A Type of the Cauchy-Euler Equations: Distinct Complex Roots

Vol. 19, No. 1, 2019, 79-88

ISSN: 2279-087X (P), 2279-0888(online) Published on 13 March 2019

www.researchmathsci.org

DOI: http://dx.doi.org/10.22457/apam.601v19n1a10

79

Annals of

A Type of the Cauchy-Euler Equations:

Distinct Complex Roots

Gumpon Sritanratana and Atiwath Chanram1

Department of Mathematics Rajabhat Mahasarakham University,

Mahasarakham 44000, Thailand. E-mail: [email protected]

1

Corresponding author. E-mail: [email protected]

Received 2 February 2019; accepted 12 March 2019

Abstract. In this research, the authors give the family of all homogeneous Cauchy-Euler

equations such that each equation has general solution depending on distinct complex numbers and their conjugates.

Keywords: Linear Differential Equations, Cauchy-Euler equations. AMS Mathematics Subject Classification (2010): 47E05

1. Introduction

Consider a homogeneous Cauchy-Euler equation of order n of the form

₁ ... _{1 0} 0.

1 1

1 + + + =

+ − −₋

− a y

dx dy x a dy

y d x a dx

y d x

a _n

n n n n n n

n (1.1)

where a₀,a₁,...,a_n are real numbers with a_n ≠0. Details for methods to find solutions of the equation (1.1) was explained in [2, 3, 4, 9]. Moreover, Sabuwala and Leon [7] studied the particular solution for the most general n-th order Euler differential equation when the non-homogeneity is a polynomial. They found a formula which can be used to compute the unknown coefficients in the form of the particular solution. It is well known that the general solution of (1.1) can be found from the characteristic equation

∑

∏

= =

= + + −

n

j j

i

j m i a

a

1 1

0 0

) 1

( (1.2)

of the linear ordinary differential equation with constant coefficients

, 0 1

1 1

0 =

  

  

+

   

 

+ −

∑

_∏

= =

y a i

dt d a

n

j j

i j

where t=ln x. In general, the general solution of any homogeneous Cauchy-Euler

equations depends on zeros of the polynomial

∑

∏

= =

+ + −

n

j j

i

j m i a

a

1 1

0.

) 1

(

80

(

)

∑

= −

+

= k

j

j j j

j x c x

c x

y j

1

2 1

2 sin(

β

ln ) cos(

β

ln )

α

is the general solution of (1.1) on (0,∞) where

α

_j and

β

_jare real and imaginary parts

of distinctzj with

β

j ≠0 for all j=1,...,k.

2. Preliminary

In this section, we shall give the related basic notions that can be found in [1, 5, 8].

Let n∈ℕ_, ∈

n

a a

a₀, ₁,..., ℝ _with ≠_0.

n

a An ordinary differential equation of the

form

1 1 0 0

1

1 + + + =

+ − −₋ a y

dx dy a dy

y d a dx

y d

a _n

n

n n n

n ⋯ (2.1)

is said to be a homogeneous linear ordinary differential equation with constant coefficients. By a transformation y=emx, where mis a suitable number, the equation (2.1) is transformed into the polynomial equation

0 ... 1 0 1

1 + + + =

+ −

−m am a

a m

a n n

n

n ,

which is said to be the characteristic equation of (2.1).

A linear ordinary differential equation form

0

0 1

1 1 1

1 + + + =

+ − −₋

− a y

dx dy x a dy

y d x a dx

y d x

a _n

n n n n n n

n ⋯ (2.2)

is called a homogeneous Cauchy-Euler equation.

