Explain the determine red and all theorem is there Case INone of the roots in the set {ri} occurs in the set {s;}.For this case, equation (4.135) is the general solution of the tth-orderequationg(E)yk = 0.(4.136)The functions v", i = 1,2, ..., t, are those that appear in Rk plus all thoseof the (finite) families of the individual terms that compose R.134Difference EquationsCase IISome of the roots in the set {ri} occur in the set {s;}.In this situation, the set {ri}+{s;}ity than the two individual sets of roots. To proceed, determine the generalsolution of equation (4.131a), drop all the functions y that appear in y,and use the remaining functions to find the proper form for the particularsolutions.now contains roots of higher multiplic-This procedure for obtaining particular solutions to the inhomogeneousequation (4.125) can be summarized as follows:(i) Construct the family of Rk.(ii) If the family contains no terms of the homogeneous solution, then writethe particular solution yof that family. Determine the constants of combination such that theinhomogeneous difference equation is identically satisfied.(P)as a linear combination of the members(iii) If the family contains terms of the homogeneous solution, then multiplyeach member of the family by the smallest integral power of k for whichall such terms are removed. The particular solution y can then bewritten as a linear combination of the members of this modified family.Again, determine the constants of combination such that the inhomoge-neous difference equation is identically satisfied.The following examples will illustrate the use of the method of undeter-mined coefficients.First, assume that g(E) is a nullifying operator of R on the right-handside of equation (4.125). Applying g(E) to both sides of equation (4.125) givesg(E)f(E)yk = 0,(4.131a)%3DwboreYk+na1Yk+n–1 + + anyk =Rk,an #0,(4.125) 4.5.2 Example BConsider the equationYk+2 – 6yk+1 + 8yk2+ 3k2 – 5. 3k.(4.146)The characteristic equation isp2 – 6r + 8 = (r – 2)(r – 4) = 0,(4.147)which leads to the following solution of the homogeneous equation:(H)Y = c12* + c24*,(4.148)where c and c2 are arbitrary constants. The families of the terms in R are- [1],k? - [1, k, k?],3k – [3*].(4.149)The combined family is [1, k, k2, 3k] and contains no members that occur inthe homogeneous solution. Therefore, the particular solution takes the form(P)y = A+ Bk + Ck² + D3k,(4.150)where A, B, C, and D are constants to be determined. Substitution of equation(4.150) into (4.146) and simplifying the resulting expression gives(3A – 4B – 2C) + (3B – 8C')k + 3Ck? – D3k = 2 + 3k2 – 5 · 3k.(4.151)

Name: Explain the determine red and all theorem is there Case INone of the r
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Description: Explain the determine red and all theorem is there Case INone of the roots in the set {ri} occurs in the set {s;}.For this case, equation (4.135) is the general solution of the tth-orderequationg(E)yk = 0.(4.136)The functions v", i = 1,2, ..., t, are

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Explain the determine red and all theorem is there Case INone of the roots in the set {ri} occurs in the set {s;}.For this case, equation (4.135) is the general solution of the tth-orderequationg(E)yk = 0.(4.136)The functions v&#34;, i = 1,2, ..., t, are those that appear in Rk plus all thoseof the (finite) families of the individual terms that compose R.134Difference EquationsCase IISome of the roots in the set {ri} occur in the set {s;}.In this situation, the set {ri}+{s;}ity than the two individual sets of roots. To proceed, determine the generalsolution of equation (4.131a), drop all the functions y that appear in y,and use the remaining functions to find the proper form for the particularsolutions.now contains roots of higher multiplic-This procedure for obtaining particular solutions to the inhomogeneousequation (4.125) can be summarized as follows:(i) Construct the family of Rk.(ii) If the family contains no terms of the homogeneous solution, then writethe particular solution yof that family. Determine the constants of combination such that theinhomogeneous difference equation is identically satisfied.(P)as a linear combination of the members(iii) If the family contains terms of the homogeneous solution, then multiplyeach member of the family by the smallest integral power of k for whichall such terms are removed. The particular solution y can then bewritten as a linear combination of the members of this modified family.Again, determine the constants of combination such that the inhomoge-neous difference equation is identically satisfied.The following examples will illustrate the use of the method of undeter-mined coefficients.First, assume that g(E) is a nullifying operator of R on the right-handside of equation (4.125). Applying g(E) to both sides of equation (4.125) givesg(E)f(E)yk = 0,(4.131a)%3DwboreYk+na1Yk+n–1 + + anyk =Rk,an #0,(4.125) 4.5.2 Example BConsider the equationYk+2 – 6yk+1 + 8yk2+ 3k2 – 5. 3k.(4.146)The characteristic equation isp2 – 6r + 8 = (r – 2)(r – 4) = 0,(4.147)which leads to the following solution of the homogeneous equation:(H)Y = c12* + c24*,(4.148)where c and c2 are arbitrary constants. The families of the terms in R are- [1],k? - [1, k, k?],3k – [3*].(4.149)The combined family is [1, k, k2, 3k] and contains no members that occur inthe homogeneous solution. Therefore, the particular solution takes the form(P)y = A+ Bk + Ck² + D3k,(4.150)where A, B, C, and D are constants to be determined. Substitution of equation(4.150) into (4.146) and simplifying the resulting expression gives(3A – 4B – 2C) + (3B – 8C')k + 3Ck? – D3k = 2 + 3k2 – 5 · 3k.(4.151)

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