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 ?