Theorem 2.1. [5] Let k be a positive integer, n=2k and

α

1,

α

2,...,

α

k,

β

1,

β

2,..., k

k,a0,a1,...,a2

β

be real numbers with an ≠0.Then the distinct complex numbers

,

1 1 1 i

z =

α

+

β

z₂ =

α

₂+

β

₂i,...,z_k =

α

_k+

β

_ki, z_k+1=z₁, z_k+2 =z₂,..., z_2k =z_k are

distinct 2k zeroes of the polynomial 1 0

1 1

0 =

  

  

+

   

 

+ −

∑

_∏

= =

y a i

dt d a

n

j j

i

j if and only if

∑

(

)

= −

+

= k

j

j j j

j x c x

c x

y j

1

2 1

2 sin(

β

ln ) cos(

β

ln )

α

(2.3)

is the general solution of homogeneous Cauchy-Euler equation (2.2) on an open interval

), , 0

( ∞ where c1,c2,...,c2k are arbitrary constants.

Definition 2.1. [6] For each j,k∈ℕ with j≤k we define

{

k

}

N_k:= 1,2,..., ,

k j j

k

j aa a a a a N

P_, :={ _{1 2⋯} : ₁, ₂,..., ∈ and a₁<a₂ <_⋯<a_j}, ,

: ,

,

∑

∈

=

k j P p k

j p

81

and for every integers j,k with k> j we define Nk,j:=0.

Lemma 2.1. [6] Let n∈ℕ _{with n}≥_{2, m}∈ℂ _{and a ,...,1 an}∈ℝ _{with an} ≠_{0. Then}

∑

∏

= =

+ + −

j

i n

j

j m i a

a

1

0 1

) 1

( ( 1) .

0

0 1

, 1

1

∑

∑

= +− − +−

−

=

− _{− +}

+

= j

i

j i n j i n i i n

j j n n

nm m N a a

a (2.4)

The following theorem is a direct consequence of Theorem 2.1 and Lemma 2.1.

Theorem 2.2. Let k be a positive integer, n=2k and

α

1,

α

2,...,

α

k,

β

1,

β

2,...,

β

k, k

a a

a0, 1,..., 2 be real numbers with an ≠0. Then the distinct complex numbers ,

1 1 1 i

z =

α

+

β

z₂ =

α

₂+

β

₂i,...,z_k =

α

_k+

β

_ki,

z

_k+1

=

z

₁

,

z

_k+2

=

z

₂

,...,

z

_2k

=

z

_k are

distinct 2k zeroes of the polynomial

∑

∑

= +− − +−

−

=

− _{− +}

+ j

i

j i n j i n i i n

j j n n

nm m N a a

a

0

0 1

, 1

1

) 1

( if

and only if (2.3) is the general solution of homogeneous Cauchy-Euler equation (2.2) on an open interval (0,∞), where c1,c2,...,c2k are arbitrary constants.

3. Main theorems

In this section, before proceeding to our main results, the following terminologies and concepts are required.

Definition 3.1. Let n be a positive integer and z1,z2,...,zn be distinct complex numbers.

Then for every integer j,k with 1≤ j≤k≤n we define

} ,..., ,

{ _{1 2 k}

k z z z

C = ,

: } ,..., , {{ 1 2 ,k j

j a a a

S = a1,a2,...,aj∈Ck and a1,a2,...,aj are distinct},

j S

a a k

j aa a

C

k j j

⋯

∑

∈

=

, 1,..., }

{

2 1

, , C0,0=C0,j =1,

and C_k,j =0 for all j<k≤n.

Form above definition, it is important to note that

k k

k zz z

C, = 1 2⋯ and Ck,k =Ck+1,k+1

for every positive integer k Furthermore, we have the following applicable lemma. .

Lemma 3.1. For every positive n. Ifz1,z2,...,zn are distinct complex numbers and }

,..., ,

{ _{1 2 s}

s z z z

C = for all positive integer s with s≤n, then for every positive integer

j

i,

with

i

≤

j

≤

n

,

.

, 1 , 1 1

,j j i j i j

i z C C

82

Proof: We shall proof by mathematical induction on n . For n=2, let

i,

j

be positive integers such that

i,

j

with i≤ j≤2 and

z

₁

, z

₂are complex numbers and

C

₂

=

{

z

₁

,

z

₂

}.

