1
$\begingroup$

I am aiming to explicitly write the matrix-vector representation of this system: $$\begin{aligned}y'_1 &= 5y_2 - y_1 + y_3\\ y_2' &= 3y_1 - y_2 + t^2\\ y_3' &= y_3 - ty_2\end{aligned}$$

This is what I have so far: $\left[\begin{matrix}y_1'\\ y_2'\\y_3'\end{matrix}\right]=\left[\begin{matrix}-1 & \;\;5 &\; 1\\ \;\;\;3 & -1 & \;0 \\ \;\;\;0 & \;\;? &\; 1 \end{matrix}\right]\cdot \left[\begin{matrix}0\\t^2\\?\end{matrix}\right]$

Just not sure how to attack $-ty_2.$
Any help would be appreciated, thanks guys.

$\endgroup$
3
  • $\begingroup$ is it t2 as $t^2$? $\endgroup$
    – janmarqz
    Commented Oct 17, 2015 at 1:46
  • $\begingroup$ yes t^2 is the value $\endgroup$
    – Simon
    Commented Oct 17, 2015 at 3:11
  • 1
    $\begingroup$ Thank you, much appreciated $\endgroup$
    – Simon
    Commented Oct 17, 2015 at 7:59

2 Answers 2

2
$\begingroup$

Just set $$ \left(\begin{array}{c} y'_1\\ y'_2\\ y'_3 \end{array} \right)= \left(\begin{array}{ccc} -1&5&1\\ 3&-1&0\\ 0&-t&1 \end{array} \right) \left(\begin{array}{c} y_1\\ y_2\\ y_3 \end{array} \right)+ \left(\begin{array}{c} 0\\ t^2\\ 0 \end{array} \right). $$

$\endgroup$
3
  • $\begingroup$ If you dont mind me asking do you use software to insert formulas in the forum? $\endgroup$
    – Simon
    Commented Oct 17, 2015 at 8:39
  • $\begingroup$ all is here, and you can visit meta.math.stackexchange.com/questions/5020/… , but also don't be afraid to press the edit button and see how anyone can write with $\LaTeX$ $\endgroup$
    – janmarqz
    Commented Oct 17, 2015 at 11:40
  • $\begingroup$ @Simon - a quick way to display a vector is \pmatrix{x \\ y \\ z} which renders to $$\pmatrix{x \\ y \\z}$$ In all honesty, find posts with math, right click on them and select show math as TeX. This is the best way to learn. $\endgroup$ Commented Dec 14, 2018 at 21:27
1
$\begingroup$

You have the following system of odes:

$\begin{align*} y_1'&= -y_1+5y_2+y_3\\ y_2'&=3y_1-y_2+t^2\\ y_3'&=y_3-ty_2 \end{align*}$

Then make $u=\begin{bmatrix} y_1&y_2&y_3 \end{bmatrix}^T$ and you get

$$u'= \begin{bmatrix}-1&5&1\\ 3&-1&0\\ 0&-t&1 \end{bmatrix}u+ \begin{bmatrix}0\\t^2\\0 \end{bmatrix}$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .