I have
$$x_1(t+1) = (1-m \rho_1)x_1(t) + n\rho_2 x_2(t) + h1$$
$$x_2(t+1) = (1-m \rho_2)x_2(t) + n\rho_1 x_1(t) + h2$$
Suppose $x_1(0)$ and $x_2(0)$ are known. How can I find the analytical form of $x_1(t)$ and $x_2(t)$? I know without the recursion it is
$$ x_1(t) = (1-m\rho_1)^t x_1(0) + \frac{1-(1-m\rho_1)^t}{m\rho_1}h_1 $$
but the recursive form makes it too complicated.
PS
Actually I want to compute the derivative $\frac{\partial R}{\partial \rho_1}$ with $R=\sum_t^T x_1(t)$. Perhaps some terms could be removed from the beginning?
Many thanks!
Define generating functions $X_1(t) = \sum_{t \ge 0} x_1(t) z^t$ and $X_2(t) = \sum_{t \ge 0} x_2(t) z^t$, mutiply your recurrences by $z^t$ and sum over $t \ge 0$. Then recognize the resulting sums: \begin{align} \frac{X_1(z) - x_1(0)}{z} &= (1 - m \rho_1) X_1(z) + n \rho_2 X_2(z) + \frac{h_1}{1 - z} \\ \frac{X_2(z) - x_2(0)}{z} &= (1 - m \rho_2) X_1(z) + n \rho_1 X_2(z) + \frac{h_2}{1 - z} \end{align} Solve this for $X_1(z)$ and $X_2(z)$, express them as partial fractions and use the generalized binomial theorem to get the coefficients: $$ \binom{-n}{k} = \binom{k + n - 1}{n - 1} (-1)^k $$ A computer algebra system, like maxima, is a must if you are prone to stupid mistakes (like I am).