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 ?