powers. 

3. Methods and Materials 

Consider the general second order two point boundary value problem of ordinary differential equation 



American Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2019) Volume 51, No  1, pp 225-234 

229 
 

𝑦𝑦′′(𝑥𝑥) + 𝑝𝑝(𝑥𝑥)𝑦𝑦′(𝑥𝑥) + 𝑓𝑓(𝑥𝑥,𝑦𝑦) = 𝑔𝑔(𝑥𝑥) , 0 ≤ 𝑥𝑥 ≤ 1                                             

Subject to the boundary conditions: 

𝑦𝑦(0) = 𝛼𝛼   ,𝑦𝑦(1) = 𝛽𝛽       

As in [2], let us approximate 𝑦𝑦(𝑥𝑥), 𝑝𝑝(𝑥𝑥), 𝑓𝑓(𝑥𝑥,𝑦𝑦) and 𝑔𝑔(𝑥𝑥) by the shifted Legendre polynomials as 

𝑦𝑦(𝑥𝑥) = ∑ 𝑐𝑐𝑟𝑟𝑃𝑃𝑟𝑟(𝑥𝑥)𝑚𝑚
𝑟𝑟=0 = 𝐶𝐶𝑇𝑇𝜑𝜑(𝑥𝑥),                                                                                                        (3.1) 

𝑝𝑝(𝑥𝑥) = ∑ 𝑝𝑝𝑟𝑟𝑃𝑃𝑟𝑟(𝑥𝑥)𝑚𝑚
𝑟𝑟=0 = 𝑃𝑃𝑇𝑇𝜑𝜑(𝑥𝑥) ,                                                                                                      (3.2) 

𝑓𝑓(𝑥𝑥,𝑦𝑦) = 𝑓𝑓(𝑥𝑥,𝐶𝐶𝑇𝑇𝜑𝜑(𝑥𝑥)),                                                                                                                     (3.3) 

𝑔𝑔(𝑥𝑥) = ∑ 𝑔𝑔𝑟𝑟𝑃𝑃𝑟𝑟(𝑥𝑥)𝑚𝑚
𝑟𝑟=0 = 𝐺𝐺𝑇𝑇𝜑𝜑(𝑥𝑥)                                                                                                        (3.4) 

Where the unknowns are 𝐶𝐶 = [𝑐𝑐0, 𝑐𝑐1, .  .  .  , 𝑐𝑐𝑚𝑚]𝑇𝑇 

Using Legendre operational matrix of differentiation, equation (1.1) can be written as 

𝐶𝐶𝑇𝑇𝐷𝐷2𝜑𝜑(𝑥𝑥) + 𝑃𝑃𝑇𝑇𝐷𝐷1𝜑𝜑(𝑥𝑥) + 𝑓𝑓(𝑥𝑥,𝐶𝐶𝑇𝑇𝜑𝜑(𝑥𝑥)) ≈ 𝐺𝐺𝑇𝑇𝜑𝜑(𝑥𝑥)                                                                        (3.5) 

The residual 𝑅𝑅𝑚𝑚(𝑥𝑥) for equation (3.5) can be written as 

𝑅𝑅𝑚𝑚(𝑥𝑥) = 𝐶𝐶𝑇𝑇𝐷𝐷2𝜑𝜑(𝑥𝑥) + 𝑃𝑃𝑇𝑇𝐷𝐷1𝜑𝜑(𝑥𝑥) + 𝑓𝑓�𝑥𝑥,𝐶𝐶𝑇𝑇𝜑𝜑(𝑥𝑥)� − 𝐺𝐺𝑇𝑇𝜑𝜑(𝑥𝑥)                                                        (3.6) 

Applying typical Tau method, which is used in the sense of particular form of the Petrov-Galerkin method as 

cited in [2], [5], equation (3.5) can be transformed into 𝑚𝑚 − 1 linear or nonlinear equations by applying 

