"2nd Order to Matrix: Find Fundamental Matrix

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    2nd order Matrix
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Homework Help Overview

The problem involves converting a second-order linear differential equation, specifically \(x'' + 3x' + 2x = 0\), into a matrix form to find the fundamental matrix. The subject area is differential equations and systems of equations.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss how to express the second-order differential equation in terms of a first-order system. There are attempts to define the derivatives in terms of non-derivatives and express the system as a matrix equation. Some participants question the correctness of their matrix representations and eigenvalues.

Discussion Status

The discussion is ongoing, with participants exploring different representations of the system and questioning their assumptions about the matrix forms and eigenvalues. Some guidance has been offered regarding the structure of the matrix, but there is no explicit consensus on the correct formulation yet.

Contextual Notes

There are indications of confusion regarding the conversion process from the differential equation to the matrix form, as well as discrepancies in the eigenvalues derived from different matrix representations. Participants also mention a lack of clarity on the method used to derive the matrix from the original equation.

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

<br /> \begin{bmatrix}<br /> 0 &amp; 2\\ <br /> -3 &amp; -2 <br /> \end{bmatrix}<br /> <br />

But the eiganvalues for this do not work.
 
freezer said:
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.
freezer said:
<br /> \begin{bmatrix}<br /> 0 &amp; 2\\ <br /> -3 &amp; -2 <br /> \end{bmatrix}<br /> <br />

But the eiganvalues for this do not work.
 
freezer said:
x1' = 0x2 + 2x1
x2' = -3x2 -2x1

<br /> \begin{bmatrix}<br /> 0 &amp; 2\\ <br /> -3 &amp; -2 <br /> \end{bmatrix}<br />

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.
 
Okay,

then

<br /> <br /> \begin{bmatrix}<br /> 2 &amp; 0\\ <br /> -3 &amp; -2 <br /> \end{bmatrix}<br /> <br />

then

<br /> (2-\lambda )(-2-\lambda )<br /> <br />

I need a lamda = 2 and lamda = 1.

I will need

<br /> \begin{bmatrix}<br /> 1 &amp; 0\\ <br /> -3 &amp; 2 <br /> \end{bmatrix}<br />

but I don't see how i get from the original DE to this matrix
 
Recheck your equations.
 
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.
 
freezer said:
x1' = 0x2 + 2x1

freezer said:
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.
 
  • #10
x'' + 3x' + 2x = 0

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

r= -2 r = -1

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

<br /> <br /> \begin{bmatrix}<br /> 0 &amp; 1\\ <br /> -3 &amp; -2<br /> \end{bmatrix}<br /> <br />

Then,

<br /> \begin{bmatrix}<br /> 0 &amp;1 \\ <br /> -3&amp;-2 <br /> \end{bmatrix}\begin{bmatrix}<br /> x_2\\ <br /> x_1<br /> \end{bmatrix}= \begin{bmatrix}<br /> x_1 \\ <br /> -3x_2 -2x1<br /> \end{bmatrix}<br /> <br />

Okay I see now.

So if I had:

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

x1' = x2
x2' = x3
x3' = -x'' - 3x' - 2x
 
  • #11
So to complete the problem

<br /> <br /> \begin{bmatrix}<br /> 2 &amp; 1\\ <br /> -3 &amp;0 <br /> \end{bmatrix}\begin{bmatrix}<br /> a\\ <br /> b<br /> \end{bmatrix}<br /> =<br /> \begin{bmatrix}<br /> 0\\ <br /> 0<br /> <br /> \end{bmatrix}<br /> and<br /> \begin{bmatrix}<br /> 1 &amp;1 \\ <br /> -3&amp; -1<br /> \end{bmatrix}\begin{bmatrix}<br /> a\\ <br /> b<br /> \end{bmatrix}=\begin{bmatrix}<br /> 0\\ <br /> 0<br /> \end{bmatrix}<br /> <br />


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
Okay, found the error.

Should be:

<br /> <br /> \begin{bmatrix}<br /> 0 &amp; 1\\ <br /> -2 &amp;-3 <br /> \end{bmatrix}<br /> <br />

That works out to be

<br /> <br /> C_1\begin{bmatrix}<br /> -1\\ <br /> 1<br /> \end{bmatrix}<br /> e^{-t}+C_2\begin{bmatrix}<br /> 1\\ <br /> -2<br /> \end{bmatrix}e^{-2t}<br /> <br /> <br />

Would you agree?
 
  • #13
freezer said:
Okay, found the error.

Should be:

<br /> <br /> \begin{bmatrix}<br /> 0 &amp; 1\\ <br /> -2 &amp;-3 <br /> \end{bmatrix}<br /> <br />

That works out to be

<br /> <br /> C_1\begin{bmatrix}<br /> -1\\ <br /> 1<br /> \end{bmatrix}<br /> e^{-t}+C_2\begin{bmatrix}<br /> 1\\ <br /> -2<br /> \end{bmatrix}e^{-2t}

Would you agree?

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

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