Question

Third Order ODE

Answer 1

My question is when I look at these solutions. Yeah, these are solutions. Basically I just don't understand how they transform the equation into the matrix form, especially the 3X3 matris part. I do not how to find it.

Answer 2

First of all, notice that the given equation
\[y''' + 2y'+y=0\]

is same as
\[y''' = -2y'-y\]

Answer 3

But that's just one equation with 3 unknowns : \(y,~y',~y''\)

Answer 4

If you had 3 equations, you would be able to write the system as \(y' = Ay\), where A is a "square" matrix,then use matrix methods to solve it.

Answer 5

But all you have is just one equation,
so you can't express the coefficients of the system as a square matrix, yet.

Answer 6

We want 3 equations, so lets cookup two dummy equations in our system :

\[y'=y'\]

and

\[y''=y''\]

Answer 7

Our system would then be
\[y' = y'\tag{1}\]
\[y''=y''\tag{2}\]
\[y'''=-2y'-y\tag{3}\]

Answer 8

Nothing fancy, I have just added two dummy equations so that I get a 3x3 system.

Answer 9

Let me know once you're happy with everything so far

Answer 10

thank you. Now I understand now. It's much easier to understand when you write down equation 1,2 and 3. Now I realise how easy is it.

Answer 11

Its really easy if you're good at linear algebra :)

Let \[Y = \begin{bmatrix}y\\y'\\y''\end{bmatrix} \]

Answer 12

then we get
\[Y' = \begin{bmatrix}y\\y'\\y''\end{bmatrix}' \]

Answer 13

\[y' =\color{blue}{0}y+\color{blue}{1} y'+\color{blue}{0}y''\tag{1}\]
\[y''=\color{blue}{0}y+\color{blue}{0}y'+\color{blue}{1} y''\tag{2}\]
\[y'''=\color{blue}{-1}y\color{blue}{-2}y'+\color{blue}{0}y''\tag{3}\]

Answer 14

In matrix form it is
\[ \begin{bmatrix}y\\y'\\y''\end{bmatrix}' =\begin{bmatrix}\color{blue}{0}&\color{blue}{1}&\color{blue}{0}\\\color{blue}{0}&\color{blue}{0}&\color{blue}{1}\\\color{blue}{-1}&\color{blue}{-2}&\color{blue}{0}'\end{bmatrix}' \begin{bmatrix}y\\y'\\y''\end{bmatrix}\]

Answer 15

\[Y' = \color{blue}{A}Y\]

Answer 16

Cheers. :)