"2nd Order to Matrix: Find Fundamental Matrix

[freezer]

Homework Statement

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

Find fundamental matrix

Homework Equations

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

The Attempt at a Solution

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

The eiganvalues should be 1 and 2

 

[vela]

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.

 

[freezer]

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

[tex]
\begin{bmatrix}
0 & 2\\
-3 & -2
\end{bmatrix}

[/tex]

But the eiganvalues for this do not work.

 

[Mark44]

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

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.

[tex]
\begin{bmatrix}
0 & 2\\
-3 & -2
\end{bmatrix}

[/tex]

But the eiganvalues for this do not work.

 

[vela]

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

[tex]
\begin{bmatrix}
0 & 2\\
-3 & -2
\end{bmatrix}
[/tex]

But the eigenvalues for this do not work.

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.

 

[freezer]

Okay,

then

[tex]

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

[/tex]

then

[tex]
(2-\lambda )(-2-\lambda )

[/tex]

I need a lamda = 2 and lamda = 1.

I will need

[tex]
\begin{bmatrix}
1 & 0\\
-3 & 2
\end{bmatrix}
[/tex]

but I don't see how i get from the original DE to this matrix

 

[vela]

Recheck your equations.

 

[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.

 

[vela]

x1' = 0x2 + 2x1

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.

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.

 

[freezer]

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

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

r= -2 r = -1

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

[tex]

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

[/tex]

Then,

[tex]
\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}

[/tex]

Okay I see now.

So if I had:

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

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

 

[freezer]

So to complete the problem

[tex]

\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}

[/tex]


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.

 

[freezer]

Okay, found the error.

Should be:

[tex]

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

[/tex]

That works out to be

[tex]

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


[/tex]

Would you agree?

 

[Dick]

Okay, found the error.

Should be:

[tex]

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

[/tex]

That works out to be

[tex]

C_1\begin{bmatrix}
-1\\
1
\end{bmatrix}
e^{-t}+C_2\begin{bmatrix}
1\\
-2
\end{bmatrix}e^{-2t}[/tex]

Would you agree?

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