Problem
Given an equation $x_t -2.5x_{t-1} + x_{t-2} = u_t$ where $u_t$ is white noise.
Does this equation have a solution of form $ x_t = \dots + \alpha_{-1}y_{t+1} + \alpha_0 u_t + \alpha_{1}y_{t-1} + \dots$ ?
If does, find $\alpha_{-1}, \alpha_{0},\alpha_{1}$
My Ideas
If we solve the characteristic polynomial correspond $ b^t -2.5b^{t-1} + b^{t-2} = 0$ we get that $b = 1/2, 2, 0$. According to some (unfortunately, I don't remember the its name) when we have both $|b|$ both $<1$ and $>1$< out equation does have solution of form $\dots + \alpha_{-1}y_{t+1} + \alpha_0 u_t + \alpha_{1}y_{t-1} + \dots$, however I do not understand how to evaluate $\alpha_{-1}, \alpha_{0},\alpha_{1}$.
The idea with characteristic polynomial is correct, however we need some other approaches to evaluate the coefficients.
Let us use the following operators: $$ L:\quad L\cdot y_t = y_{t-1}, \quad L^2 y_t = y_{t-2},\quad ...$$ $$F: \quad L^{-1} = F$$ Then $$u_t = x_t -2.5x_{t-1} + x_{t-2} = x_t -2.5L\cdot x_t + L^2 x_t =$$ $$= x_t(L-2)(L-\frac{1}{2}) = (x_t - 2Lx_t)(1-\frac{L}{2}) $$ Note, $x_t = Fx_{t-1}$. From that follows $$ -2x_{t-1} = u_t \cdot \frac{1}{1-\frac{F}{2}} \cdot \frac{1}{1-\frac{L}{2}} = $$ $$ = u_t \sum_{i=0}^\infty \bigg(\frac{F}{2}\bigg)^i \sum_{i=0}^\infty \bigg(\frac{L}{2}\bigg)^i$$ Note, $F^m\cdot L^m = 1$ and $F^m\cdot L^{m+k} = L^k$. Then we need to regroup the summands in the resulting sum (after the product) such that $L^0, L^1, L^2$ are left.
$$ = \dots + u_tL^2\sum_{i=1}^\infty \bigg(\frac{1}{4}\bigg)^i + u_tL^1\sum_{i=1}^\infty \bigg(\frac{1}{4}\bigg)^i + u_0/2L + u_tL^0\sum_{i=0}^\infty \bigg(\frac{1}{4}\bigg)^i + \dots = $$ $$ = \dots + \frac{1}{3}u_{t-2} + \frac{2}{3}u_{t-1} + \frac{4}{3}u_{t}+\dots$$ Hence, $$\alpha_0 = - \frac{1}{3},$$ $$\alpha_{-1} = - \frac{2}{3},$$ $$\alpha_1 = - \frac{1}{6}$$