Then we have 3 cases: i= j=1, i=1 and j=2, and i= j=2. It follow that

,

1 , 1 1 1 1 , 1 1 1 0 ,

1 zC z C

C + _{− −} = = C_1,1+z₂C_1−1,2−1=z₁+z₂=C_1,2 and C2,1+z2C2₋1,2₋1=

,

2 , 2 2 1 z C

z ⋅ = this implies the lemma is true for n=2.

Next, we let k be arbitrary positive integer with k≥2. Suppose that this lemma is true for n=k, that is for every distinct complex numbers z1,z2,...,zk and Cs={z1,

} ,...,

2 zs

z for all positive integer s with

s

≤

n

,

for every positive integers i,j with

k j

i≤ ≤ , C_i,j−1+z_jC_i−1,j−1 =C_i,j. Let z1,z2,...,zk+1 be distinct complex numbers and

} ,..., ,

{ 1 2 s

s z z z

C = for all positive integer s with s≤k+1. Let i,j be positive integers with i≤ j≤k+1. We shall proof that

.

, 1 , 1 1

,j j i j i j

i z C C

C ₋ + _{− −} =

Since z1,z2,...,zk are complex numbers, from the inductive hypothesis, we obtain .

, 1 , 1 1

,j j i j i j

i z C C

C ₋ + _{− −} =

for every positive integers i,j with i≤ j≤k, and so we only prove that

.

, 1 , 1 1

,j j i j i j

i z C C

C ₋ + _{− −} =

for all i≤k+1. Let i be a positive integer with i≤k+1 and let

Ri,k+1 ={{a1,a2,...,ai−1,zk+1}:a1,a2,...,ai−1∈Ck and a1,a2,...,ai−1,zk+1are distinct}.

We claim that Si,k∩Ri,k₊1= ∅ and Si,k∪Ri,k+1 =Si,k+1. For Si,k∩Ri,k+1 = ∅,

suppose that Si,k ∩Ri,k₊1 ≠ ∅. Let {a1,a2,...,ai}∈Si,k∩Ri,k+1. Then {a1,a2,...,ai}∈

k i

S_, and {a₁,a₂,...,a_i}∈R_i,k+1, and thus a₁,a₂,...,a_i∈C_k and a₁,a₂,...,a_i are distinct and there exists j∈Ni such that aj =zk+1. Since as∈Ck for every s∈Ni, we obtain

1

+

≠ k

s z

a which is impossible. Hence S_i,k∩R_i,k+1= ∅.

Next, we shall prove that S_i,k∪R_i,k+1 =S_i,k+1. Let {a₁,a₂,...,a_i}∈S_i,k∪R_i,k+1. Then {a1,a2,...,ai}∈Si,k or {a1,a2,...,ai}∈Ri,k+1.

Case 1. Suppose that{a1,a2,...,ai}∈Si,k. Then a1,a2,...,ai∈Ck and a1,a2,...,ai

are distinct. Since C_k ⊆C_k+1, a₁,a₂,...,a_i∈C_k+1.It follows that {a1,a2,...,ai}∈Si,k+1.

Case 2. Suppose that{a1,a2,...,ai}∈Ri,k+1.Then a1,a2,...,aj−1,aj+1,..., ai∈Ck

are distinct element in C_k and there exists j∈N_i such that a_j =z_k+1 and thus a₁,...,a_i

.

1

+

∈Ck Hence {a1,a2,...,ai}∈Si,k+1.

83

Now we let {a₁,a₂,...,a_i}∈S_i,k+1. Then a₁,a₂,...,ai∈Ck₊₁ and a1,a2,...,ai are

distinct.

Case 1. Suppose that there exists j∈N_i such that a_j =z_k+1. Since a1,a2,...,a_i

are distinct complex numbers. Then {a1,a2,...,ai}∈Ri,k+1 and so {a1,a2,...,ai} ∈Si,k∪ .

1 ,k+ i

R

Case 2. Suppose that a_j ≠z_k+1for all j∈N_i. Since for everyj∈Ni, ai∈Ck+1,

we obtain a_i∈C_k+1. Hence {a₁,a₂,...,a_i}∈S_i,k and thus {a₁,a₂,...,a_i} ∈S_i,k∪R_i,k+1.

