Discrete Mathematics Assignment | Problems

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Running head: PROBLEMS ON DISCRETE MATHEMATICSPROBLEMS ON DISCRETE MATHEMATICSName of the StudentName of the UniversityAuthor Note
1PROBLEMS ON DISCRETE MATHEMATICS1. (a)The given difference equation is,un=32nun1+3(2n+11)and the initial value,u0=1.Let us assume thatun=dnbelonging to the set of Natural numbers.Then, d =32nd+3(2n+11).Or, d =3(2n+11)132nFor the homogeneous part,un32nun1=0.Its solution isun=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+11)132n...(1)Since,u0=1,1 =C+9113Or,1 =C+82Or, 1 =C4Thus,C=5.Substituting the value of C in (1), the general solution can be obtained asun=5(9n2)+3(2n+11)132n.
2PROBLEMS ON DISCRETE MATHEMATICS1. (b)The given difference equation is,un=6un15un232n+44and the initial values are ,u0=9,u1=25.For the homogeneous part,un=6un15un2Considerun=Awn.In that case,Awn=6Awn15Awn2Or, 1 =6w5w2Or,w2=6w5Or,w26w5=0Or,(w3)(w1)=0Thus,w=51.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 asun=A1+5nA2+f(n)Or,un=A1+5nA2+a+bn+cn2; wheref(n)=¿a+bn+cn2Now substitutingf(n)in the given equation,a+bn+cn2=6[a+b(n1)+c(n1)2]- 5[a+b(n2)+c(n2)2] -32n+44Or,a+bn+cn2=6a+6b+6c5(a2b+4c)+[b(a2c)5(b4c)]+n2c32n+44By comparing we get,a=0+4b14c+44
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