Vol 2013

ISSN 2307-7743 http://scienceasia.asia

_______________

2010 Mathematics Subject Classification: 34C07.

Key words and phrases: Mutua’s method, particular polynomial.

© 2013 Science Asia 1 / 9

MUTUA’S METHOD: PARTICULAR POLYNOMIAL AND COEFFICIENTS OF

PARTICULAR INTEGRALS OF THE FORM

e Ax

SAMUEL FELIX MUTUA

Abstract. In many problems involving solutions to ordinary differential equations, students and researchers in the field of applied mathematics are faced with challenges of obtaining the particular integrals. In particular, when the defined problem is formulated wit h non-homogeneous Ordinary Differential Equations with constant coefficients, Methods of Undetermined coefficients prove to be lengthy. Solutions to this will be reached by use of a polynomial, which we shall call particular polynomial whose values at given instants will be used to determine the coefficients of these particular integrals. We will define the polynomial p_q(t) of degree q for q  n ,n _{been the order of the given ODE, by}

) )(

)...( )(

( )

( _{1 2 q 1 q}

q t t m t m t m t m

p     _  . m_s being the root of the characteristic function all not

equal to c (in the exponential). If k are the number of roots all equal to c, then the particular integral is,

cx k

k n

p x e

c P k y

) ( !

1

 

n k  0,1,2,..., _.

1.0 INTRODUCTION

Previously the methods of undetermined coefficients and variation of parameters have been used as techniques for solving these particular integrals. The former involves differentiating the assumed particular integral, substituting to the given ODE and then equating the LHS and RHS coefficients to solve for values of the coefficients. In the latter method, the constants are replaced with parameters (functions) which are varied using the given ODE to solve them. Use of the particular polynomial will simplify this process of obtaining such particular integrals. This will prove helpful as it will give results without having to go through the entire complex algorithms encountered in the past.

2.0 MUTUA’S METHOD

2.1 PRELIMINARIES

Consider the second order ODE

x

e y dx dy

dx y

d 2

2 2

2

3  



0 2 3

2 _{  }

m m

Which has roots m₁  1 and m₂  2 and the complementary function is

x x

c Ae Be

y    2

To solve for the particular integral using method of undetermined coefficients, since 2 is not a root we let

x

p Ae

y  2 ,y p Ae x

2 1 '

2

 , x

p Ae

y ''  22 2

Substituting back to the given ODE yields

x x x

x x

e Ae Ae

Ae

Ae 2 1 2 2 2 2

2

12 2

) 2 ( 3 )

2

(     (RHS)



12 1  A

 x

p e

y 2

12 1 

Now if we critically look at the coefficient

12 1

, its denominator has a very close

correlation with the value of characteristic function f(m) m2 3m  2 atm  2

12 2 ) 2 ( 3 2 ) 2

(  2   

f

) 2 ( 1

12 1

f

A  



2.2 THE PARTICULAR POLYNOMIAL

Now, we consider a general linear ODE of order n with constant coefficients,

cx n

n

n n n

n a y e

dx dy a dx

y d a dx

y d

a     

 

 1 1 0

1

1 ... (1)

Where a_n  0 and c is a constant,

We have the characteristic function

0 1 1

1 ...

)

(m a m a m a m a

f  _n n  _n n    ₍₂₎

We define a polynomial p_q(t) of degree q for q  n by

) )(

)...( )(

( )

( _{1 2 q 1 q}

q t t m t m t m t m

Where m₁,m₂,...,m_q is a subset of the roots of f (m) all c.

We shall call this particular polynomial as it will be used to determine the particular integral.

CASE1: If all the roots of the characteristic function are not equal toc, we find that 0

)

(c 

f

And therefore we take our particular polynomial to be

(4)

Now, the coefficient A of the particular integral, y_p  Aecx becomes

) ( 1

c p

A  i.e. to say

that Particular integral cx

p e

c p y

) ( 1 

(5)

CASE2:

If m₁  m₂ ...  m_k  c and m_k1,..., m_n  c

Then we realise that;

(6) Yields f(c)  0,Since the first k factors become zero.

In this case the coefficient of particular integral cannot be

) ( 1

c f

. Clearly it can be seen

that cx

e will be absorbed in the complementary function.

Also, cx cx k cx

e x e x

xe , 2 ,..., 1 are contained in the complementary and they cannot form the

particular integral.

