How do they got the solution for this $2\times 2$ matrix? 
For $y'=\frac{1}{x}\begin{pmatrix}0 &1 \\ 2 & -1\end{pmatrix}y$ the solutions are given by $y_1=\begin{pmatrix}x \\ x\end{pmatrix},y_2=\begin{pmatrix}x^{-2} \\ -2x^{-2}\end{pmatrix}$ with $x\in \Bbb R,y\in \Bbb R^2$

I would like to know how they got this solutions.
I tried the following:$$det\frac{1}{x}\begin{pmatrix}0-\lambda &1 \\ 2 & -(1+\lambda)\end{pmatrix}=\frac{1}{x}(\lambda^2+\lambda-2)$$
$\Rightarrow \lambda_{1,2}=-\frac{1}{2}+/-\frac{3}{2}\Rightarrow\lambda_1=-2 , \lambda_2=1$
Than $$ker\begin{pmatrix}-1&1 \\ 2 & -2\end{pmatrix}\Rightarrow v_1=\begin{pmatrix}1 \\ 1\end{pmatrix}\Rightarrow y_1=e^x\begin{pmatrix}1 \\ 1\end{pmatrix}$$
also $$ker\begin{pmatrix}2&1 \\ 2 & 1\end{pmatrix}\Rightarrow v_1=\begin{pmatrix}1 \\ -2\end{pmatrix}\Rightarrow y_2=e^{-2x}\begin{pmatrix}1 \\ -2\end{pmatrix}$$
I guess this way is wrong 
Edit:
$y_1'=\frac{1}{x}y_2 $ and $y_2'=\frac{1}{x}(2y_1-y_2)$

Let $x = e^u$. Then we have 
$$ \def\d#1#2{\frac{d#1}{d#2}}
   \d yu = \d yx \cdot \d xu = y' \cdot x $$
Hence, your equation reads 
$$ \d yu = \begin{pmatrix} 0 & 1 \\ 2 & -1 \end{pmatrix}y $$
Now solve this linear system as usual and resubstitue $e^u$ by $x$, giving the stated solutions.