From both cases, we obtain S_i,k+1⊆S_i,k∪R_i,k+1. It follows that Si,k∪ .

1 , 1 ,k+ = ik+

i S

R Hence

=

+ k+ i− k

k

i z C

C_{, 1 1,}

∑

− − ∈

− +

+

k i i S a a a

i k

k

i z aa a

C

, 1 1 2 1, ,..., }

{

1 2 1 1

, ⋯

∑

∑

− − ∈

+ − ∈

+ =

k i i k

i

i a a a S

k i S

a a a

i a a a z

a a a

, 1 1 2 1 ,

2

1 { , ,..., }

1 1 2 1 }

,..., , {

2

1 ⋯ ⋯ .

Let zk₊1Si₋1,k ={{a1,a2,...,ai−1,zk+1}:{a1,a2,...,ai−1}∈Si−1,k}. Then

k i k k

i z C

C_, + _{+1 −1,}

∑

∑

− + +

− ∈

+ − ∈

+ =

k i k k i k

i

i a a a z z S

k i S

a a a

i aa a z

a a a

, 1 1 1 1 2 1 ,

2

1 { , ,..., , }

1 1 2 1 }

,..., , {

2

1 ⋯ ⋯

∑

∑

− +

− ∈

− ∈

+ =

k i k i i k

i

i a a a a z S

i i S

a a a

i aa a a

a a a

, 1 1 1 2 1 ,

2

1 { , ,..., , }

1 2 1 }

,..., , {

2

1 ⋯ ⋯ .

Since {a1,a2,...,ai−1,ai}∈zk+1Si−1,k

⇔ there exists j∈N_i such that a_j =z_k+1 and

a1,a2,...,aj−1,aj+1,...,ai−1,ai∈Ck

⇔ {a1,a2,...,aj−1,zk+1,aj+1,...,ai−1,ai}∈Rk+1

⇔ {a1,a2,...,ai−1,ai}∈Ri,k+1,

we obtain zk+1Si−1,k =Rk+1. Therefore

k i k k

i z C

C_, + _{+1 −1,}

∑

∑

+

∈ ∈

+ =

1 2

1 ,

2

1 { , ,..., }

2 1 }

,..., , {

2

1 .

k i k

i

i a a a R i S

a a a

i aa a

a a

a _{⋯ ⋯}

Since Si,k∩Ri,k₊1= ∅ and Si,k∪Ri,k+1 =Si,k+1,we obtain

k i k k

i z C

C_, + _{+1 −1,}

∑

+

∪ ∈

=

1 , , 2 1, ,..., }

{

2 1

k i k i i S R a a a

i

a a

a _⋯

∑

+

∈

=

1 , 2 1, ,..., }

{

2 1

k i i S a a a

i

a a

a _⋯ =C_i,k+1.

Hence Ci,k+zk+1Ci−1,k =Ci,k+1.

By mathematical induction, this Lemma is true for all positive integer n. □

Lemma 3.2. Let n be a positive integer. If z1,z2,...,zn are distinct complex numbers and },

,..., ,

{ 1 2 n

n z z z

84 . ) 1 ( ) 1 ( ) ( 1 1 0 , ,

∏

∑

= − = − _{+ −} − = − n i n i n n n i n n i i

i C m C

z

m (3.2)

Proof: We prove by mathematical induction. For n=1, let z₁ be a complex number. Then since C_0,1 =1 and C_1,1 =z₁,

1

1

1

)

(m z m z

i

i = −

−

∏

= 1 , 1 1 0 1 1 , 0 0

)

1

(

)

1

(

−

C

m

+

−

C

=

− 1 , 1 1 0 0 1 1

, ( 1)

) 1

( C m C

i

i i

i + −

− =

∑

= − . ) 1 ( ) 1 ( _, 1 0

, nn n n i i n n i i C m

C + −

−

=

∑

−

=

−

This implies that (3.2) is true for n=1.