〈𝑅𝑅𝑚𝑚(𝑥𝑥),𝑃𝑃𝑟𝑟(𝑥𝑥)〉 = ∫ 𝑅𝑅𝑚𝑚(𝑥𝑥)𝑃𝑃𝑟𝑟(𝑥𝑥)1
0 𝑑𝑑𝑥𝑥 = 0;     𝑟𝑟 = 0,1,2, . . . ,𝑚𝑚 − 2                                                    (3.7) 

The boundary conditions are given by 

𝑦𝑦(0) = 𝐶𝐶𝑇𝑇𝜑𝜑(0) = 𝑑𝑑0 ,  𝑦𝑦(1) = 𝐶𝐶𝑇𝑇𝜑𝜑(1) = 𝑑𝑑1                                                                                   (3.8) 

Equations (3.7)  and (3.8)  generate 𝑚𝑚 + 1 linear or nonlinear systems of algebraic equations. After solving 

these equations we obtain the unknowns in vector 𝐶𝐶 and use them to find 𝑦𝑦�(𝑥𝑥). 

4. Numerical Examples 

Example 1:  Consider the second order boundary value problem 

𝑦𝑦′′ + 𝑦𝑦′ = 𝑥𝑥                                                                                                                                          (4.1) 



American Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2019) Volume 51, No  1, pp 225-234 

230 
 

With boundary conditions  

𝑦𝑦(0) = 0 ,    𝑦𝑦(1) = 1                                                                                                                         (4.2) 

The exact solution is  𝑦𝑦(𝑥𝑥) = 𝑥𝑥 

Substituting equations (4.2) in equation (2.7) we get 

𝑐𝑐0 − 𝑐𝑐1 + 𝑐𝑐2 = 0
𝑐𝑐0 + 𝑐𝑐1 + 𝑐𝑐2 = 1                                                                                                                                  (4.3) 

For 𝑚𝑚 = 2 from equation (3.7) we get  

𝑐𝑐1 + 13𝑐𝑐2 = 1 
2
                                                                                                                                     (4.4) 

Solving a 3 × 3 system of algebraic equations (4.3) and (4.4) to gives  

𝑐𝑐0 = 1
2

 , 𝑐𝑐1 = 1
2
    and   𝑐𝑐2 = 0 

Thus the approximate solution is now becomes 𝑦𝑦�(𝑥𝑥) = 𝑥𝑥 which is the exact solution. 

 

Figure 1: Graphical illustration of example 1 

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

y

Comparison with different models

 

 
Exact solution
approximate solution with current method
approximate solution with FDM
error with current method
error with FDM



American Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2019) Volume 51, No  1, pp 225-234 

231 
 

Example 2: Consider the second order boundary value problem 

𝑦𝑦′′ + 𝑦𝑦′ + 𝜋𝜋2𝑦𝑦 = −𝜋𝜋sin (𝜋𝜋𝑥𝑥) ,                                                                                                            (4.5) 

With boundary conditions  

𝑦𝑦(0) = 1  ,     𝑦𝑦(1) = −1                                                                                                                     (4.6) 

The exact solution is  𝑦𝑦(𝑥𝑥) = cos (𝜋𝜋𝑥𝑥) 

 

Substituting equations (4.6) in equation (2.7) we get 

𝑐𝑐0 − 𝑐𝑐1 + 𝑐𝑐2 = 1
𝑐𝑐0 + 𝑐𝑐1 + 𝑐𝑐2 = −1                                                                                                                               (4.7) 

For 𝑚𝑚 = 2 from equation (3.7) we get  

𝜋𝜋2𝑐𝑐0 + 2𝑐𝑐1 + 12𝑐𝑐2 = −2                                                                                                                    (4.8) 

Solving a 3 × 3 system of algebraic equations (4.7) and (4.8) to gives  

𝑐𝑐0 = 0 , 𝑐𝑐1 = −1   and   𝑐𝑐2 = 0 

Thus the approximate solution is now becomes  𝑦𝑦�(𝑥𝑥) = −2𝑥𝑥 + 1 

 

Figure 2: Graphical illustration of example 2 

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

y

Comparison with different models

 

