2-order difference equation as a system of 1-order difference equations

Question

I've been given the relation:

(1) $Y(t)=C(t)+I(t)+G(t)$

I've been given these asumptions 1 to describe (1):

(2a) $C(t)=bY(t-1)$

(2b) $I(t)=k(C(t)-C(t-1)$

(2c) $G(t)=G_0$

$k>0$ and $0<b<1$

Although (2c) isn't a difference equation it is my initial value, making the 2 order difference equation inhomogenous:

(3) $Y(t)=(bk+b)Y(t-1)-kbY(t-2)+G(0)$

My question is, how do i describe (3) as a system of 1 order difference equations?

I am to represent it as a vector system:

(4) $\begin{bmatrix} Y_1(t+1)\\ Y_2(t+2) \end{bmatrix} = \begin{bmatrix} &&&\\ &&& \end{bmatrix} \begin{bmatrix} Y_1(t)\\ Y_2(t) \end{bmatrix} + \begin{bmatrix} 0\\ G(0) \end{bmatrix}$.

Answer 1

You define $Y_2(t)=Y(t), Y_1(t)=Y(t-1)$. The point is to make it only one time step, so you rewrite $(3)$ as $Y_1(t)=(bk+b)Y_1(t-1)-kbY_2(t-1)+G(0)$

Your matrix equation should be $$\begin{bmatrix} Y_1(t+1)\\ Y_2(t+1) \end{bmatrix} = \begin{bmatrix} &&&\\ &&& \end{bmatrix} \begin{bmatrix} Y_1(t)\\ Y_2(t) \end{bmatrix} + \begin{bmatrix} G(0)\\0 \end{bmatrix}$$ Your empty box will be a $2 \times 2$ matrix that reflects the new version of $(3)$, so $$\begin{bmatrix} Y_1(t+1)\\ Y_2(t+1) \end{bmatrix} = \begin{bmatrix} bk+b&-kb\\ 1&0 \end{bmatrix} \begin{bmatrix} Y_1(t)\\ Y_2(t) \end{bmatrix} + \begin{bmatrix} G(0)\\0 \end{bmatrix}$$