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Discrete Mathematics Assignment | Problems

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Added on  2020-05-28

Discrete Mathematics Assignment | Problems

   Added on 2020-05-28

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Running head: PROBLEMS ON DISCRETE MATHEMATICSPROBLEMS ON DISCRETE MATHEMATICSName of the StudentName of the UniversityAuthor Note
Discrete Mathematics Assignment | Problems_1
1PROBLEMS ON DISCRETE MATHEMATICS1. (a) The given difference equation is, un=32nun−1+3(2n+1−1) and the initial value, u0=1.Let us assume that un=d∀n belonging to the set of Natural numbers.Then, d = 32nd+3(2n+1−1).Or, d = 3(2n+1−1)1−32nFor the homogeneous part, un−32nun−1=0.Its solution is un=C¿, for some C belonging to the set of Real numbers.Thus the general solution is given by (Atici and Eloe 2015)un=9n2C+3(2n+1−1)1−32n ...(1)Since, u0=1, 1 = C+9−11−3Or, 1 = C+8−2Or, 1 = C−4Thus, C=5.Substituting the value of C in (1), the general solution can be obtained as un=5(9n2)+3(2n+1−1)1−32n.
Discrete Mathematics Assignment | Problems_2
2PROBLEMS ON DISCRETE MATHEMATICS1. (b)The given difference equation is, un=6un−1−5un−2−32n+44 and the initial values are ,u0=9, u1=25.For the homogeneous part, un=6un−1−5un−2Consider un=Awn.In that case,Awn=6Awn−1−5Awn−2Or, 1 = 6w−5w2Or, w2=6w−5Or, w2−6w−5=0Or, (w−3)(w−1)=0Thus, w=5∨1.Therefore, solution of the homogeneous part isA1(w1)n+A2(w2)n= A1(1)n+A2(5)n=A1+5nA2Now we consider the general solution as un=A1+5nA2+f(n)Or, un=A1+5nA2+a+bn+cn2 ; where f(n)=¿a+bn+cn2Now substituting f(n) in the given equation, a+bn+cn2=6[a+b(n−1)+c(n−1)2] - 5[a+b(n−2)+c(n−2)2] - 32n+44Or, a+bn+cn2=6a+6b+6c−5(a−2b+4c)+[b(a−2c)−5(b−4c)]+n2c−32n+44By comparing we get, a=0+4b−14c+44
Discrete Mathematics Assignment | Problems_3

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