Thread: 1. order Difference equations

Hi

I've just started this class "math methods". The introduction was not very soft. I'm a bit confused about the solutions to 1. order difference equation.

Is this correct?:

The solution to is where a and c are functions of n.


The solution to is where a and c are constants.

Hey Lindskou.

I figure you are having trouble, so I'll start you with a hint.

For the first one you have y(n+1) = a(n)y(n) + c(n). Now going along you have
y(n+2)
= a(n+1)y(n+1) + c(n+1)
= a(n+1)[a(n)y(n) + c(n)] + c(n+1)
= a(n+1)a(n)y(n) + c(n)a(n+1) + c(n+1)

y(n+3)
= a(n+3)y(n+3) + c(n+3)
= a(n+3)[y(n+2)a(n+2) + c(n+2)] + c(n+3)
= a(n+3)a(n+2)a(n+1)a(n)y(n) + c(n)a(n+3)a(n+2)a(n+1) + a(n+3)a(n+2)c(n+1) + a(n+3)c(n+2) + c(n+3)

Now if the a's and c's are constant you got in the above

= a^3y(n) + ca^3 + ca^2 + ac + c
= a^3y(n) + c[a^3 + a^2 + a + 1]

which is approaching that formula.

For back-substitution you should use y(n) (from the series) with y(n+1) (use n = n+1 instead of a normal n) and then see if you can re-arrange the expression so that you get the solution for y(n+1) = a(n)y(n) + c(n) and this is simply expanding the series out on both sides and collecting terms.

Thank you.

I think you wrote the sequence y(n+3) a bit odd though. Isn't it y(n+3) = a(n+2)y(n+2) + c(n+2)

Well, I know the pattern for the solution, - I get the first term now, but the second one is quite complicated I think:

http://latex.codecogs.com/gif.latex?y(n)=\left(\prod_{k=0}^{n-1}a(k)%20\right%20)y_0+\sum_{k=0}^{n-1}a^{k}\left[%20\left(%20\prod_{j=k+1}^{n-1}a(j)\right)c(k)%20\right%20]

Please fix up your latex.

Sorry. I'll try once more.