In this case we assume a particular integral of the form k cx

p Ax e

y  ₍₇₎

Remark

From the differentiation of k cx

e

x (by product rule) where k  zwe realise that as the

order of differentiation increases, in some of the terms the value of finite k diminishes and ultimately tends to zero. The first term with k =0 is in the kth ordered derivative

and will be of the form cx

Be where

! k

B  ,

It must be noted that if the method of undetermined coefficient was to be used the resulting th

k ordered derivative needs to be differentiated (n  k) more times so that )

)...( )(

)...( )(

( )

(m m m₁ m m₂ m m_k m m_{k 1} m m_n

f      _ 

0 1 1

1 ...

)

(t a m a m a m a

upon substitution to the given ODE and equating the LHS and RHS coefficients particular integral is obtained.

Now;

When k 1, ecx is absorbed in the complementary function.

We take cx

p Axe

y  and the particular polynomial as

) )...(

)( (

)

(m m m₂ m m₃ m m_n

p     (8)

Where p(c)  0

Using the knowledge of integration by parts (its reverse), we arrive at the following.

The coefficient A as

) ( . 1

1

c p

When k  2 , ecx,xecx are absorbed in the complementary function.

We take cx

p Ax e

y  2

(9)

Once again p(c) 0

The coefficient A becomes

) ( ! 2

1

) ( . 1 . 2

1

c p c

p

A  

For k  3,4,..., q  n

The corresponding coefficients of A are;

) ( !

1 ,..., ) ( ! 4

1 , ) ( ! 3

1

4

3 c P c q P c

P_{n n nq}

And the particular integrals are cx cx q cx

e Ax e

Ax e

Ax 3 , 4 ,..., 1 , with A_s as defined above. In

general, the particular integral is given by;

cx k

k n

p x e

c P k y

) ( !

1

 

n

k  0,1,2,..., (10)

Equation (10) illustrates MUTUA’S METHOD of finding the particular integral.

Note: If the RHS of equation (1) is multiplied by a constant, then equation (10) is also multiplied by the same constant when evaluating the particular integral.

) )...(

)( (

)

(m m m₃ m m₄ m m_n

3.0 APPLICATION

The above method can be used in modelling phenomenon in which the governing mathematical equations are linear non-homogeneous Ordinary Differential Equation with constant coefficients, the RHS being an exponential function.

3.1 NUMERICAL TEST CASES

EXAMPLE 1

Solve x

e y dx dy

dx y

d 3

2 2

3 5

2   

Solution

The characteristic function is 2m2  5m 3  2(m  3)(m 1 2) 0

2 1 ,

3 ₂

1   m 

m

 Complementary function x x

c Ae Be

y  3  1 2

Here all roots are not equal to 3

Particular polynomialP(m)  2m2 5m 3, P(3) 2(3)25(3)3  30

Thus x

p e

y 3

30 1



And the complete general solution of the given ODE is

x x

x p

c y Ae Be e

y

y 3 1/2 3

30 1  

 

 

EXAMPLE 2

Solve x

e y dx dy

dx y d

dx y

d 2

2 2

3 3

4

4  

 

The characteristic function is m3 m2 4m  4  (m 1)(m 2)(m  2)  0

2 ,

2 ,

1 _{2 3}

1  m  m  

m

 Complementary function x x x

c Ae Be Ce

y   2  2

Here m₂  2  c

Thus x x

p xe xe

y 2 2

4 1

4 !. 1

1





And the complete general solution of the given ODE is

x x

x x

p

c y Ae Be Ce xe

y

y 2 2 2

4 1  

  

 

EXAMPLE 3

Solve x

e y dx dy

dx y d

dx y d

4 3

3 2 2

3 3

    

The characteristic function is m3 3m2 3m 1 (m 1)3  0

1 , 1 ,

1 _{2 3}

1  m  m 

m

 Complementary function x x x

c Ae Bxe Cx e

y    2

Here m₁  m₂  m₃ 1 c

Particular polynomialP(m) 1, P(1) 1

Thus x x

p x e x e

y 3 3

3 2

1 !. 3

4





And the complete general solution of the given ODE is

x x

x x

p

c y Ae Bxe Cx e x e

y

y 2 3

3 2  

   