Next, we let k be a positive integer. Suppose that if z1,z2,...,zk are distinct

complex numbers and Ck ={z1,z2,...,zk}, then

. ) 1 ( ) 1 ( ) ( 1 0 , , 1

∑

∏

− = − = − + − = − n i k k k i k n i i k i

i C m C

z m

Let z1,z2,...,zk,zk+1 be distinct complex numbers and Ck+1 ={z1,z2,...,zk+1}. Since

∏

∏

= + + = − − = − k i i k k i

i m z m z

z m 1 1 1 1 ), ( ) ( ) (

by the inductive hypothesis, we obtain

, ) 1 ( ) 1 ( ) ( 1 0 , , 1

∑

∏

− = − = − + − = − k i k k k i k k i i k i

i C m C

z m and thus

∏

+ = − 1 1 ) ( k i i z m       − + − − =

∑

− = − + 1 0 , ,

1) ( 1) ( 1)

( k i k k k i k k i i

k C m C

z m

∑

∑

− = − = + − − + _{− −} − = 1 0 1 0 1 , 1

, ( 1)

) 1 ( k i k i k i k k i i i k k i i z m C m

C k kk

k k k k C z m

C , ( 1) 1 ,

) 1 (− − − ₊ +

∑

∑

− = + − − = − + + _{+ − − −} = 2 0 1 , 1 1 1 , 1 ) 1 ( ) 1 ( k i k i k k i i k i i k k i i k z m C m C m

( 1)

(

)

( 1) 1, 1.

1 ,

, 1

1 + +

+ −

+ + + −

−

+ k k

k k k k k k k C m C C z

By Lemma 3.1, we obtain zk+1Ck−1,k+Ck,k =Ck,k+1 and therefore

∏

+ = − 1 1 ) ( k i i z

m

∑

∑

− = + − − = − + + _{+ − − −} = 2 0 1 , 1 1 1 , 1 ) 1 ( ) 1 ( k i k i k k i i k i i k k i i k z m C m C m . ) 1 ( ) 1

(− _{, +1} + − +1 _{+1, +1}

+ k k

k k k k C m C

Let j=i−1. Then i= j+1 and thus

∏

+ = − 1 1 ) ( k i i z

m

∑

∑

85

1, 1

1 1

, ( 1)

) 1

(− ₊ + − + _{+ +}

+ k k

k k k k C m C

∑

∑

− = − + + − = − + + + _{+ − + −} = 2 0 , 1 1 2 0 , 1 1 1 ) 1 ( ) 1 ( k i i k k i k i k i i k k i i k m C z m C m

+(−1)kC_k,k+1m+(−1)k+1C_k+1,k+1

1 , 1 1 1 , 2 0 1 , 1 1 1 ) 1 ( ) 1 ( ) 1 ( ₊ + _{+ +} − = − + + + + _{+ − + − + −}

=

∑

k k

k k k k k i i k k i i k C m C m C m

by Lemma 3.1. Let j=i+1. Then i= j−1 and so

∏

+ = − 1 1 ) ( k i i z

m 1, 1

1 1 , 1 1 1 1 , 1 ) 1 ( ) 1 ( ) 1 ( ₊ + _{+ +} − = − + + + _{+ − + − + −}

=

∑

k k

k k k k k j j k k j j k C m C m C m 1 , 1 1 1 , 1 1 1 1 , 1 ) 1 ( ) 1 ( ) 1 ( ₊ + _{+ +} − = − + + + _{+ − + − + −}

=

∑

k k

k k k k k i i k k j i k C m C m C m

1, 1

1 0

1 1

, ( 1)

) 1 ( + _{+ +} = − + + + − −

=

∑

k k

k k i i k k i i C m

C □

The following corollary is a direct consequence of Lemma 3.2.

Corollary 3.1. Let n be a positive integer. If z₁,z₂,...,z_n are distinct complex numbers and Cn ={z1,z2,...,zn}, then z1,z2,...,zn are zeroes of the polynomial

( 1) ( 1) _, .

