STUDY THE BEHAVIOR OF THE SOLUTION AND ASYMPTOTIC BEHAVIORS OF EIGENVALUES OF A SIX ORDER BOUNDARY VALUE PROBLEM

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STUDY THE BEHAVIOR OF THE SOLUTION AND ASYMPTOTIC BEHAVIORS OF EIGENVALUES OF A SIX ORDER BOUNDARY VALUE

PROBLEM

Karwan H.F. Jwamer & Aryan Ali M.

Department of Mathematics, Faculty of Science and Science Education, School of Science, University of Sulaimani, Kurdistan Region, Sulaimani, Iraq

[email protected], [email protected]

ABSTRACT

In this paper we study the behavior of the solution and asymptotic behavior of eigenvalues of a six order boundary value problem of the form:

−𝑦⁶ 𝑥 + 𝑝4 𝑥 𝑦⁴ 𝑥 + 𝑝3 𝑥 𝑦^′′ 𝑥 + 𝑝2 𝑥 𝑦^′ 𝑥 + 𝑝1 𝑥 𝑦 𝑥 = 𝜆⁶𝜌 𝑥 𝑦 𝑥 , 0 < 𝑥 < 𝑎,

With boundary conditions

𝑈𝑗 𝑦 = 𝑦^{(𝑗 )} 0 = 0 , 𝑗 = 0,1,2,3 𝑈𝑗 𝑦 = 𝑦^{(5−𝑗 )} 𝑎, 𝜆 = 0 , 𝑗 = 4,5,

Where 𝑝₄ 𝑥 , 𝑝₃ 𝑥 , 𝑝₂ 𝑥 , 𝑝₁ 𝑥 and 𝜌 𝑥 are real functions and 𝜌 𝑥 > 0, and 𝜆 is spectral parameter in which𝜆 = 𝜍 + 𝑖𝜏 , 𝜍, 𝜏 ∈ 𝑅, 𝑖 = −1 , 𝜏 ≠ 0

Keywords: Asymptotic behavior of eigenvalues, boundary value problems, continuous functions.

2000 MR Subject Classification: 34B15, 34L20

1. INTRODUCTION

The investigation of boundary value problems for which the eigenvalues parameter appears in both the equation and boundary conditions originates from the works of G.D. Birkhoff [2-3]. There are many papers and books, where the spectral properties of such problems are investigates (see, for example [1] and [4-9] ). But in this paper we study the behavior of the solution and asymptotic behaviors of eigenvalues of a six order boundary value problem𝑆1 which is defined by:

−𝑦⁶ 𝑥 + 𝑝₄ 𝑥 𝑦⁴ 𝑥 + 𝑝₃ 𝑥 𝑦^′′ 𝑥 + 𝑝₂ 𝑥 𝑦^′ 𝑥 + 𝑝₁ 𝑥 𝑦 𝑥 = 𝜆⁶𝜌 𝑥 𝑦 𝑥 , 0 < 𝑥 < 𝑎, (1.1) with boundary conditions

𝑈_𝑗 𝑦 = 𝑦^{(𝑗 )} 0 = 0 , 𝑗 = 0,1,2,3 𝑈_𝑗 𝑦 = 𝑦^{(5−𝑗 )} 𝑎, 𝜆 = 0 , 𝑗 = 4,5,

where 𝑝₄ 𝑥 , 𝑝₃ 𝑥 , 𝑝₂ 𝑥 , 𝑝₁ 𝑥 and 𝜌 𝑥 are real functions and 𝜌 𝑥 > 0, and 𝜆 is spectral parameter in which𝜆 = 𝜍 + 𝑖𝜏 , 𝜍, 𝜏 ∈ 𝑅, 𝑖 = −1 , 𝜏 ≠ 0. Here we assume that 𝑝₄ 𝑥 ∈ 𝐶⁴ 0, 𝑎 , 𝑝₃ 𝑥 ∈ 𝐶² 0, 𝑎 , 𝑝₂ 𝑥 ∈ 𝐶¹ 0, 𝑎 , 𝑝₁ 𝑥 ∈ 𝐶 0, 𝑎 , 𝜌 𝑥 ∈ 𝐶⁶ 0, 𝑎 . We have introduced the sectors 𝑇_𝑘 and their conjugates𝑇 (relative to 𝑘

the x-axis). Let 𝜆 located in some fixed sector 𝑇𝑘 or 𝑇 , and let 𝑤𝑘 _𝑗’s ( j=0,1,2,3,4,5 )be different roots of unity of degree 6, and ordered so that for all 𝜆 ∈ 𝑇𝑘 (or 𝑇 ) satisfied the inequality: 𝑘

Re(i𝑤𝑜𝜆) ≤Re(i𝑤1𝜆) ≤Re(i𝑤2𝜆) ≤Re(i𝑤3𝜆) ≤Re(i𝑤4𝜆) ≤Re(i𝑤5𝜆),

Numbering depends, the selected sector contains 𝜆. Entire complex plane of 𝜆 = 𝜍 + 𝑖𝜏divided into 12 sectors 𝑇𝑘

and 𝑇 in the plane 𝜆 determined by the inequalities 𝑘 ^kπ₃ ≤ arg λ ≤^kπ₃ +^π₆ ,

𝑘 = 0,1,2,3,4,5, and we assume that 𝑤_𝑘 = 1⁶ , and ∅_𝑘= 𝑖𝑤_𝑘⁶ 𝜌(𝑥) . (1.2)

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2. STUDY THEBEHAVIOR OF THE SOLUTION OF SIXTH ORDER BOUNDARY VALUE PROBLEM The aim of this section is to estimate the behavior of the solution to the given sixth order boundary value problem 𝑆₁and finding their coefficients 𝐴_𝑖 𝑥 , 𝑖 = 0,1,2,3,4,5,6.

