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Needs a explanation on how it was solved to.

Needs a explanation on how it was solved to.
Answered 1 days After Oct 21, 2022

Solution

Baljit answered on Oct 23 2022
6.3.2
) yn+ 2yn-1 =0
As we can see given equation is homogeneous difference equation
So put yn= Î»n
So
yn+ 2yn-1 =Î»n +2Î»n-1 =0
Â· Î»n-1 (Î» +2)=0
Î»n-1 â‰  0, So Î» +2=0
Â· Î»= -2
Solution of Homogenous Difference Equation is
y n=CÎ»n
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= Î»n
So
yn -5yn-1 +6yn-2 = Î»n -5Î»n-1 +6 Î»n-2 =0
Â· Î»n-2 (Î»2-5Î»+6)=0
Î»n-2 â‰  0, So Î»2-5Î»+6=0
Â· Î»2 -5Î»+6=0
Â· (Î»-3)(Î»-2)=0
Â· Î»=3,2
Solution of Homogenous Difference Equation is
y n=c1 (Î»1)n +c2 (Î»2)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= Î»n
So
4yn +8yn-1 +3yn-2 = 4Î»n +8Î»n-1 +3Î»n-2 =0
Â· Î»n-2 (4Î»2+8Î»+3)=0
Î»n-2 â‰  0, So 4Î»2+8Î»+3=0
Â· 4Î»2+8Î»+3=0
Â· 2Î»(2Î»+3)+(2Î»+3)=0
Â· (2Î»+1)(2Î»+3)=0
Â· Î»= ,
Solution of Homogenous Difference Equation is
y n=c1 (Î»1)n +c2 (Î»2)n
Therefore Solution of Given Equation is
y n=c1 ()n +c2...
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