Explain the determine 4.5.2 Еxample BConsider the equationYk+2 – 6yk+1 + 8yk = 2+ 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)c12* + c24*,(4.148)whereC1andC2 are arbitrary constants. The families of the terms in Rp are→ [1],k² → [1, k, k²],3k → [3k).(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(Р)= 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(ЗА — 4B — 2C) + (3B — 8C)k + ЗСК? — D3*= 2+ 3k2 – 5 · 3k.(4.151)136Difference EquationsEquating the coefficients of the linearly independent terms on both sides tozero givesЗА4B — 2С — 2,ЗВ - 8С — 0,ЗС %3 3, D % 5,(4.152)which has the solution44/9,= /4, C = 1, D= 5.D = 5.(4.153)Consequently, the particular solution is,(Р)= 44/9 + 8/3k + k² +5. 3k,(4.154)and the general solution to equation (4.146) isYk = c12* + c24k + 44/9 + 8/3k + k² + 5· 3k.(4.155)

Name: Explain the determine 4.5.2 Еxample BConsider the equationYk+2 – 6yk+1
Uploaded: 2024-03-20T06:47:08.000Z
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Description: Explain the determine 4.5.2 Еxample BConsider the equationYk+2 – 6yk+1 + 8yk = 2+ 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)c12* + c2