Theorem2.1:

Suppose 𝜆 ∈ 𝑇𝑘or 𝜆 ∈ 𝑇 (𝑘 = 0,1,2,3,4,5) and ∅𝑘 _𝑘(𝑘 = 0,1,2,3,4,5) satisfies the inequality (1.2), then there exist linear independent solutions 𝑦𝑘 𝑥, 𝜆 , (𝑘 = 0,1,2,3,4,5) of equation (1.1) regular with sufficient large 𝜆 such that when 𝑗 = 0,1,2,3,4,5 uniformly in 0 ≤ 𝑥 ≤ 𝑎 satisfy the relation

𝑦_𝑘^𝑗 𝑥, 𝜆 = ∅_𝑘(𝑥) 𝜆 ^𝑗𝑒^{𝜆 ∅}⁰^𝑥 ^𝑘^{𝑡 𝑑𝑡} 𝐴𝑖 𝑥 𝜆^𝑖

6

𝑖=0

+ 𝑂 1 𝜆⁷ , such that

𝐴𝑜 𝑥 = 𝜌 𝑥 ⁻¹²⁵ , 𝐴1 𝑥 = 𝐴_𝑜 𝑥 −5

2 𝐴𝑜′′ 𝑡

∅𝑘 − 10∅_𝑘^′

∅_𝑘²𝐴𝑜′ 𝑡 − 15 2

∅_𝑘^′ ²

∅_𝑘³ +10 3

∅_𝑘^′′

∅_𝑘²−𝑝4 𝑡

6∅𝑘 𝐴𝑜 𝑡

𝑥

0

𝑑𝑡 𝐴𝑜 𝑡 , 𝐴₂ 𝑥 = 𝐴_𝑜 𝑥 [ −5

2 𝐴1′′ 𝑡

∅_𝑘 − 10∅_𝑘^′

∅_𝑘²𝐴₁^′ 𝑡 − 15 2

∅_𝑘^′ ²

∅_𝑘³ +10 3

∅_𝑘^′′

∅_𝑘²−𝑝4 𝑡

6∅_𝑘 𝐴₁ 𝑡 −10 3

𝐴𝑜3 𝑡

∅_𝑘²

𝑥

0

− 15∅_𝑘^′

∅_𝑘³𝐴^′′_𝑜 𝑡

− 10∅𝑘′′

∅_𝑘³+ 15 ∅_𝑘^′ ²

∅_𝑘⁴ −2 3

𝑝4 𝑡

∅_𝑘² 𝐴𝑜′ 𝑡

−( 5 2

(∅_𝑘³

∅_𝑘³ + 10∅_𝑘^′∅_𝑘^′′

∅_𝑘⁴ +5 2

∅_𝑘^′ ³

∅_𝑘⁵ −∅_𝑘^′

∅_𝑘³𝑝4 𝑡 ) 𝐴_𝑜 𝑡 ] 𝑑𝑡 𝐴𝑜 𝑡 ,

𝐴3 𝑥 = 𝐴_𝑜 𝑥 [ −5 2

𝐴₂^′′ 𝑡

∅𝑘 − 10∅_𝑘^′

∅_𝑘²𝐴2′ 𝑡 − 15 2

∅_𝑘^′ ²

∅_𝑘³ +10 3

∅_𝑘^′′

∅_𝑘²−𝑝₄ 𝑡

6∅𝑘 𝐴2 𝑡 −10 3

𝐴₁³ 𝑡

∅_𝑘²

𝑥

0

− 15∅_𝑘^′

∅_𝑘³𝐴1′′ 𝑡

− 10∅_𝑘^′′

∅_𝑘³+ 15 ∅_𝑘^′ ²

∅_𝑘⁴ −2 3

𝑝4 𝑡

∅_𝑘² 𝐴₁^′ 𝑡

−( 5 2

∅𝑘 3

∅_𝑘³ + 10∅𝑘′∅𝑘′′

∅_𝑘⁴ +5 2

∅_𝑘^′ ³

∅_𝑘⁵ −∅𝑘′

∅_𝑘³𝑝₄ 𝑡 ) 𝐴₁ 𝑡 − 5 2

𝐴_𝑜⁽⁴⁾ 𝑡

∅_𝑘³ − 10∅𝑘′

∅_𝑘⁴𝐴_𝑜³ 𝑡

− 10∅𝑘′′

∅_𝑘⁴+15 2

∅_𝑘^′ ²

∅_𝑘⁵ −𝑝4 𝑡

∅_𝑘³ 𝐴′′𝑜 𝑡 − 5 ∅𝑘 3

∅_𝑘⁴ + 10∅𝑘′∅𝑘′′

∅_𝑘⁵ − 2∅𝑘′

∅_𝑘⁴𝑝4 𝑡 𝐴𝑜′ 𝑡

−( ∅𝑘 4

∅_𝑘⁴ +5 2

∅𝑘′∅_𝑘³

∅_𝑘⁵ +5 3

∅_𝑘^′′ ²

∅_𝑘⁵ −1 2

∅_𝑘^′ ²

∅_𝑘⁵ 𝑝₄ 𝑡 − −2 3

∅𝑘′′

∅_𝑘⁴𝑝₄ 𝑡 −𝑝3 𝑡

6∅_𝑘³ ) 𝐴_𝑜 𝑡 ] 𝑑𝑡 𝐴_𝑜 𝑡 ,

