solution of system of differential equation Consider the system of ODE in $R^2$ ,
$Y'$ =  AY , ${\bf Y(0)} = \pmatrix{0\cr 1\cr}$ , t>0 
where  A = $\pmatrix{ -1 & 1\cr 0 & -1 \cr}$
${\bf Y(t)}$ = $\pmatrix{y_1(t)\cr y_2(t)\cr}$ 
Then
 $y_1(t)$ and  $y_2(t)$ are monotonically decreasing for t >1. 
I know how to solve linear system but here problem is A
  is not diagonalisable and eign vector is $\pmatrix{1\cr 0\cr}$ for -1 then how to get  $y_1(t)$ and  $y_2(t)$ . if there is another way to find information about  $y_1(t)$ and  $y_2(t)$ without solving. 
 A: From the second row you know that
$$y_2'(t) = -y_2(t) \quad\text{ and }\quad y(0)=1,$$
which is quite immediate. Substitute that solution for $y_2$ in the first equation:
$$y_1'(t)+y_1(t)=y_2(t)=e^{-t}.$$
Finally, you can use the integrating factor $e^t$:
$$\left(e^{t} y_1(t)\right)' = e^t\left(y_1'(t)+y_1(t)\right) = 1.$$
I hope you can finish from there.
A: 
"but here problem is A is not diagonalisable and eign vector is $\pmatrix{1\cr 0\cr}$ for -1 then how to get  $y_1(t)$ and  $y_2(t)$ "

You can proceed that way as well. Here, eigenvector $K=\left(\begin{matrix} 1 \\ 0\end{matrix}\right)$. Take any $\eta$ such that $(A-\lambda I)\eta=K$ as the second eigenvector, say $\eta=\left(\begin{matrix} 0 \\ 1\end{matrix}\right)$
Now since we have equal roots, we have $Y(t)=C_1 K e^{-t}+C_2e^{-t}(tK+\eta)$, apply initial condition and solve to get $Y(t)=\left(\begin{matrix} te^{-t} \\ e^{-t}\end{matrix}\right)$
Consider $y_1=te^{-t}$. $y_1'=0\implies t=1,y_1''(1)<0\implies y_1$ has a maximum at $t=1$, combined with $y_1(0)=0,y_1(t)>0$ for $t>0$, we can make the necessary conclusions. 
