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.

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!!
