We consider:
$$y''' - 2y'' + 2y' - y = 0$$
The real solution to this equation is:
$$y(x) = c_3e^{x} + c_2e^{x/2}sin\left(\frac{\sqrt{3}x}{2}\right) + c_1e^{x/2}cos\left(\frac{\sqrt{3}x}{2}\right)$$
How do we now represent it as a fundamental- system/matrix ?
On
Proceeding from
$$X' = Ax = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & -2 & 2 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$$
we find the eigenvalues $\lambda_1 = 1, \lambda_2 = \frac{1}{2}\left(1 + i\sqrt{3}\right), \lambda_3 = \frac{1}{2}\left(1 - i\sqrt{3}\right)$, which correspond to the following eigenvectors: $$v_1 = \begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix}, v_2 = \begin{pmatrix}\frac{1}{2}\left(-1 - i\sqrt{3}\right)\\ \frac{1}{2}\left(1 - i\sqrt{3}\right)\\ 1\end{pmatrix}, v_3 = \begin{pmatrix}\frac{1}{2}\left(-1 + i\sqrt{3}\right)\\ \frac{1}{2}\left(1 + i\sqrt{3}\right)\\ 1\end{pmatrix}$$
The Real Fundamental System is then:
$$y_1(x) = e^x\begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix} \\y_2(x) = e^{\frac{1}{2}x}\begin{pmatrix}\frac{1}{2}(-\cos(\frac{\sqrt{3}x}{2})+\sqrt{3}\sin(\frac{\sqrt{3}x}{2}))\\ \frac{1}{2}(\cos(\frac{\sqrt{3}x}{2})+\sqrt{3}\sin(\frac{\sqrt{3}x}{2}))\\ 1\end{pmatrix} \\y_3(x) = e^{\frac{1}{2}x}\begin{pmatrix}\frac{1}{2}(-\cos(\frac{\sqrt{3}x}{2})+\sqrt{3}\sin(\frac{\sqrt{3}x}{2}))\\ \frac{1}{2}(\cos(\frac{\sqrt{3}x}{2})+\sqrt{3}\sin(\frac{\sqrt{3}x}{2}))\\ 1\end{pmatrix}$$
Tell me if there's something wrong or another and better representation for the Real Fundamental System.
Write your DEQ as a System of First Order Equations, find eigenvalues / eigenvectors and proceed in the usual way.
We have $$y''' - 2y'' + 2y' - y = 0$$
To write it as a system of first order equations we let $x_1 = y$, so
$$\begin{align} x_1 ' &= y' = x_2 \\ x_2' &= y'' = x_3 \\ x_3' &= y''' = 2y'' - 2 y' + y = 2 x_3 - 2 x_2 + x_1 \end{align}$$
In matrix form, we have
$$X' = Ax = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & -2 & 2 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$$
Can you proceed?