EXAMPLE 4

Solve x

e y dx dy

dx y

d 7

2 2

5

2   



Solution

The characteristic function is m2  2m 5  (m 1 2i)(m 12i) 0

i m

i

m₁  12 , ₂  1 2

 Complementary function y_c  Acos x  Bsin x

Here all roots are not equal to -7

Thus x

p e

y 7

40

1 _



And the complete general solution of the given ODE is

x p

c y A x B x e

y

y 7

40 1 sin

cos   

  

Note: The above procedure can also be used when c

EXAMPLE 5

Solve x

e y dx dy

dx y

d _{1 3}

2 2

2 6

5   



Solution

The characteristic function is m2 5m 6  (m 3)(m  2)  0

3 ,

2 ₂

1  m 

m

 Complementary function x x

c Ae Be

y  2  3

Here all roots are not equal to -1/3

Particular polynomialP(m)  m2 5m 6 , P(1/3) (1/3)2 5(1/3) 6  70 /9

Thus x x

p e e

y 1/3 1/3

35 9 )

70 9 (

2   



And the complete general solution of the given ODE is

x x

x p

c y Ae Be e

y

y 2 3 1/3

35

9 _

 

  

COMPARISON OF MUTUA’S METHOD OF PARTICULAR POLYNOMIAL WITH THE METHOD OF UNDETERMINED COEFFICIENTS

Suppose we redo example 3 with the METHOD OF UNDETERMINED COEFFICIENTS

Solve x

e y dx dy

dx y d

dx y d

4 3

3 2 2

3 3

    

The characteristic function is m3 3m2 3m 1 (m 1)3  0

1 , 1 ,

1 _{2 3}

1  m  m 

 Complementary function x x x c Ae Bxe Cx e

y    2

To solve for the particular integral, we assume a particular integral of the form.

x

p Ax e

y  3

Where Ais a constant to be determined. Differentiating the above equation three times

we get,

x x

p Ax e Ax e

y'  3 2  3

x x

x x

x x

x

p Axe Ax e Ax e Ax e Axe Ax e Ax e

y''  6 3 2 3 2  3  6  6 2  3

x x

x x

x x

x x

x x

p Ae Axe Axe Ax e Ax e Ax e Ae Axe Ax e Ax e

y'''  6 6 12 6 2 3 2  3  6 18 9 2  3

Substituting into the given ODE with have

x x x

x x

x x

x x

x

e e Ax Ax

e Ax e

Ax e Ax Axe

e Ax e Ax Axe

Ae 18 9 ) 3(6 6 ) 3(3 ) 4

6

(   2  2   2  3  2  3  3 

Or

x x

e

Ae 4

6 



3 2  A

x

p x e

y 3

3 2  

And the complete general solution of the given ODE is

x x

x x

p

c y Ae Bxe Cx e x e

y

y 2 3

3 2  

   

3.2 CONCLUSION

Clearly from the above example we conclude that MUTUA’s method of particular polynomial arrives to the required particular integral of the form n cx

e

Ax faster than the

already existing alternatives. In particular when the ODE in question is of high order a lot of time is spent on differentiation. Thus MUTUA’S method is of great significance since it simplifies the computation process involved in obtaining particular integrals. This is an important aspect to any mathematical formula.

3.3 ACKNOWLEDGMENT

variable inputs, Mr. Raymond Musyoka Munyao (student) and Mr. Ibare John Kamau (fellow student) who helped in proof reading and editing of this work .

REFERENCES

[1] Glyn James (2007): Modern Engineering Mathematics, 4th edition, Pearson Prentice Hall.

[2] K.A Stroud and Dexter J.Booth (2003): Advanced Engineering Mathematics, 4th edition, Palgrave Macmillan, 175 Fifth Avenue, New York.

[3] K.A Stroud and Dexter J.Booth (2007): Engineering Mathematics, 6th edition, Palgrave Macmillan, 175 Fifth Avenue, New York.

[4] K.S.Bhamra (2010): Partial Differential Equation, PHI Learning Private Limited, New Delphi.

[5] Schaum’s (2009): Ordinary Differential Equation, McGraw-Hill Companies, United States of America. [6] Thomas George B (2008): Thomas Calculus, 11th edition, New Delhi: Person Education.

SAMUEL FELIX MUTUA

is a 23 years old, 4th year student and a young potential Mathematician pursuing BSC: Mathematics and Computer Science, from