Find an explicit conjugacy between their flows. $X' = \begin{bmatrix}-1& 1 \\ 0 & 2\end{bmatrix}X$ and $Y' $... Find an explicit conjugacy between their flows. 
$X' = \begin{bmatrix}-1& 1 \\ 0 & 2\end{bmatrix}X$  and   $Y' = \begin{bmatrix}1& 0 \\ 1 & -2\end{bmatrix}Y$ 
I have no clue how to started ....please help me started  
 A: we know that solution of system  
$x'=Ax$
is following
$u_1*e^{\lambda*t}$
where $u_1$-eigenvector and $\lambda$-eigenvalue,so  you may need to find using eigenvector/eigenvalue solution of both system,general solution

where $\lambda_1, \lambda_2, ... \lambda_n$
are the eigenvalues of 
$\mathbf{A}$;
$\mathbf{u}_1, \mathbf{u}_2, ...  \mathbf{u}_n$ 
are the respective eigenvectors of $\mathbf{A}$ and $c_1, c_2, ....  c_n $
are constants.you may continue from this
this may help you
http://tutorial.math.lamar.edu/Classes/DE/HOHomogeneousDE.aspx
http://www.math.byu.edu/~grant/courses/m634/f99/lec29.pdf
you may  look on this


EDITED:
eigenvalue of the first system is following
A=[-1 1;0 2]

A =

    -1     1
     0     2

>> [V d]=eig(A)

V =

    1.0000    0.3162
         0    0.9487


d =

    -1     0
     0     2

there  diagonal elements of $d$ are eigenvalue and  column element of  V are eigenvector,for system $y$
y=[1 0;0 -2]

y =

     1     0
     0    -2

>> [V1 D1]=eig(y)

V1 =

     0     1
     1     0


D1 =

    -2     0
     0     1

now apply formula please
