Problem involving system of differential equations Solve following system of diferential equations$$\begin{cases}
\frac{ds}{dt}=y+z\\  \frac{dy}{dt}=s+z\\ \frac{dz}{dt}=z-s.
\end{cases}.$$
I tried many tehniques without any success. I would appreciate some help with this problem. One of my tries
$$\frac{dz}{dt}=z-s\Rightarrow \frac{e^{-t}dz}{ds}=ze^{-t}-se^{-t}\Rightarrow -\frac{d(e^{-t})}{dt}\frac{dz}{dt}=-z\frac{d(e^{-t})}{dt}+s\frac{d(e^{-t})}{dt}\Rightarrow$$
$$\Rightarrow \frac{d}{dt}\left( \frac{y}{e^t}\right)=-\frac{z}{e^t}\Rightarrow \int d\left(\frac{y}{e^t}\right)=-\int\frac{z}{e^t}dt\Rightarrow y=e^t\left(-\int\frac{z}{e^t}dt+C\right) $$
Inserting this into the first equations doesn't lead to anything pleasant.

Let be $\pmb{x}=\pmatrix{s\\ y\\ z}$ and $\pmb{A}=\pmatrix{0 & 1 & 1\\
1 & 0 & 1\\
-1 & 0 & 1
}$ so that the system of differential equations becomes $\pmb{x}'(t)=\pmb{A}\pmb{x}(t)$.
This differential equation has the following general solution
$$
\pmb{x}(t) = c_1 e^{\lambda_1 t} \pmb{u}_1 + c_2 e^{\lambda_2 t} \pmb{u}_2 + c_3 e^{\lambda_3 t} \pmb{u}_3
$$
where $\lambda_1=1+i,\,\lambda_2=1-i,\,\lambda_3=-1$ and 
$\pmb{u}_1=\pmatrix{-i\\ -i\\ 1},\,\pmb{u}_2=\pmatrix{i\\ i\\ 1},\,\pmb{u}_3=\pmatrix{2\\ -3\\ 1}$
