Find a solution $\Phi$ of the system \begin{align*} y_1'&=-y_1\\ y_2'&=y_1+y_2\\ \end{align*}
satisfying the initial condition $\Phi(0)=(2,1)$.
Thanks for the help. I found $y_1$ but cannot get $y_2$ to work correctly for $y'_2$.
I got $y_1=2 \Bbb e^{-t}$, but for $y_2$ I kept getting $(1+2t) \Bbb e ^{-t}$ which doesn't work because the derivative is $-y_1 -y_2$ instead of $y1 +y2$.
You have $y_1 = 2 e^{-t}$, which is correct. Then rearranging the equation for $y_2$ gives you \begin{align*} y_2' - y_2 = 2 e^{-t}. \end{align*} Multiply the equation through by the integrating factor $e^{-t}$ and you get \begin{align*} ( y_2 e^{-t} )' = 2 e^{-2t}. \end{align*} Integrate the above expression from $0$ to $t$ and use the initial condition $y_2(0) = 1$ to arrive at \begin{align*} y_2(t) e^{-t} - 1 = \int_0^t 2 e^{-2s} ds = - e^{-2s}|_0^t = 1 - e^{-2t}. \end{align*} Thus, you arrive at (after simplifying) \begin{align*} y_2(t) = 2 e^t - e^{-t}. \end{align*} You can now check to see that $y_1$ and $y_2$ satisfy your given system.