Finding a particular solution of a recurrence relation

Question

Consider the following recurrence relation $$X_{n+1} - \rho X_n = W_{n+1}, \ \ \ \ \ \ \ \ \ \ \ n\geq 0$$ with initial condition $X_0 = Y$ and $\rho \in (0,1)$. I would like to find a particular solution of this relation, but I'm stuck. I have found the homogeneous solution: $X_n = Y\rho^n$. I thought maybe that the particular solution is in the form $X_n = c\sum_{i =1}^nW_i$ for some constant $c$. But I believe that I cannot solve it since I obtained the following expression if I substitute my guess into the recurrence relation: $$c\sum_{i=1}^{n+1}W_i - \rho\sum_{i=1}^nW_i = cW_{n+1} + c(1-\rho)\sum_{i=1}^nW_i.$$ I believe that there exists no $c$ such that $c = 1$ and $c(1-\rho) = 0$. Anyonelse have an idea?

Answer 1

Here is my proposition to solve that recurrence relation :

Let $ n\in\mathbb{N} $, we have that : $$ \left(\forall k< n\right),\ X_{n-k}-\rho X_{n-k-1}=W_{n-k} $$

Multiplying everything by $ \rho^{k} $, we get : $$ \left(\forall k<n\right),\ \rho^{k}X_{n-k}-\rho^{k+1}X_{n-k-1}=\rho^{k}W_{n-k} $$

Summing up gives the following : $$ \sum_{k=0}^{n-1}{\left(\rho^{k}X_{n-k}-\rho^{k+1}X_{n-k-1}\right)}=\sum_{k=0}^{n-1}{\rho^{k}W_{n-k}} $$

$$ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ X_{n}-\rho^{n}X_{0}=\sum_{k=1}^{n}{\rho^{n-k}W_{k}} $$

Thus, $$ \left(\forall n\in\mathbb{N}\right),\ X_{n}=\rho^{n}Y+\sum_{k=1}^{n}{\rho^{n-k}W_{k}} $$

Answer 2

For an equation like this, I think the most practical solution is to make use of a "summing factor" (the analogous of the integration factor). So, given $$ X_{\,n + 1} - \rho X_{\,n} = W_{\,n + 1} $$ we divide both sides by $\rho ^{n+1}$ to obtain $$ {{X_{\,n + 1} } \over {\rho ^{\,n + 1} }} - {{\rho X_{\,n} } \over {\rho ^{\,n + 1} }} = {{X_{\,n + 1} } \over {\rho ^{\,n + 1} }} - {{X_{\,n} } \over {\rho ^{\,n} }} = {{W_{\,n + 1} } \over {\rho ^{\,n + 1} }} $$ and pass to solve $$ x_{\,n + 1} - x_{\,n} = w_{\,n + 1} $$