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I am going over some lecture notes and there is the following exercise:

Solve $(k+1)^{2}y(k+1)-k^{2}y(k)=1$ with the initial condition $y(1)=0$

where $k$ it for the time, hence not constant.

The solution defines $z(k):=k^{2}y(k)$ and this gives the linear with fixed coefficients equation: $z(k+1)-z(k)=1$ with $z(1)=0$

My question is this: How do I know how to choose $z(k)$ s.t I will get a linear equation with fixed coefficients ? is there some calculation that may lead me to such $z(k)$ or is it just a guess ?

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    The Intuition is trying to rearrange things to get things into a form resembling Andre's answer. If it can be done, it is likely (but perhaps not always) entirely straight forward manner. The big trick is just knowing to try.2012-10-15

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Well, precisely the same substitution would be made for $af(k+1)y(k+1)+bf(k)y(k)=c,$ where $f(j)$ is any fixed function which is nowhere $0$, and $a$, $b$, $c$ are constants.

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    @belgi: yes, that is correct.2012-10-15