𝐴4 𝑥 = 𝐴_𝑜 𝑥 [ −5 2

𝐴₃^′′ 𝑡

∅𝑘 − 10∅_𝑘^′

∅_𝑘²𝐴3′ 𝑡 − 15 2

∅_𝑘^′ ²

∅_𝑘³ +10 3

∅_𝑘^′′

∅_𝑘²−𝑝₄ 𝑡 6∅𝑘 𝐴3 𝑡

𝑥

0

−10 3

𝐴₂³ 𝑡

∅_𝑘² − 15∅_𝑘^′

∅_𝑘³𝐴2′′ 𝑡 − 10∅_𝑘^′′

∅_𝑘³+ 15 ∅_𝑘^′ ²

∅_𝑘⁴ −2 3

𝑝₄ 𝑡

∅_𝑘² 𝐴2′ 𝑡 − ( 5 2

∅_𝑘³

∅_𝑘³ +10∅_𝑘^′∅_𝑘^′′

∅_𝑘⁴ +5 2

∅_𝑘^′ ³

∅_𝑘⁵ −∅_𝑘^′

∅_𝑘³𝑝4 𝑡 ) 𝐴₂ 𝑡 − 5 2

𝐴₁⁴ 𝑡

∅_𝑘³ − 10∅_𝑘^′

∅_𝑘⁴𝐴₁³ 𝑡 − (10∅_𝑘^′′

∅_𝑘⁴

+15 2

∅_𝑘^′ ²

∅_𝑘⁵ −𝑝4 𝑡

∅_𝑘³ )𝐴₁^′′ 𝑡 − 5 ∅_𝑘³

∅_𝑘⁴ + 10∅_𝑘^′∅_𝑘^′′

∅_𝑘⁵ − 2∅_𝑘^′

∅_𝑘⁴𝑝₄ 𝑡 𝐴₁^′ 𝑡 − ( ∅_𝑘⁴

∅_𝑘⁴

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2

∅_𝑘^′∅_𝑘³

∅_𝑘⁵ +5 3

∅_𝑘^′′ ²

∅_𝑘⁵ −1 2

∅_𝑘^′ ²

∅_𝑘⁵ 𝑝4 𝑡 −2 3

∅_𝑘^′′

∅_𝑘⁴𝑝4 𝑡 −𝑝3 𝑡

6∅_𝑘³ ) 𝐴1 𝑡 −𝐴⁵_𝑜 𝑡

∅_𝑘⁴

−5 2

∅_𝑘^′

∅_𝑘⁵𝐴_𝑜⁴ 𝑡 − 10 3

∅_𝑘^′′

∅_𝑘⁵−2 3

𝑝4 𝑡

∅_𝑘⁴ 𝐴_𝑜³ 𝑡 − 5 2

∅_𝑘³

∅_𝑘⁵ −∅_𝑘^′

∅_𝑘⁵𝑝₄ 𝑡 𝐴_𝑜^′′ 𝑡 − ( ∅_𝑘⁴

∅_𝑘⁵

−2 3

∅𝑘′′

∅_𝑘⁵𝑝₄ 𝑡 −1 3

𝑝3 𝑡

∅_𝑘⁴ )𝐴^′_𝑜 𝑡 − ( ∅𝑘 5

6∅_𝑘⁵ −∅_𝑘³

6∅_𝑘⁵𝑝₄ 𝑡 − ∅𝑘′

6∅_𝑘⁵𝑝₃ 𝑡 −𝑝2 𝑡

6∅_𝑘⁴ ) 𝐴_𝑜 𝑡 ] 𝑑𝑡

𝐴𝑜 𝑡 ,

𝐴5 𝑥 = 𝐴_𝑜 𝑥 [ −5 2

𝐴₄^′′ 𝑡

∅𝑘 − 10∅_𝑘^′

∅_𝑘²𝐴4′ 𝑡 − 15 2

∅_𝑘^′ ²

∅_𝑘³ +10 3

∅_𝑘^′′

∅_𝑘²−𝑝₄ 𝑡 6∅𝑘 𝐴4 𝑡

𝑥

0

−10 3

𝐴₃³ 𝑡

∅_𝑘² − 15∅𝑘′

∅_𝑘³𝐴₃^′′ 𝑡 − 10∅𝑘′′

∅_𝑘³+ 15 ∅_𝑘^′ ²

∅_𝑘⁴ −2 3

𝑝4 𝑡

∅_𝑘² 𝐴₃^′ 𝑡 − ( 5 2

∅𝑘 3

∅_𝑘³ +10∅_𝑘^′∅_𝑘^′′

∅_𝑘⁴ +5 2

∅_𝑘^′ ³

∅_𝑘⁵ −∅_𝑘^′

∅_𝑘³𝑝4 𝑡 ) 𝐴₃ 𝑡 − 5 2

𝐴₂⁴ 𝑡

∅_𝑘³ − 10∅_𝑘^′

∅_𝑘⁴𝐴₂³ 𝑡 − (10∅_𝑘^′′

∅_𝑘⁴

+15 2

∅_𝑘^′ ²

∅_𝑘⁵ −𝑝4 𝑡

∅_𝑘³ )𝐴′′2 𝑡 − 5 ∅_𝑘³

∅_𝑘⁴ + 10∅_𝑘^′∅_𝑘^′′

∅_𝑘⁵ − 2∅_𝑘^′

∅_𝑘⁴𝑝4 𝑡 𝐴₂^′ 𝑡 − ( ∅_𝑘⁴

∅_𝑘⁴

+5 2

∅_𝑘^′∅_𝑘³

∅_𝑘⁵ +5 3

∅_𝑘^′′ ²

