show me the steps of determine blue and the inf is here. And that's it Theorem 13.Let {xn} be a solution of Eq.(1). Then thefollowingstatementsaretrue:(i)N>0, the intial conditionsSupposeb <dandforsomeXN-1+1,..., XN-k+1, ..., XN-1, XN Eare valid, then for b + e and d + be, we have theinequality(A+B+C+D) +b(b-e)'(44)(df– be)< Xn < (A+B+C+D)+forallN.(ii)N>0, the intial conditionsSupposeb >dandforsomeXN-1+1,.. XN-k+1, ·…,XN-1, XN E |1,are valid, then for b + e and d + be, we have theinequality(A+B+C+D)+ a s Xm s%(A+B+C+D)+ T(f-be)(45)for all n> N.Proof.First of all, if for some N>0, < XN <1 and b#e,we havebxN-kXN+1= AXN+ BXN-k+Cxn–1+DxN-o +dxN-k- exN-1bxN-k N that theinequality (44) is valid. Hence, the proof of part (i) iscompleted.Similarly, if 1 < XN <, then we can prove part (ii)which is omitted here for convenience. Thus, the proof isnow completed.O The objective of this article is to investigate somequalitative behavior of the solutions of the nonlineardifference equationbxn-kXn+1 = Axn+ Bx–k+Cxn-1+ Dxn-o+dxn-k– exŋ-n= 0,1,2,.....(1)where the coefficients A, B, C, D, b, d, e E (0,0), whilek, 1 and o are positive integers. The initial conditionsX_g,..., X_1,.., X_k, ….., X_1, X are arbitrary positive realnumbers such that k <1< 0. Note that the special casesof Eq.(1) have been studied in [1] when B=C= D=0,and k= 0,1= 1, b is replaced by – b and in [27] whenB=C= D=0, and k= 0, b is replaced by[33] when B = C = D = 0, 1= 0 and in [32] whenA=C= D=0, 1=0, b is replaced by – b.b and in%3|

To get the inequality (48). XN+1≥bdA+B+C+D+bXN−kdXN−k−eXN−1 .......................48…

Theorem 13.Let {xn} be a solution of Eq.(1). Then the following statements are true: (i) N>0, the intial conditions Suppose b < d and for some XN-1+1, XN-k+1,•…•, XN-1, XN E ...) are valid, then for b+ e and ď + be, we have the inequality $(A+B+C+D)+ S Xn S (A+B+C+D)+• b (df–be) (b-e) (44) for all n N. (ii) N>0, the intial conditions Suppose b > d and for some XN-1+1,·…, XN–k+1, •……, XN–1, XN E 6. are valid, then for b + e and d # be, we have the inequality (A+B+C+D)+ s Xns (A+B+C+D)+ (f-be)' (45) for all n> N. Proof.First of all, if for some N>0, < XN S1 and b#e, we have bxN-k XN+1 = AXN+ BXN–k+CxN–1+Dxŋ-o+ dxN-k- exN-1 bxN-k

Theorem 13.Let {xn} be a solution of Eq.(1). Then the following statements are true: (i) N>0, the intial conditions Suppose b < d and for some XN-1+1, XN-k+1,•…•, XN-1, XN E ...) are valid, then for b+ e and ď + be, we have the inequality $(A+B+C+D)+ S Xn S (A+B+C+D)+• b (df–be) (b-e) (44) for all n N. (ii) N>0, the intial conditions Suppose b > d and for some XN-1+1,·…, XN–k+1, •……, XN–1, XN E 6. are valid, then for b + e and d # be, we have the inequality (A+B+C+D)+ s Xns (A+B+C+D)+ (f-be)' (45) for all n> N. Proof.First of all, if for some N>0, < XN S1 and b#e, we have bxN-k XN+1 = AXN+ BXN–k+CxN–1+Dxŋ-o+ dxN-k- exN-1 bxN-k

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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