Needs a explanation on how it was solved to.

Question

Baljit · Accepted Answer

6.3.2  
b)  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= ,

Needs a explanation on how it was solved to.

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