∅_𝑘⁵ −1 2

∅_𝑘^′ ²

∅_𝑘⁵ 𝑝4 𝑡 −2 3

∅_𝑘^′′

∅_𝑘⁴𝑝4 𝑡 −𝑝₃ 𝑡

6∅_𝑘³ ) 𝐴2 𝑡 −𝐴₁⁵ 𝑡

∅_𝑘⁴

−5 2

∅_𝑘^′

∅_𝑘⁵𝐴₁⁴ 𝑡 − 10 3

∅_𝑘^′′

∅_𝑘⁵−2 3

𝑝4 𝑡

∅_𝑘⁴ 𝐴₁³ 𝑡 − 5 2

∅_𝑘³

∅_𝑘⁵ −∅_𝑘^′

∅_𝑘⁵𝑝4 𝑡 𝐴₁^′′ 𝑡 − ( ∅_𝑘⁴

∅_𝑘⁵

−2 3

∅_𝑘^′′

∅_𝑘⁵𝑝4 𝑡 −1 3

𝑝3 𝑡

∅_𝑘⁴ )𝐴1′ 𝑡 − ∅_𝑘⁵ 6∅_𝑘⁵ −∅_𝑘³

6∅_𝑘⁵𝑝4 𝑡 − ∅_𝑘^′

6∅_𝑘⁵𝑝3 𝑡 −𝑝2 𝑡 6∅_𝑘⁴ 𝐴1 𝑡

−𝐴𝑜 6 𝑡

6∅_𝑘⁵ +𝑝4 𝑡

6∅_𝑘⁵ 𝐴_𝑜⁴ 𝑡 +𝑝3 𝑡

6∅_𝑘⁵ 𝐴^′′_𝑜 𝑡 +𝑝2 𝑡

6∅_𝑘⁵ 𝐴^′_𝑜 𝑡 +𝑝1 𝑡

6∅_𝑘⁵ 𝐴_𝑜 𝑡 ] 𝑑𝑡 𝐴_𝑜 𝑡 ,

𝐴6 𝑥 = 𝐴_𝑜 𝑥 [ −5 2

𝐴₅^′′ 𝑡

∅𝑘 − 10∅_𝑘^′

∅_𝑘²𝐴5′ 𝑡 − 15 2

∅_𝑘^′ ²

∅_𝑘³ +10 3

∅_𝑘^′′

∅_𝑘²−𝑝₄ 𝑡 6∅𝑘 𝐴5 𝑡

𝑥

0

−10 3

𝐴₄⁽³⁾ 𝑡

∅_𝑘² − 15∅_𝑘^′

∅_𝑘³𝐴4′′ 𝑡 − 10∅_𝑘^′′

∅_𝑘³+ 15 ∅_𝑘^′ ²

∅_𝑘⁴ −2 3

𝑝₄ 𝑡

∅_𝑘² 𝐴4′ 𝑡 − ( 5 2

∅_𝑘³

∅_𝑘³ +10∅_𝑘^′∅_𝑘^′′

∅_𝑘⁴ +5 2

∅_𝑘^′ ³

∅_𝑘⁵ −∅_𝑘^′

∅_𝑘³𝑝4 𝑡 ) 𝐴₄ 𝑡 − 5 2

𝐴₃⁴ 𝑡

∅_𝑘³ − 10∅_𝑘^′

∅_𝑘⁴𝐴₃³ 𝑡 − (10∅_𝑘^′′

∅_𝑘⁴

+15 2

∅_𝑘^′ ²

∅_𝑘⁵ −𝑝4 𝑡

∅_𝑘³ )𝐴^′′₃ 𝑡 − 5 ∅_𝑘³

∅_𝑘⁴ + 10∅_𝑘^′∅_𝑘^′′

∅_𝑘⁵ − 2∅_𝑘^′

∅_𝑘⁴𝑝₄ 𝑡 𝐴₃^′ 𝑡 − ( ∅_𝑘⁴

∅_𝑘⁴

+5 2

∅_𝑘^′∅_𝑘³

∅_𝑘⁵ +5 3

∅_𝑘^′′ ²

∅_𝑘⁵ −1 2

∅_𝑘^′ ²

∅_𝑘⁵ 𝑝4 𝑡 −2 3

∅_𝑘^′′

∅_𝑘⁴𝑝4 𝑡 −𝑝3 𝑡

6∅_𝑘³ ) 𝐴3 𝑡 −𝐴₂⁵ 𝑡

∅_𝑘⁴

−5 2

∅_𝑘^′

∅_𝑘⁵𝐴₂⁴ 𝑡 − 10 3

∅_𝑘^′′

∅_𝑘⁵−2 3

𝑝4 𝑡

∅_𝑘⁴ 𝐴₂³ 𝑡 − 5 2

∅_𝑘³

∅_𝑘⁵ −∅_𝑘^′

∅_𝑘⁵𝑝₄ 𝑡 𝐴₂^′′ 𝑡 − ( ∅_𝑘⁴

∅_𝑘⁵

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−2 3

∅_𝑘^′′

∅_𝑘⁵𝑝4 𝑡 −1 3

𝑝₃ 𝑡

∅_𝑘⁴ )𝐴′2 𝑡 − ∅_𝑘⁵ 6∅_𝑘⁵ −∅_𝑘³

6∅_𝑘⁵𝑝4 𝑡 − ∅_𝑘^′

6∅_𝑘⁵𝑝3 𝑡 −𝑝₂ 𝑡

6∅_𝑘⁴ 𝐴2 𝑡

−𝐴₁⁶ 𝑡

6∅_𝑘⁵ +𝑝4 𝑡

6∅_𝑘⁵ 𝐴₁⁴ 𝑡 +𝑝3 𝑡

6∅_𝑘⁵ 𝐴1′′ 𝑡 +𝑝2 𝑡

6∅_𝑘⁵ 𝐴1′ 𝑡 +𝑝1 𝑡

6∅_𝑘⁵ 𝐴1 𝑡 ] 𝑑𝑡 𝐴𝑜 𝑡 . Proof:

