Hi, I have a question regarding the solution of(adsbygoogle = window.adsbygoogle || []).push({}); lineardifference equations. Provided that the 'RHS' is not too complicated, will the same "method of undetermined coefficients" fordifferentialequations work fordifferenceequations.

For example consider the second order difference equation [tex]x_{n + 2} + 4x_{n + 1} + 4x_n = n[/tex]. The homogeneous solution should be [tex]x_n = A\left( { - 2} \right)^n + Bn\left( { - 2} \right)^n [/tex]. If this was adifferentialequation the trial function would be initally be y = Cx + D. But since something with an x is already in the homogeneous solution we would multiply by x^2 and the trial function would be y = Cx^3 + Dx^2. (I think that's what would be done in the method of undetermined coefficients...I usually don't use that procedure to solve DEs so I'm not entirely sure)

In the case of this difference equation the 'guess' particular solution would initially be x_n = C + Dn. But an 'n' is already in the homogeneous equation, so, generally, in such a situation would I multiply the trial function by some power of n? If so how would I decide which power of n to multiply the trial function by? Any help would be great thanks.

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# Homework Help: Difference equations - just a little clarification wanted

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