1

1

, nn n n i i n n i i C m

C + −

−

∑

− =

−

(3.3)

Lemma 3.3. Let n be a positive integer. If z1,z2,...,zn are distinct complex numbers

and C_n ={z₁,z₂,...,z_n}, then

0 ( 1) n,n, n

C

a = − an =1 and

∑

− = + − + + − − _{+ −} −

= n j

i j i j i i i n j n j n

j C N a

a

1

1 , 1 , ( 1)

) 1

( (3.4)

for every j=1,2,...,n−1 if and only if

0 0 1 , 1 1 ) 1

( N a a

m m a j i j i n j i n i i n j j n n

n +

∑

∑

− +

= +− − +− − = − . ) 1 ( ) 1 ( , 1 0

, nn n n i i n n i i C m

C + −

−

=

∑

−

=

−

(3.5)

Proof: Suppose that (3.4) is true for all j=1,2,...,n−1. Then for any j=1,2,...,n−1,

, ) 1 ( ) 1 ( , 1 1

, in j n j n i j i j i i i

j N a C

a − − = + − + − = − +

∑

that is for every s=1,2,...,n−1,

. ) 1 ( ) 1 ( , 1 1

, in s n s n i s i s i i i

s N a C

a − − = + − + − = − +

∑

86

. ) 1 ( )

1

( _,

1

1

, in

j j

i

j i n j i n i i j

n N a C

a +

∑

− = −

= +− − +−

−

Since N_0,n−j−1=1,

n i j j

i

j i n j i n i i j

n j

n a N a C

N ,

1

1 , 1

, 0 0

) 1 ( )

1 ( )

1

(− +

∑

− = −

= +− − +−

− − −

and thus

, ) 1 ( )

1

( _,

0

1

, in

j j

i

j i n j i n i i

C a

N = −

−

∑

= +− − +−

for every j=1,2,...,n−1. Multiplying both sides of this equation by mn−j, we obtain

. )

1 ( )

1

( ,

0

1 ,

j n n i j j

i

j i n j i n i i j

n

m C a

N

m −

= +− − +−

−

∑

_{− = −}

Adding this n−1 equations, we have

∑

∑

= +− − +−

−

=

− j ₋

i

j i n j i n i i n

j j n

a N

m

0

1 , 1

1

) 1

( ( 1) .

1

1

,

∑

− =

−

−

= n

j

j n n i j

m C

That is

0 0

1 , 1

1

) 1

( N a a

m m

a

j

i

j i n j i n i i n

j j n n

n +

∑

∑

− +

= +− − +−

−

= −

. ) 1 ( )

1

( _,

1

0

, nn n n

i

i n n i i

C m

C + −

−

=

∑

−

=

−

Conversely, assume that the equation (3.5) is true. Since 1,m,m2,...,mn

are linearly independent, by undetermined coefficients, we obtain

, 1

=

n

a nn

n

C

a0 =(−1) ,

and for every j=1,2,...,n−1,

∑

= +− − +−

−

j

i

j i n j i n i i

a N

0

1 ,

) 1

( ( 1) ,

1

1

,

∑

− =

−

=n

j

n j j

C

that is

∑

= +− − +−

− + −

j

i

j i n j i n i i j

n N a

a

1

1 ,

) 1

( =(−1)jC_j,n,

Let s=n− j. Then for every s=1,2,...,n−1,

∑

−

= + − +

−

+n s

i

s i s i i i

s N a

a

1

1 ,

) 1

( n sn

s n

C ,

) 1 (− − ₋

=

and thus for every j=1,2,...,n−1,

+

−

=

− ₋

n j n j n

j

C

a

(

1

)

_, ( 1) .