From [9], he proved the solutions of equation (1.1) for sufficient large 𝜆 which are can be written of the form

𝑦𝑘 𝑥, 𝜆 = 𝑒^{𝜆 ∅}⁰^𝑥 ^𝑘^{𝑡 𝑑𝑡} 𝐴𝑖 𝑥 𝜆^𝑖

6

𝑖=0

+ 𝑂 1

𝜆⁷ ( 2.1)

By derivation (2.1) up to sixth order with respect to x, the following relations are obtained:

𝑦^′ 𝑥 = (∅_𝑘(𝑥) 𝜆) 𝑒^{𝜆 ∅}⁰^𝑥 ^𝑘^{𝑡 𝑑𝑡} 𝐴𝑜 𝑥 + _𝜆_𝑖+1¹ ( 𝐴𝑖+1 𝑥 +^𝐴^𝑖^′_∅^𝑥

𝑘

5𝑖=0 ) + 𝑂 _𝜆¹₇ , ( 2.2) 𝑦^′′ 𝑥 = ∅_𝑘 𝑥 𝜆 ²𝑒^{𝜆 ∅}⁰^𝑥 ^𝑘^{𝑡 𝑑𝑡} [ 𝐴_𝑜 𝑥 +1

𝜆 𝐴₁ 𝑥 + 3𝐴^′𝑂 𝑥

∅_𝑘 + 3∅^′

∅²𝐴_𝑜 𝑥 +

1

𝜆^𝑖+2( 𝐴_𝑖+2 𝑥 + 2^𝐴^𝑖+1^′_∅^𝑥

𝑘

4𝑖=0 +^∅_∅₂^′𝐴_𝑖+1 𝑥 +^𝐴^′^′_∅^𝑖^𝑥

𝑘2 ) + 𝑂 _𝜆¹₇ ] , ( 2.3)

𝑦³ 𝑥 = (∅𝑘(𝑥) 𝜆)³𝑒^{𝜆 ∅}⁰^𝑥 ^𝑘^{𝑡 𝑑𝑡} [ 𝐴𝑜 𝑥 +1

𝜆 𝐴1 𝑥 + 3𝐴^′𝑂 𝑥

∅_𝑘 + 3∅^′

∅²𝐴𝑜 𝑥 + 1

𝜆² 𝐴2 𝑥 + 3𝐴^′₁ 𝑥

∅𝑘 + 3∅^′

∅²𝐴1 𝑥 + 3𝐴_𝑜 𝑥

∅² + 3∅^′

∅³𝐴𝑜′ 𝑥 +∅^{′ ′}

∅³𝐴𝑜 𝑥 + 1

𝜆^𝑖+3( 𝐴𝑖+3 𝑥 + 3𝐴^′_𝑖+2 𝑥

∅𝑘 + 3∅^′

∅²𝐴𝑖+2 𝑥 + 3𝐴_𝑖+1 𝑥

∅² + 3∅^′

∅³𝐴_𝑖+1^′ 𝑥 +∅ ^′′

∅³ 𝐴𝑖+1 𝑥

3

𝑖=0

+^𝐴^𝑖⁽³⁾_∅₃^𝑥 ) + 𝑂 _𝜆¹₇ ], ( 2.4)

𝑦⁴ 𝑥 = (∅_𝑘(𝑥) 𝜆)⁴𝑒^{𝜆 ∅}⁰^𝑥 ^𝑘^{𝑡 𝑑𝑡} [ 𝐴𝑜 𝑥 +1

𝜆 𝐴1 𝑥 + 4𝐴^′₀ 𝑥

∅𝑘 + 6∅^′

∅²𝐴𝑜 𝑥 + 1

𝜆² 𝐴₂ 𝑥 + 4𝐴^′1 𝑥

∅𝑘 + 6∅^′

∅²𝐴₁ 𝑥 + 6𝐴^′′𝑜 𝑥

∅_𝑘² + 12∅^′

∅³𝐴_𝑜^′ 𝑥 + (3(∅^′)²

∅⁴ + 4∅′^′

∅³)𝐴_𝑜 𝑥 + 1

𝜆³ ( 𝐴3 𝑥 + 4𝐴^′₂ 𝑥

∅𝑘 + 6∅^′

∅²𝐴2 𝑥 + 6𝐴^′′₁ 𝑥

∅_𝑘² + 12∅^′

∅³𝐴1′ 𝑥 + (3(∅^′)²

∅⁴ + 4∅^′′

∅³ )𝐴1 𝑥 + 4𝐴⁽³⁾𝑜 𝑥

∅_𝑘³ + 6∅^′

∅⁴𝐴^′′_𝑜 𝑥 + 4∅^′′

∅⁴ 𝐴_𝑜^′ 𝑥 +∅⁽³⁾

∅⁴ 𝐴_𝑜 𝑥 ) + 1 𝜆^𝑖+4

2

𝑖=0

(𝐴_𝑖+4 𝑥 + 4𝐴^′𝑖+3 𝑥

∅_𝑘 + 6∅^′

∅²𝐴𝑖+3 𝑥 + 6𝐴^′′𝑖+2 𝑥

∅_𝑘² + 12∅^′

∅³𝐴𝑖+2′ 𝑥 + (3(∅^′)²

∅⁴ + 4∅^′′

∅³ )𝐴𝑖+2 𝑥 + 4𝐴⁽³⁾𝑖+1 𝑥

∅_𝑘³

+6_∅^∅₄^′ 𝐴_𝑖+1^′′ 𝑥 + 4^∅_∅^′′₄ 𝐴_𝑖+1^′ 𝑥 +^∅⁽³⁾_∅₄ 𝐴𝑖+1 𝑥 +^𝐴^𝑖⁽⁴⁾_∅₄^𝑥+ 𝑂 _𝜆¹₇ ], ( 2.5)

