Non-homogeneous Difference Equations

  • Thread starter Daniel323
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  • #1
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Hi everyone, this is my first post but I've been reading through the forum for a quite sometime now. I've recently began trying to do some difference equations, for the most I've been doing good, but just a few things have puzzled me.
I have two questions relating to the particular solutions of non-homogeneous equation.
First one is, if the equation is x(k+1) - x(k) = 8log2(k) and I wanted to get the particular solution:
f(x) = 8long2(k) but what is particular solution? I know for example if f(x) = 8e^k it would be c*e^k.

Secondly if the equation is x(3k) = 2x(k) for 1,3,9,27,..., with x(1) = 1, how do I get this into the standard form? Or is it already in the standard form.
My first step for this one was:
let k = 2^m and x(k) = y(m) hence,
3k = 2^(m+2) and x(3k) = y(m+2)
But have no idea of this is the right approach.

Sorry I know they are stupid questions, but they've really stumped me. Any help would be great. Thanks.
 
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Answers and Replies

  • #2
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Any ideas anyone?
 
  • #3
arildno
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Well, let us exponentiate, and we get:
[tex]\frac{y_{k+1}}{y_{k}}=k^{8}, y_{k}=2^{x_{k}}[/tex]

Furthermore, introduce [tex]z_{k}=\sqrt[8]{y_{k}}[/tex], whereby we get the difference equation:
[tex]z_{k+1}-kz_{k}=0[/tex]
The solution of this is well-known:
[tex]z_{k}=B(k-1)![/tex]
Thus, we get:
[tex]y_{k}=C((k-1)!)^{8}[/tex]
And the particular solution for x becomes:
[tex]x_{k}=D+8\log_{2}((k-1)!)[/tex]

Also note that this includes the homogenous solution, i.e a constant, so this is, as it happens, the general solution.
 
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  • #4
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Ah I see, I'm still a bit confused but that helped a great deal. Thank you arildno, I appreciate it.
 

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