# 2nd order to matrix

1. Dec 14, 2013

### freezer

1. The problem statement, all variables and given/known data

x'' + 3x' + 2x = 0

Find fundamental matrix

2. Relevant equations

x = x1
x' = x2 = x1'
x'' = x3 = x2' = x1''

3. The attempt at a solution

Not sure how to convert this to a matrix...

The eiganvalues should be 1 and 2

2. Dec 14, 2013

### vela

Staff Emeritus
You want to write the derivatives in terms of non-derivatives. You have, so far,
\begin{align*}
x_1' &= x_2 \\
x_2' &= x''
\end{align*} Use the differential equation to express x'' in terms of x1 and x2. You can then express this system as a matrix equation
$$\begin{pmatrix} x_1' \\ x_2' \end{pmatrix} = A \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$$ with the appropriate A.

3. Dec 14, 2013

### freezer

x1' = 0x2 + 2x1
x2' = -3x2 -2x1

$$\begin{bmatrix} 0 & 2\\ -3 & -2 \end{bmatrix}$$

But the eiganvalues for this do not work.

4. Dec 14, 2013

### Staff: Mentor

Why did you write them this way?
I would write the system like this:
x1' = -2x1 + 0x2
x2' = -2x1 - 3x2

That will make a difference in how your matrix appears.

5. Dec 14, 2013

### vela

Staff Emeritus
If you multiply the system out with your matrix, you get
$$\begin{bmatrix} 0 & 2\\ -3 & -2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 2x_2 \\ -3x_1-2 x_2 \end{bmatrix}$$ which isn't what you want.

6. Dec 14, 2013

### freezer

Okay,

then

$$\begin{bmatrix} 2 & 0\\ -3 & -2 \end{bmatrix}$$

then

$$(2-\lambda )(-2-\lambda )$$

I need a lamda = 2 and lamda = 1.

I will need

$$\begin{bmatrix} 1 & 0\\ -3 & 2 \end{bmatrix}$$

but I dont see how i get from the original DE to this matrix

7. Dec 14, 2013

### vela

Staff Emeritus
Recheck your equations.

8. Dec 14, 2013

### freezer

Is there a proper name for this method that i can lookup a lesson on how to do this? The professor went over this the last few minuets of class and just y = x1, x2 = x1', x3 = x2'.... then just built the matrix mentally but I am not not seeing the process.

9. Dec 14, 2013

### vela

Staff Emeritus
No, there's not a name for this because it's trivial to do. You're really overthinking this.

http://tutorial.math.lamar.edu/Classes/DE/SystemsDE.aspx

Look at the equation you wrote above and look at the second equation your professor wrote. They're supposed to be the same.

10. Dec 14, 2013

### freezer

x'' + 3x' + 2x = 0

r^2 + 3r + 2
(r+2)(r+1)

r= -2 r = -1

x1' = x2
x2' = -3x2 -2x1

$$\begin{bmatrix} 0 & 1\\ -3 & -2 \end{bmatrix}$$

Then,

$$\begin{bmatrix} 0 &1 \\ -3&-2 \end{bmatrix}\begin{bmatrix} x_2\\ x_1 \end{bmatrix}= \begin{bmatrix} x_1 \\ -3x_2 -2x1 \end{bmatrix}$$

Okay I see now.

So if I had:

x''' + x'' + 3x' + 2x = 0

x1' = x2
x2' = x3
x3' = -x'' - 3x' - 2x

11. Dec 14, 2013

### freezer

So to complete the problem

$$\begin{bmatrix} 2 & 1\\ -3 &0 \end{bmatrix}\begin{bmatrix} a\\ b \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix} and \begin{bmatrix} 1 &1 \\ -3& -1 \end{bmatrix}\begin{bmatrix} a\\ b \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix}$$

2a + b = 0
-3a=0

and

a + b = 0
-3a-b = 0

Unfortunately, these don't seem correct so I am guessing i still have an error.

12. Dec 14, 2013

### freezer

Okay, found the error.

Should be:

$$\begin{bmatrix} 0 & 1\\ -2 &-3 \end{bmatrix}$$

That works out to be

$$C_1\begin{bmatrix} -1\\ 1 \end{bmatrix} e^{-t}+C_2\begin{bmatrix} 1\\ -2 \end{bmatrix}e^{-2t}$$

Would you agree?

13. Dec 14, 2013

### Dick

Now that looks reasonable. Looking back that wasn't so hard, was it?

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