𝑦⁵ 𝑥 = (∅_𝑘(𝑥) 𝜆)⁵𝑒^{𝜆 ∅}⁰^𝑥 ^𝑘^{𝑡 𝑑𝑡} [ 𝐴_𝑜 𝑥 +1

𝜆 𝐴₁ 𝑥 + 5𝐴^′𝑂 𝑥

∅_𝑘 + 10∅^′

∅²𝐴_𝑜 𝑥 +

(5)

794 1

𝜆²( 𝐴2 𝑥 + 5𝐴^′₁ 𝑥

∅𝑘 + 10∅^′

∅²𝐴1 𝑥 + 10𝐴^′′_𝑜 𝑥

∅_𝑘² + 30∅^′

∅³𝐴𝑜′ 𝑥 + (15(∅^′)²

∅⁴ + 10∅ ^′′

∅³ ) 𝐴𝑜 𝑥 ) + 1

𝜆³ ( 𝐴3 𝑥 + 5𝐴^′₂ 𝑥

∅𝑘 + 10∅^′

∅²𝐴2 𝑥 + 10𝐴^′′₁ 𝑥

∅_𝑘² + 30∅^′

∅³𝐴1′ 𝑥 + (15(∅^′)²

∅⁴ +10∅^′′

∅³ )𝐴1 𝑥 + 10𝐴³_𝑜 𝑥

∅_𝑘³ + 30∅^′

∅⁴𝐴𝑜′′ 𝑥 + 20∅^′′

∅⁴ 𝐴𝑜′ 𝑥 + 5∅³

∅⁴ 𝐴𝑜 𝑥 +15(∅^′)²

∅⁵ 𝐴_𝑜^′ 𝑥 + 10∅^′∅^′′

∅⁵ 𝐴_𝑜 𝑥 ) + 1

𝜆⁴ ( 𝐴₄ 𝑥 + 5𝐴^′3 𝑥

∅_𝑘 + 10∅^′

∅²𝐴₃ 𝑥 + 10𝐴^′′2 𝑥

∅_𝑘² +30∅^′

∅³𝐴′2 𝑥 + (15(∅^′)²

∅⁴ + 10∅^′′

∅³ )𝐴2 𝑥 + 10𝐴³1 𝑥

∅_𝑘³ + 30∅^′

∅⁴𝐴1′′ 𝑥 + 20∅^′′

∅⁴ 𝐴1′ 𝑥 +5∅³

∅⁴ 𝐴1 𝑥 + 15(∅^′)²

∅⁵ 𝐴1′ 𝑥 + 10∅^′∅^′′

∅⁵ 𝐴1 𝑥 + 5𝐴⁴_𝑜 𝑥

∅_𝑘⁴ + 10∅^′

∅⁵𝐴_𝑜⁽³⁾ 𝑥 + 10∅^′′

∅⁵ 𝐴𝑜′′ 𝑥 + 5∅³

∅⁵ 𝐴′𝑜 𝑥 ∅⁴

∅⁵ 𝐴𝑜 𝑥 ) + 1 𝜆^𝑖+5

1

𝑖=0

( 𝐴𝑖+5 𝑥 + 5𝐴^′_𝑖+4 𝑥

∅𝑘

+10∅^′

∅²𝐴_𝑖+4 𝑥 + 10𝐴^′′𝑖+3 𝑥

∅_𝑘² + 30∅^′

∅³𝐴^′_𝑖+3 𝑥 + (15(∅^′)²

∅⁴ + 10∅^′′

∅³ )𝐴_𝑖+3 𝑥 +10𝐴³𝑖+2 𝑥

∅_𝑘³ + 30∅^′

∅⁴𝐴_𝑖+2^′′ 𝑥 + 20∅^′′

∅⁴ 𝐴_𝑖+2^′ 𝑥 + 5∅³

∅⁴ 𝐴_𝑖+2 𝑥 + 15(∅^′)²

∅⁵ 𝐴_𝑖+2^′ 𝑥 +10∅^′∅^′′

∅⁵ 𝐴𝑖+2 𝑥 + 5𝐴⁴_𝑖+1 𝑥

∅_𝑘⁴ + 10∅^′

∅⁵𝐴_𝑖+1⁽³⁾ 𝑥 + 30∅^′

∅³𝐴^′₂ 𝑥 + 10∅^′′

∅⁵ 𝐴_𝑖+1^′′ 𝑥 +5∅³