1

1 , 1

∑

−

= + − +

+

−

j n

i

j i j i i i

a N

87

Corollary 3.2. Let k be a positive integer and

α

1,

α

2,...,

α

k,

β

1,

β

2,...,

β

k,a0,a1,...,a2k

be real numbers with an ≠0. If z1,z2,...,z2k are distinct complex numbers and C2k = }

,..., ,

{z1 z2 z2k such that z1=

α

1+

β

1i, z2 =

α

2+

β

2i,...,zk =

α

k +

β

ki,

z

k+1

=

z

1

,

,...,

2 2 z

z_k+ = z_2k =z_k. Then

( 1) 2 ,2 ,

2 0 k k

k

C

a = − a_2k =1 and

∑

−

= + − +

+ −

− _{+ −}

−

= k j

i

j i j i i i k

j k j k

j C N a

a

2

1

1 , 1 2

, 2 2

) 1 ( )

1

( (3.6)

for every j=1,2,...,n−1 if and only if z1,z2,...,z2k are the zeroes of polynomial

( 1) 0.

0

2 1 2 , 1

2

1 2 2

2 m m N a a

a

j

i

j i k j i k i i k

j

j k k

k +

∑

∑

− +

= +− − +−

−

= −

(3.7)

Proof: By Lemma 3.3 with n=2k, the formulas (3.6) is true for every j=1,2,...,2k−1

if and only if

+

∑

∑

− + =

= +− − +−

−

= −

0 0

2 1 2 , 1

2

1 2 2

2 m m ( 1) N a a

a

j

i

j i k j i k i i k

j

j k k

k ( 1) ( 1) 2 ,2 .

2 1

2

0

2 2

, k k k k

i

i k k i i

C m

C + −

−

∑

− =

−

By Corollary 3.1 with n=2k, we obtain

. )

1 ( )

1 ( ) (

2

1

1 2

0

2 , 2 2 2

2 ,

∏

∑

=

−

=

− _{+ −}

− = −

k

i

k

i

k k k i

k k i i

i C m C

z m

and thus (3.6) is true for every j=1,2,...,n−1 if and only if

∏

=

= −

n

i

i

z m

1

)

( ( 1) ₀.

0

1 , 1

1

a a

N m

m a

j

i

j i n j i n i i n

j j n n

n +

∑

∑

− +

= +− − +−

−

= −

Thus (3.6) holds for every j=1,2,...,n−1, if and only if z1,z2,...,zn are zeroes of the

polynomial (3.7). □

The following main theorem is a direct consequence of Corollary 3.2 and

Theorem 2.2 with n=2k.

Theorem 3.1. Let k be a positive integer and

α

₁,

α

₂,...,

α

_k,

β

₁,

β

₂,...,

β

_k,a₀,a₁,...,a_2k

be real numbers with a_2k ≠0. If z1,z2,...,z2k are distinct complex numbers and }

,..., ,

{ 1 2 2 2k z z z k

C = such that z₁ =

α

₁+

β

₁i, z2 =

α

2+

β

2i,...,zk =

α

k +

β

ki,

,

1 1

z

z

k+

=

z

k+2

=

z

2

,...,

z

2k

=

z

k

.

Then (3.6) is true for every j=1,2,...,2k−1 if and

only if

∑

(

)

= −

+

= k

j

j j j

j x c x

c x

y j

1

2 1

2 sin(

β

ln ) cos(

β

ln )

α

(3.8)

is the general solution of homogeneous Cauchy-Euler equation

0

0 1

1 2

1 2 1 2 1 2 2 2 2

2 + ₋ + + + =

− −

− a y

dx dy x a dx

y d x a dx

y d x

a _k

k k k k k k

88

on an open interval (0,∞), where c1,c2,...,c2k are arbitrary constants.

The application of Theorem 3.1 is to establish a Cauchy-Euler equation of order

n together with its general solution on (0,∞), of the form (3.8).

4. Conclusion

We give every Cauchy-Euler differential equation from its general solution that depends only on a given finite numbers of distinct complex numbers. In the future, we will devote our attention to the family of all Cauchy-Euler differential equations that have general solutions depending only on a complex number.

Acknowledgements. The authors are thankful to the reviewers for valuable comments and

suggestions on the manuscript and thank the Faculty of Science and Technology, Rajabhat Mahasarakham University, Mahasarakham, Thailand for financial support.

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