# How to choose substitution to make the difference equation linear with fixed coefficients?

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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It can't always be done, and I doubt there is a systematic method for when it can, just as there is no sure fire way to make u-substitutions in integrals. However, just as for integrals, if you can do a substitution, experience will often let you guess the right one. – Aaron Oct 15 '12 at 17:26
@Aaron - true, but if I would of asked about some substitution in integrals usually there is some intuition behind it. this is not about the general case, but on this example – Belgi Oct 15 '12 at 17:32
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. – Aaron Oct 15 '12 at 17:51

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.
It is worth adding that you can do similar substitutions for the more general form $ag(k+1,y(k+1)))+bg(k,y(k))=c$. For example, $k+1/y(k+1)+k/y(k)=1$ would use the substitution $x(k)=k/y(k)$. – Aaron Oct 15 '12 at 18:19
@Aaron where $g$ is any fixed function with two variables ? – Belgi Oct 15 '12 at 18:38