I want to solve a simple first-order recurrence relation using the characteristic roots of the associated polynomial equation. I know there are simpler ways of doing it (iteration, telescoping) but I think it is interesting to try solving it using the associated polynomial equation in order to understand generating series.
I also do not want to use symbolic differentiation to reduce my relation to the homogeneous case. I want to explicitly work on the particular solution for the non-homogeneous case.
Given a sequence $\left(u_n\right)=\begin{cases}
u_0=1 \\
u_n=1.5u_{n-1}+1
\end{cases}$
The associated polynomial equation for the homogeneous part $u_n=1.5u_{n-1}$ is $x=1.5$ and we have the root for the homogeneous part.
We should have the general solution in the form $u_n=\left\{u_n^H+u_n^P\right\}$ where $\left\{u_n^H\right\}$ is the homogeneous solution and $\left\{u_n^P\right\}$ is the particular solution.
From there, how can I determine the particular solution?
Best Answer
One approach is to say your work suggests $u_n=k\times 1.5^n+c$
and you know $u_0=1$ and $u_1=2.5$
so substituting gives $1=k+c$ and $2.5 = k\times 1.5 +c$ and thus $k = 3$ and $c=-2$
with the final result $$u_n = 3 \times 1.5^n - 2$$
If you prefer the "particular solution" route, you need to solve $c \times 1.5 +1 = c$ and so $c=-2$ which with $u_0=1$ gives the same result