 
Exact solution
approximate solution with current method
approximate solution with FDM
error with current method
error with FDM



American Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2019) Volume 51, No  1, pp 225-234 

232 
 

Example 3:  Consider the boundary value problem [4]  

𝑦𝑦′′ + 𝑦𝑦2 = 𝑥𝑥4 + 2 ,                                                                                                                              (4.9) 

With boundary conditions  

𝑦𝑦(0) = 0   ,𝑦𝑦(1) = 1                                                                                                                        (4.10) 

The exact solution is 𝑦𝑦(𝑥𝑥) = x2 

Substituting equations (4.10) in equation (2.7) we get 

𝑐𝑐0 − 𝑐𝑐1 + 𝑐𝑐2 = 0
𝑐𝑐0 + 𝑐𝑐1 + 𝑐𝑐2 = 1                                                                                                                               (4.11) 

For 𝑚𝑚 = 2 from equation (3.7) we get  

𝑐𝑐02 + 1
3
𝑐𝑐12 + 1

5
𝑐𝑐22 + 12𝑐𝑐2 = 11

5
                                                                                                        (4.12) 

Solving a 3 × 3 system of algebraic equations (4.11) and (4.12) gives  

𝑐𝑐0 = 1
3

 , 𝑐𝑐1 = 1
2
    and   𝑐𝑐2 = 1

6
 

Hence the approximate solution is now becomes  𝑦𝑦�(𝑥𝑥) = x2 which is the exact solution. 

 

Figure 3: Graphical illustration of example 3 

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

y

Comparison with current model

 

 
Exact solution
approximate solution with current method
error with current method



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Example 4:  Consider the boundary value problem   

𝑦𝑦′′ + ln 𝑥𝑥𝑦𝑦′ + 𝑦𝑦2 = 2 + 2𝑥𝑥 ln 𝑥𝑥 + 𝑥𝑥4,                                                                                               (4.13) 

With boundary conditions  

𝑦𝑦(0) = 0 ,    𝑦𝑦(1) = 1                                                                                                                       (4.14) 

The exact solution is  𝑦𝑦(𝑥𝑥) = x2 

Substituting equations (4.14) in equation (2.7) we get 

𝑐𝑐0 − 𝑐𝑐1 + 𝑐𝑐2 = 0
𝑐𝑐0 + 𝑐𝑐1 + 𝑐𝑐2 = 1                                                                                                                               (4.15) 

For 𝑚𝑚 = 2 from equation (3.7) we get  

𝑐𝑐02 + 1
3
𝑐𝑐12 + 1

5
𝑐𝑐22 − 2𝑐𝑐1 + 15𝑐𝑐2 = 17

10
                                                                                             (4.16) 

Solving a 3 × 3 system of algebraic equations (4.15) and (4.16) gives  

𝑐𝑐0 = 1
3

 , 𝑐𝑐1 = 1
2
    and   𝑐𝑐2 = 1

6
 

Hence the approximate solution is now becomes  𝑦𝑦�(𝑥𝑥) = x2 which is the exact solution. 

 

Figure 4: Graphical illustration of example 4 

5. Conclusion and Recommendations 

For problems in examples 1,3 & 4 where the exact solution are polynomials, the method produces the exact 

solution itself by just taking 𝑚𝑚 =  2, and using only first few shifted Legendre polynomials. This makes the 

method more accurate than others. For problem in example 2, where the exact solution is not polynomial a 

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

y

Comparison with current model

 

 
Exact solution
approximate solution with current method
error with current method



American Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2019) Volume 51, No  1, pp 225-234 

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better approximation to the exact solution is found when compared to the finite difference method. The other 

advantage of this method over the other methods like spline methods and finite difference method is that it 

needs less computational time and effort.  In the feature, the method can be extended to find the solutions of 

second order linear and nonlinear two point boundary value problems of partial differential equations.  

Acknowledgements 

The author would like to express his sincere gratitude to the anonymous referees for their exciting comments 

which helped him in improving this manuscript significantly. He is also thankful to the previous authors since 

their efforts have been used as references in the study. 

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