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Eigenvalues of a 5x5 Matrix, continued

  1. May 18, 2010 #1
    1. The problem statement, all variables and given/known data
    The https://www.physicsforums.com/showthread.php?t=403476" was to determine the eigenvalues of the following matrix. View attachment 25800

    The problem of interest deals with actually finding a solution to the system above without the use of matrix methods. Image (4).jpg

    2. Relevant equations
    The template was to use the following system to solve this problem.
    [tex]X\prime = AX =>\begin{bmatrix}x_1\prime \\x_2\prime \\x_3\prime \\x_4\prime \\x_5\prime\end{bmatrix}=\begin{bmatrix}2 & 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 0 & 2\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{bmatrix}[/tex]

    3. The attempt at a solution

    So multiplying the matrix yields the following.

    [tex]x_1\prime = 2x_1 + x_2[/tex]

    [tex]x_2\prime = 2x_2[/tex]

    [tex]x_3\prime = 2x_3[/tex]

    [tex]x_4\prime = 2x_4 +x_5[/tex]

    [tex]x_5\prime = 2x_5[/tex]

    It was suggested that [tex]x_2[/tex] and [tex]x_5[/tex] be eliminated to solve this system. Any pointers on how to best do that? Can you separate the variables? For example, [tex]x_5\prime = 2x_5[/tex] is separated to

    [tex]\int dy = \int 2x_5 dx_5 [/tex]

    and becomes [tex]y = x_5^2[/tex]
     
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. May 18, 2010 #2

    Mark44

    Staff: Mentor

    No it doesn't. The differential equation you solved here does not involve y. It involved a dependent variable, x5, and an independent variable that is not shown, that I will call t.

    There is a big difference between these two differential equations:
    y' = 2x (solution y = x2 + C)
    x' = 2x (solution x = Ae2t)
     
    Last edited by a moderator: Apr 25, 2017
  4. May 18, 2010 #3

    Landau

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    @Mark: why would you quote the whole message, especially if you're the first who replies?

    @gabriels-horn: I have no idea what the exercise is. What is "a solution to the matrix"?
     
  5. May 18, 2010 #4
    I rephrased to "a solution to the system" instead. I noticed the attachment was not working but I uploaded it again and it seems to be working fine.
     
  6. May 18, 2010 #5

    Mark44

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    This is the recommended policy on this forum. Occasionally a poster will go back and edit the initial post or even delete its content. Quoting the whole message ensures that there is a record of the initial post.
     
  7. May 18, 2010 #6
    Agreed. I mixed up the variables before and changed y to t for convention. So separating the equation [tex]\frac{dx_5}{dt} = 2x_5[/tex] yields

    [tex]\int 2dt=\int \frac{dx_5}{x_5}[/tex]

    which simplifies to [tex]x_5=e^2^t[/tex]

    My question is how to proceed from this point to solve the system.
     
    Last edited: May 18, 2010
  8. May 18, 2010 #7

    Mark44

    Staff: Mentor

    It actually becomes x5 = Ae2t. Do the same for x2, and then you'll have a system of three DEs in three unknowns.
     
  9. May 18, 2010 #8
    Indeed, although don't forget the arbitrary constant. Your solution might be a class of functions (which would constitute the general solution). You should likewise solve for all the variables that are in a differential equation by themselves and then substitute.
     
  10. May 18, 2010 #9
    Solving for [tex]x_2[/tex] and [tex]x_5[/tex] and plugging them in yields the new system

    [tex]x_1\prime = 2x_1 + C_1e^2^t[/tex]

    [tex]x_3\prime = 2x_3[/tex]

    [tex]x_4\prime = 2x_4 +C_2e^2^t[/tex]

    I changed A to C since A has been used already. Any tips on how to proceed from here without using matrix methods as the problem says?
     
    Last edited: May 18, 2010
  11. May 18, 2010 #10

    Mark44

    Staff: Mentor

    The constants should be different, say C1 and C2. In the remaining differential equations, two are nonhomogeneous and one is homogeneous. Each equation involves only one dependent variable, so they should all be fairly easy to solve.
     
  12. May 18, 2010 #11
    Solving the system yields the following solutions,

    [tex]x_1 = c_1te^2^t[/tex]

    [tex]x_3 = c_2e^2^t[/tex]

    [tex]x_4 = c_3te^2^t[/tex]

    which is starting to look like the general solution that I need in matrix form, i.e.
    [tex]X_G = c_1\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{bmatrix}te^2^t+c_2\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{bmatrix}2e^2^t+c_3\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{bmatrix}te^2^t[/tex]

    Does this look right? By using this method, how do you fill the entries for the matrices?
     
    Last edited: May 18, 2010
  13. May 18, 2010 #12

    Mark44

    Staff: Mentor

    The LaTeX above is screwed up.

    Don't forget to put in x2 and x5. You can always check you solution by verifying that <x1', x2', x3', x4', x5'>T = A<x1, x2, x3, x4, x5>T.
     
  14. May 18, 2010 #13
    Corrected as above.
     
  15. May 18, 2010 #14

    vela

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    Your solutions for x1 and x4 aren't complete. Those are just the particular solutions; they're missing the homogeneous part of the general solution.
     
  16. May 18, 2010 #15

    Mark44

    Staff: Mentor

    These (above) are not right. Go back to post #9 and solve those three DEs. Also, you need to include x_2 and x_5.
    This isn't right either. There should be no x on the right side. After all, you're trying to show what x (the vector) is in terms of the independent variable t.
     
  17. May 18, 2010 #16
    I solved [tex]x_1\prime = 2x_1 + C_1e^2^t[/tex] by rearranging to

    [tex]x_1\prime - 2x_1 = C_1e^2^t[/tex]

    Finding [tex]p(t) = -2[/tex] and [tex]h(t) = e^2^t[/tex] allowed me to solve for [tex]\mu(t)[/tex], which gave the above solution. I tacked on the constant at the end to appear more like the usual solutions do for systems of linear ODE's.

    Sorry for the confusion as to putting x's in the general solution above, they were meant to be constant entries and in no way related to the x values that I solved for earlier.
     
    Last edited: May 18, 2010
  18. May 18, 2010 #17

    vela

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    Well, you're making mistakes somewhere because the solutions you came up with aren't complete solutions. It sounds like you're using the integrating factor approach, which if you apply it correctly, should yield the complete solution. Show us your calculations.
     
  19. May 18, 2010 #18

    Mark44

    Staff: Mentor

    But if x1' - 2x1 = c1e2t, the general solution is NOT x1(t) = Cte2t, which is what you have. There is another term. The same is true for x4.
     
  20. May 18, 2010 #19
    You're right, I ignored the second term. Here is the work,

    [tex]\mu(t) = e^\int^p^(^t^)^d^t \to e^-^2^t[/tex] which gets put into the integrating factor equation, yielding

    [tex]\mu(t)=\frac{\int e^-^2^t e^2^tdt+ C}{e^-^2^t}[/tex] which simplifies to

    [tex]\mu(t)=te^2^t + Ce^2^t[/tex]
     
    Last edited: May 18, 2010
  21. May 18, 2010 #20

    Mark44

    Staff: Mentor

    So, what do you have now for x_1(t), x_2(t), ... , x_5(t)?
     
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