Need help with difference equation

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Hello, Consider the difference equation:

y(n + 2) + y(n + 1) + y(n) = 0, y(1) = y(2) = 1.

Compute y(3), y(4), y(5), y(6), y(7).

I understand how this can be done on paper, but it is not too easy to count. How it can be done in Matlab? My solution:

y(n)=c*z^n

c*z^(n+2)-c*z^(n+1) - c*z^n=0

c*z^n*(z^2-z-1)=0

z1=(1+sqrt(5))/2

z2=(1-sqrt(5))/2

y(n) = c1*((1+sqrt(5))/2)^n + c2 * ((1+sqrt(5))/2)^n

y(1)=c1*((1+sqrt(5))/2) + c1*((1+sqrt(5))/2) =1

y(2)=c1*((1+sqrt(5))/2)^2 + c2 * ((1+sqrt(5))/2)^2=1

Further solving the system can obtain the unknown constants ... but they are too attractive for mental calculations.

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Answer 1

Azzi Abdelmalek 2012-10-27

编辑：Azzi Abdelmalek 2012-10-27

1 个投票

It's easier to solve it in discret time domain

y(1)=1
y(2)=1
for n=3:7
y(n)=-y(n-1)-y(n-2) % equivalent to y(n+2)=-y(n+1)-y(n)
end

Alex 2012-10-27

Thank you very much, really easy))