∅⁵ 𝐴_𝑖+1^′ 𝑥 +∅⁴

∅⁵ 𝐴_𝑖+1 𝑥 +𝐴_𝑖⁽⁵⁾ 𝑥

∅⁵ ) + 𝑂 1 𝜆⁷ ],

𝑦⁶ 𝑥 = (∅𝑘(𝑥) 𝜆)⁶𝑒^{𝜆 ∅}⁰^𝑥 ^𝑘^{𝑡 𝑑𝑡} [ 𝐴𝑜 𝑥 +1

𝜆 𝐴1 𝑥 + 6𝐴^′𝑂 𝑥

∅_𝑘 + 15∅^′

∅²𝐴𝑜 𝑥 + 1

𝜆²( 𝐴2 𝑥 + 6𝐴^′1 𝑥

∅𝑘 + 15∅^′

∅²𝐴1 𝑥 + 15𝐴^′′𝑜 𝑥

∅_𝑘² + 60∅^′

∅³𝐴𝑜′ 𝑥 + (45(∅^′)²

∅⁴ + 20∅ ^′′

∅³ ) 𝐴_𝑜 𝑥 ) + 1

𝜆³ ( 𝐴₃ 𝑥 + 6𝐴^′2 𝑥

∅𝑘 + 15∅^′

∅²𝐴₂ 𝑥 + 15𝐴^′′1 𝑥

∅_𝑘² + 60∅^′

∅³𝐴₁^′ 𝑥 + (45(∅^′)²

∅⁴ + 20∅^′′

∅³ )𝐴₁ 𝑥 + 20𝐴³𝑜 𝑥

∅_𝑘³ + 90∅^′

∅⁴𝐴_𝑜^′′ 𝑥 + 60∅^′′

∅⁴ 𝐴_𝑜^′ 𝑥 + 15∅³

∅⁴ 𝐴_𝑜 𝑥 +90(∅^′)²

∅⁵ 𝐴𝑜′ 𝑥 + 60∅^′∅^′′

∅⁵ 𝐴𝑜 𝑥 + 15(∅^′)³

∅⁶ 𝐴𝑜 𝑥 ) + 1

𝜆⁴ ( 𝐴4 𝑥 + 6𝐴^′3 𝑥

∅𝑘

+15∅^′

∅²𝐴3 𝑥 + 15𝐴^′′₂ 𝑥

∅_𝑘² + 60∅^′

∅³𝐴₂^′ 𝑥 + (45(∅^′)²

∅⁴ + 20∅^′′

∅³ )𝐴2 𝑥 + 20𝐴³₁ 𝑥

∅_𝑘³ + 90∅^′

∅⁴𝐴₁^′′ 𝑥 + 60∅^′′

∅⁴ 𝐴₁^′ 𝑥 + 15∅³

∅⁴ 𝐴₁ 𝑥 + 90(∅^′)²

∅⁵ 𝐴₁^′ 𝑥 + 60∅^′∅^′′

∅⁵ 𝐴₁ 𝑥 + 15(∅^′)³

∅⁶ 𝐴₁ 𝑥 + 15𝐴⁴𝑜 𝑥

∅_𝑘⁴ + 60∅^′

∅⁵𝐴_𝑜⁽³⁾ 𝑥 + 60∅^′′

∅⁵ 𝐴_𝑜^′′ 𝑥 + 30∅³

∅⁵ 𝐴_𝑜^′ 𝑥 +6∅⁴

∅⁵ 𝐴_𝑜 𝑥 + 45(∅^′)²

∅⁶ 𝐴_𝑜^′′ 𝑥 + 60∅^′∅^′′

∅⁶ 𝐴_𝑜^′ 𝑥 + 15∅^′∅³

∅⁶ 𝐴_𝑜 𝑥 +10 (∅′^′)²

∅⁶ 𝐴_𝑜 𝑥 ) + 1

𝜆⁵ ( 𝐴₅ 𝑥 + 6𝐴^′4 𝑥

∅_𝑘 + 15∅^′

∅²𝐴₄ 𝑥 + 15𝐴^′′3 𝑥

∅_𝑘² +60∅^′

∅³𝐴′3 𝑥 + (45(∅^′)²

∅⁴ + 20∅^′′

∅³ )𝐴3 𝑥 + 20𝐴³2 𝑥

∅_𝑘³ + 90∅^′

∅⁴𝐴2′′ 𝑥

(6)

