Given the 4th order linear ODE
[tex]u^{(4)} - u = 0[/tex]
we substitute
[tex]x_1 = u[/tex]
[tex]x_2 = {x_1}' = u'[/tex]
[tex]x_3 = {x_2}' = {x_1}'' = u''[/tex]
[tex]x_4 = {x_3}' = {x_2}'' = {x_1}''' = u'''[/tex]
Then the given equation transforms to
[tex]{x_4}' - x_1 = 0[/tex]
but we also need to relate this to the other derivative substitutions. This gives the system of differential equations
[tex]\begin{cases} {x_1}' = x_2 \\ {x_2}' = x_3 \\ {x_3}' = x_4 \\ {x_4}' = x_1 \end{cases}[/tex]
In matrix form,
[tex]\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}' = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}[/tex]
