Express differential equations as system of first order equations Express the differential equation $$y'''-6y''-y'+6y=0$$
as a system of first order equations i.e. a matrix equation of the form 
$$A(\vec x)'=0$$
where $$\vec x\text{ is the vector }\left[ \begin{array}{rrr} 
x_1  \\\
x_2\\\
x_3
\end{array} \right].$$
 A: Here how you advance, let $y'=z$, then we will have the system
$$ z''-6z'-z+6y=0 \\ y' = z $$
Again, put $ z'=w $ which results in the system
$$ w'-6w-z+6y=0 \\ z'= w \\ y' = z $$
Arranging the above equation gives
$$  y'= z \\ z'=w \\ w'=  6w + z - 6y.  $$
Now, I am sure you know how to write this in a matrix form. 
A: Assume the following  ODE :  $$y^{(n)}+a_1 y^{(n-1)}+a_2 y^{(n-2)}+...+a_n y=0$$
we want to write the ODE as a system of first order ODEs $$\mathbf {X'}=\mathbf A \mathbf X$$ 
where $\mathbf A$ is a $n$ by $n$ matrix.
What we should do now is to choose $n$ new variables , each if them in terms of $y$ and it's derivatives $y^{(k)}$ s. As you can guess there are many (actually infinitely many) imaginable forms for doing this ,and each of these forms is identified by the matrix $\mathbf A$. The simplest form to represented an ODE as a first order system is achieved  by the following assignment of variables: 
first assign to each $k$th derivative of $y$ (i.e. $y^{(k)})$  a variable $X_{k+1}$, so :
$$X_1=y$$
$$X_2=y'$$
$$X_3=y''$$
$$.$$
$$.$$
$$.$$
$$X_n=y^{(n-1)}$$
now we write the equations of system in terms of the introduced variables:
first $n-1$ equations are obvious:
$$X_1'=X_2$$
$$X_2'=X_3$$
$$.$$
$$.$$
$$.$$
$$X_{n-1}'=X_n$$
for the $n$th , use the given DE:
$$ X_n'=-a_1 X_n - a_2 X_{n-1} -...- a_n X_1$$
this way , the $\mathbf A$ matrix  will be :
$$ \begin{pmatrix} 0 & 1 & 0 & .& . & . & 0 \\ 0 & 0 & 1 & 0 & . & . & 0\\. \\.\\0&.&.&.&.&0&1\\-a_n & -a_{n-1} & -a_{n-2} & . & . & . & -a_1\\  \\ \end{pmatrix}$$
In the context of Control Theory , this form of representation is called Phase Variable Canonical Form  .You should remember that this representation is not unique and there are other forms, each identified by the $\mathbf A$ matrix, which may be more beneficial for some special purposes. (.To name some other (widely) used  models:


*

*Input Feed-Forward Canonical Form:


$$\begin{pmatrix} -a_1 & 1 & 0 & 0 &. &.& 0\\ -a_2 & 0&1&0&0&.&0\\-a_3& 0&0&1&0&.&0\\.&.&.\\.&.&.\\-a_{n-1}&0&0&.&.&.&1\\-a_n&0&0&0&.&.&0 \end{pmatrix}$$


*

*Diagonal (or Jordan) Canonical Form (which is a little bit more complicated and results in a diagonal $\mathbf A$ . ) 

A: Heres a hint the 3 substitions will look something like. $y'=x_{1}$ let $x^{'}_{1}=x_{2}$ $x^{'}_{2}=x_{3}$
A Second Hint
After the first sub we have 
$y'=x_{1}$
$x^{''}_{1}-6x^{'}_{1}-x_{1} +6y=0$
last hint after you write out all 3 of these subs you remove their labels and list them as rows of a matrix Multiply it by $<x_{1},x_{2},x_{3}>$ 
you should be able to finish it now.
