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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 $$\vec x\text{ is the vector }\left[ \begin{array}{rrr} x_1 \\\ x_2\\\ x_3 \end{array} \right].$$

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    $\begingroup$ Done. Now it's your turn. $\endgroup$
    – Artem
    May 4, 2013 at 20:21

3 Answers 3

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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.

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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 $\mathbf A$ . )
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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.

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