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.
 A: Consider the vector $\boldsymbol V=\left[\begin{array}{r}s\\y\\z\end{array}\right]$. Then your system of equations is equal to:
$$\boldsymbol V'(t)=\left[\begin{array}{r}0&1&1\\1&0&1\\-1&0&1\end{array}\right]\boldsymbol V.$$
Let's call that matrix $\boldsymbol A$. Suppose that there is a solution $\boldsymbol V=\boldsymbol Ke^{\lambda t}$. Therefore $\boldsymbol K\lambda e^{\lambda t}=\boldsymbol A\boldsymbol Ke^{\lambda t}$. This implies $\boldsymbol {AK}=\lambda \boldsymbol K$. Therefore $\boldsymbol{AK}-\lambda\boldsymbol K = \boldsymbol 0 \Longrightarrow \boldsymbol {AK}-\lambda\boldsymbol {KI}=\boldsymbol 0.$ Then $(\boldsymbol A-\lambda \boldsymbol I)\boldsymbol K= \boldsymbol 0$. This means that to find the solution you need to find eigenvalues $\lambda$ that satisfy the last equation.
This is done by taking $\det (\boldsymbol A-\lambda \boldsymbol I)=0$. Solve this polynomial to find $\lambda$ there is one of them for each variable. Careful! Some of these $\lambda$ might be equal.
Once you find these eigenvalues, solve for $\boldsymbol K$ in $(\boldsymbol A-\lambda \boldsymbol I)\boldsymbol K= \boldsymbol 0$ for each $\lambda$. Then you will find solutions to the differential equations.
This is an easy example:

Now it's your turn to work on that $3\times 3$ matrix. The problem is that you will have to deal with complex numbers. The solutions to your system could be expressed as real solutions. This is done using Euler's formula!!
A: You can use the info that NotStrang provides you and also use The Fundamental Theorem for Linear Systems, that says:
Let $A$ be $n\times{n}$ matriz. Then for a gived $\mathbf{x}_{0} \in \mathbb{R}^{n}$, the initial value problem
$$ \mathbf{\dot{x}} = A\mathbf{x}  \qquad \mathbf{x}(0) = \mathbf{x}_{0} $$
has a unique solution given by
$$\mathbf{x}(t) = e^{At}\mathbf{x}_{0}.$$
A: 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 
$\pmb{u}_1=\pmatrix{-i\\ -i\\ 1},\,\pmb{u}_2=\pmatrix{i\\ i\\ 1},\,\pmb{u}_3=\pmatrix{2\\ -3\\ 1}$
