Find the 3 real solutions to the following system of ODEs and describe the behaviour as $t$ approached infinity. $$\begin{bmatrix} x \\ y \\ z \end{bmatrix}'=\begin{bmatrix} 0 & 1 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix}$$
I solved the system of differential equations and I got this.
$$\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 0 & -1 & 1 \\ 1 & 0 & 1 \\ 1 & -1 & 0 \end{bmatrix}\begin{bmatrix} e^t & 0 & 0 \\ 0 & \cos(t) & \sin(t) \\ 0 & -\sin(t) & \cos(t) \end{bmatrix}\begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix}$$
I know that because of the complex conjugate eigenvalue pair it creates some kind of circular winding motion. Does it spiral out into the x direction?
You obtained : $$\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 0 & -1 & 1 \\ 1 & 0 & 1 \\ 1 & -1 & 0 \end{bmatrix}\begin{bmatrix} e^t & 0 & 0 \\ 0 & cos(t) & sin(t) \\ 0 & -sin(t) & cos(t) \end{bmatrix}\begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} =\begin{bmatrix} -c_2\cos(t)-c_2\sin(t) +c_3\cos(t)-c_3\sin(t) \\ c_1e^t-c_2\sin(t)+c_3\cos(t) \\ c_1e^t-c_2\cos(t)-c_3\sin(t) \end{bmatrix}$$ It seems that this result doesn't satisfy the DE : Bringing it back into the DE shows discrepancies.
Moreover, the starting vector seems to leads to a strange result : Puting $t=0$ into your solution leads to : $$\begin{bmatrix} -c_2 +c_3 \\ c_1+c_3 \\ c_1-c_2 \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$ $$c_1=c_2=c_3=0$$
My own result is : $$\begin{bmatrix} x \\ y \\ z \end{bmatrix} =\begin{bmatrix} -C_2\cos(t)+C_2\sin(t) -C_3\cos(t)-C_3\sin(t) \\ C_1e^t+C_2\sin(t)-C_3\cos(t) \\ C_1e^t-C_2\cos(t)-C_3\sin(t) \end{bmatrix}$$ Again, more convenient starting conditions have to be defined.