795 +60∅^′′

∅⁴ 𝐴₂^′ 𝑥 + 15∅³

∅⁴ 𝐴₂ 𝑥 + 90(∅^′)²

∅⁵ 𝐴₂^′ 𝑥 + 60∅^′∅^′′

∅⁵ 𝐴₂ 𝑥 + 15𝐴⁴1 𝑥

∅_𝑘⁴ +60∅^′

∅⁵𝐴₁⁽³⁾ 𝑥 + 60∅^′′

∅⁵ 𝐴₁^′′ 𝑥 + 30∅³

∅⁵ 𝐴₁^′ 𝑥 + 6∅⁴

∅⁵ 𝐴₁ 𝑥 + 6𝐴⁵𝑜 𝑥

∅_𝑘⁵ +15 (∅′ )³

∅⁶ 𝐴2 𝑥 + 45(∅^′)²

∅⁶ 𝐴1′′ 𝑥 + 60∅^′∅^′′

∅⁶ 𝐴1′ 𝑥 + 15∅^′∅³

∅⁶ 𝐴1 𝑥 +10 (∅′^′)²

∅⁶ 𝐴1 𝑥 + 15∅^′

∅⁶𝐴_𝑜⁽⁴⁾ 𝑥 + 20∅^′′

∅⁶ 𝐴_𝑜⁽³⁾ 𝑥 + 15∅³

∅⁶ 𝐴𝑜′′ 𝑥 + 6∅⁴

∅⁶ 𝐴𝑜′ 𝑥 +∅⁵

∅⁶ 𝐴𝑜 𝑥 ) + 1

𝜆⁶ ( 𝐴6 𝑥 + 6𝐴^′₅ 𝑥

∅𝑘 + 15∅^′

∅²𝐴5 𝑥 + 15𝐴^′′₄ 𝑥

∅_𝑘² + 60∅^′

∅³𝐴4′ 𝑥 + +(45(∅^′)²

∅⁴ + 20∅^′′

∅³ )𝐴₄ 𝑥 + 20𝐴³3 𝑥

∅_𝑘³ + 90∅^′

∅⁴𝐴₃^′′ 𝑥 + 60∅^′′

∅⁴ 𝐴₃^′ 𝑥 + 15∅³

∅⁴ 𝐴3 𝑥 + 90(∅^′)²

∅⁵ 𝐴3′ 𝑥 + 60∅^′∅^′′

∅⁵ 𝐴3 𝑥 + 15𝐴⁴2 𝑥

∅_𝑘⁴ + 60∅^′

∅⁵𝐴₂⁽³⁾ 𝑥 + 60∅^′′

∅⁵ 𝐴2′′ 𝑥 + 30∅³

∅⁵ 𝐴2′ 𝑥 + 6∅⁴

∅⁵ 𝐴2 𝑥 + 6𝐴₁⁽⁵⁾ 𝑥

∅⁵ + 15(∅^′)³

∅⁶ 𝐴3 𝑥 + 45(∅^′)²

∅⁶ 𝐴2′′ 𝑥 + 60∅^′∅^′′

∅⁶ 𝐴2′ 𝑥 + 15∅^′∅³

∅⁶ 𝐴2 𝑥 + 10 (∅^′′^′)²

∅⁶ 𝐴2 𝑥 + 15∅^′

∅⁶𝐴₁⁴ 𝑥 +20∅^′′

∅⁶ 𝐴₁³ 𝑥 + 15∅³

∅⁶ 𝐴1′′ 𝑥 + 6∅⁴

∅⁶ 𝐴1′ 𝑥 +∅⁵

∅⁶ 𝐴1 𝑥 +𝐴_𝑜⁶ 𝑥

∅⁶ )

+𝑂 _𝜆¹₇ ]. ( 2.6)

Now we substitute the resulting equations (2.2) - (2.6) with equation (2.1) in equation (1.1), by long computations, we obtain that

𝐴𝑜 𝑥 = 𝜌 𝑥 ⁻¹²⁵ , 𝐴1 𝑥 = 𝐴_𝑜 𝑥 −5

2 𝐴_𝑜^′′(𝑡)

∅𝑘 − 10∅_𝑘^′

∅_𝑘²𝐴𝑜′ 𝑡 − 15 2

∅_𝑘^′ ²

∅_𝑘³ +10 3

∅_𝑘^′′

∅_𝑘²−𝑝₄ 𝑡

6∅𝑘 𝐴𝑜 𝑡

𝑥

𝑑𝑡 0

𝐴𝑜 𝑡 ,

𝐴2 𝑥 = 𝐴_𝑜 𝑥 [ −5 2

𝐴₁^′′ 𝑡

∅𝑘 − 10∅_𝑘^′

∅_𝑘²𝐴1′ 𝑡 − 15 2

∅_𝑘^′ ²

∅_𝑘³ +10 3

∅_𝑘^′′

∅_𝑘²−𝑝₄ 𝑡 6∅𝑘 𝐴1 𝑡

𝑥

0

−10 3

𝐴_𝑜³ 𝑡

∅_𝑘² − 15∅_𝑘^′

∅_𝑘³𝐴𝑜′′ 𝑡 − 10∅_𝑘^′′

∅_𝑘³+ 15 ∅_𝑘^′ ²

∅_𝑘⁴ −2 3

𝑝₄ 𝑡

∅_𝑘² 𝐴𝑜′ 𝑡 − ( 5 2

(∅_𝑘³

∅_𝑘³ +10∅𝑘′∅𝑘′′

∅_𝑘⁴ +5 2

∅_𝑘^′ ³

∅_𝑘⁵ −∅𝑘′

∅_𝑘³𝑝₄ 𝑡 ) 𝐴_𝑜 𝑡 ] 𝑑𝑡 𝐴_𝑜 𝑡 ,

𝐴3 𝑥 = 𝐴_𝑜 𝑥 [ −5 2

𝐴₂^′′ 𝑡

∅𝑘 − 10∅_𝑘^′

∅_𝑘²𝐴₂^′ 𝑡 − 15 2

∅_𝑘^′ ²

∅_𝑘³ +10 3

∅_𝑘^′′

∅_𝑘²−𝑝₄ 𝑡 6∅𝑘 𝐴2 𝑡

𝑥

0

−10 3

𝐴₁³ 𝑡

∅_𝑘² − 15∅_𝑘^′

∅_𝑘³𝐴1′′ 𝑡 − 10∅_𝑘^′′

∅_𝑘³ + 15 ∅_𝑘^′ ²

∅_𝑘⁴ −2 3

𝑝₄ 𝑡

∅_𝑘² 𝐴1′ 𝑡 − ( 5 2

∅_𝑘³

∅_𝑘³ +10∅_𝑘^′∅_𝑘^′′

∅_𝑘⁴ +5 2

∅_𝑘^′ ³

∅_𝑘⁵ −∅_𝑘^′

∅_𝑘³𝑝4 𝑡 ) 𝐴₁ 𝑡 − 5 2

𝐴_𝑜⁽⁴⁾ 𝑡

∅_𝑘³ − 10∅_𝑘^′

∅_𝑘⁴𝐴_𝑜³ 𝑡