"2nd Order to Matrix: Find Fundamental Matrix

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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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