Solve $ \vec{x^{'}} = \begin{bmatrix}1 & 2 \\ 2 & 1\end{bmatrix}\vec{x}$, $ \vec{x} (0) = \begin{bmatrix}x_0 \\ y_0\end{bmatrix}$ Solve 
$$
\vec{x^{'}} = \begin{bmatrix}1 & 2 \\ 2 & 1\end{bmatrix}\vec{x},$$
$$
\vec{x} (0) = \begin{bmatrix}x_0 \\ y_0\end{bmatrix}$$
The question is the encode the solution as a flow $\phi: $${\mathbb R} \times {\mathbb R}^2 \rightarrow {\mathbb R}^2 $  
My approach is 
when I first separate the matrix into two 
$$ A = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix},$$
$$ B = \begin{bmatrix}0 & 2 \\ 2 & 0\end{bmatrix}$$
For B I found that for even matrix such as $B^2, B^4...B^{2n}$ have entries $2^{2n}$ on top left and bottom right and for odd matrix such as $B^1, B^3...B^{2n-1}$ have entries $2^{2n-1}$ on top right and bottom left ...however...I do not know how to represent them...could someone please help with that? 

Use
$$\overrightarrow{x}(t) =e^{Mt}\overrightarrow{x}(0) $$ so that $$ \overrightarrow{x}'(t) =Me^{Mt}\overrightarrow{x}(0)$$ where $M\in M_2({\bf R})$
${\bf Solution}$ :
$$ M =\left( \begin{matrix}1&2\\2&1\end{matrix}\right)$$
Diagonalize $M$ : Eigenvalues $3,\ -1$. Then $$ v=(1,1)^T,\ w=(1,-1)^T$$ so that $$ 
M =P\left( \begin{matrix}3&0\\0&-1\end{matrix}\right) P^{-1},\ P=[v\ w]$$
Hence $$ e^{tM} = P\left( \begin{matrix}e^{3t}&0\\0& e^{-t}\end{matrix}\right) P^{-1}$$
${\bf Reference}$ : 10.3 solution by diagonalization in Advanced Engineering Mathematics - 3rd - (Zill and Cullen)
