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Answered 1 days After Oct 21, 2022

Solution

Baljit answered on Oct 23 2022
49 Votes
6.3.2
) yn+ 2yn-1 =0
As we can see given equation is homogeneous difference equation
So put yn= λn
So
yn+ 2yn-1 =λn +2λn-1 =0
· λn-1 (λ +2)=0
λn-1 ≠ 0, So λ +2=0
· λ= -2
Solution of Homogenous Difference Equation is
y n=Cλn
Therefore Solution of Given Equation is
y n=C(-2)n
d) yn -5yn-1 +6yn-2 =0
As we can see given equation is homogeneous difference equation
So put yn= λn
So
yn -5yn-1 +6yn-2 = λn -5λn-1 +6 λn-2 =0
· λn-2 (λ2-5λ+6)=0
λn-2 ≠ 0, So λ2-5λ+6=0
· λ2 -5λ+6=0
· (λ-3)(λ-2)=0
· λ=3,2
Solution of Homogenous Difference Equation is
y n=c1 (λ1)n +c2 (λ2)n
Therefore Solution of Given Equation is
y n=c1 (3)n +c2 (2)n
f) 4yn +8yn-1 +3yn-2 =0
As we can see given equation is homogeneous difference equation
So put yn= λn
So
4yn +8yn-1 +3yn-2 = 4λn +8λn-1 +3λn-2 =0
· λn-2 (4λ2+8λ+3)=0
λn-2 ≠ 0, So 4λ2+8λ+3=0
· 4λ2+8λ+3=0
· 2λ(2λ+3)+(2λ+3)=0
· (2λ+1)(2λ+3)=0
· λ= ,
Solution of Homogenous Difference Equation is
y n=c1 (λ1)n +c2 (λ2)n
Therefore Solution of Given Equation is
y n=c1 ()n +c2...
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