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General solution of a system of diff eq's.

  1. Apr 28, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the general solution of y'=Ay. Your answer must be a real-valued function.
    [tex]A=
    \begin{pmatrix}
    1 & 1\\
    0 & 1\\
    \end{pmatrix}
    [/tex]
    2. Relevant equations


    3. The attempt at a solution

    The first step would be to find the eigenvalues. I forgot the name of the term but if it is "triangular"(?) then the eigenvalues are the numbers along the diagonal (I took the determinant of the matrix and it also gave me λ=1(multiplicity 2))

    So the next step would be to find the eigenvector corresponding to λ=1.

    Plugging 1 for λ yields...

    [tex]
    \begin{pmatrix}
    0 & 1\\
    0 & 0\\
    \end{pmatrix}
    \begin{pmatrix}
    a\\
    b\\
    \end{pmatrix}=\begin{pmatrix}
    0\\
    0\\
    \end{pmatrix}
    [/tex]

    [itex]\vec{v}[/itex]=[tex]
    \begin{pmatrix}
    a\\
    b\\
    \end{pmatrix}
    [/tex]

    This gives me b=0

    So now what?

    I know what I am supposed to do. I need to solve for the vector [itex]\vec{v}[/itex] in A[itex]\vec{v}[/itex]=0, but if b=0, then would [itex]\vec{v}[/itex]=

    [tex]
    \begin{pmatrix}
    0\\
    0\\
    \end{pmatrix}?
    [/tex]

    That would make it a trivial solution, in other words, not the eigenvector.

    This seems like it is the most simple problem. The professor worked another problem almost identical to this one, but he goes so fast I can only either copy the notes or follow his process, not both. In his example, he took b=0 and somehow it became the vector

    [tex]
    \begin{pmatrix}
    1\\
    0\\
    \end{pmatrix}
    [/tex]

    I don't understand why....or maybe I wrote down the notes incorrectly.

    I also know that once I solve for the eigenvector, then I can express the solution as

    x=eλt[itex]\vec{v}[/itex]

    Which I then have to break up into a real solution and an imaginary solution. I don't see how I would even get imaginary solutions since the eigenvalues are real numbers.

    The professor and the book do not teach concepts, or attempt to help us understand what is happening with each of these processes, he just recites the process. Which doesn't give us the tools to figure things out in cases like these.

    Some guidance would be much appreciated.
     
  2. jcsd
  3. Apr 28, 2014 #2

    LCKurtz

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    You have$$
    \begin{bmatrix}
    0 & 1\\
    0 & 0
    \end{bmatrix}
    \begin{bmatrix}
    a\\ b
    \end{bmatrix}
    =\begin{bmatrix}
    b \\ 0
    \end{bmatrix}
    =\begin{bmatrix}
    0 \\ 0
    \end{bmatrix}
    $$This tells you that ##b=0## but ##a## can be anything. Any constant works so he took ##a=1## giving$$\begin{bmatrix}
    a \\
    b
    \end{bmatrix}=\begin{bmatrix}
    1\\
    0
    \end{bmatrix}$$
     
  4. Apr 28, 2014 #3
    Ok, that makes sense. So I must have missed that one sentence, IF he said it.

    So then my eigenvector is

    \begin{bmatrix}
    1\\
    0\\
    \end{bmatrix}

    All that is left is to find the second eigenvector, [itex]\vec{w}[/itex].

    Now I set (λI-A)[itex]\vec{w}[/itex]=[itex]\vec{v}[/itex]

    Solving for [itex]\vec{w}[/itex] gives me the vector

    \begin{bmatrix}
    0\\
    1\\
    \end{bmatrix}

    The solution is then y=c1et[itex]\vec{v}[/itex]+c2et[itex]\vec{w}[/itex]

    BUT....

    There is another example in my notes that says that the general solution is

    y=c1et[itex]\vec{v}[/itex]+c2et([itex]\vec{w}[/itex]+t[itex]\vec{v}[/itex])

    which only happens when I have one eigenvalue (multiplicity 2) and one eigenvector. I don't know when I would have only one eigenvector. Wouldn't I be able to find a second eigenvector, as I have here, when I find one? When does this apply?

    Thanks.
     
    Last edited: Apr 28, 2014
  5. Apr 28, 2014 #4

    LCKurtz

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    I don't think that is correct. I don't follow that about getting a second eigenvector. When you solved the equation you only got one eigenvector, even though the eigenvalue had multiplicity 2. Sometimes that happens. Actually, in the 2x2 case, it always happens except in the trivial case where the matrix A is a constant times the identity matrix.

    It applies when you only have one eigenvector, as you do in this problem. Perhaps you are confusing the condition of having one or two independent eigenvectors with whether you have two linearly independent solutions to the DE system. They aren't the same thing. In this problem you have only one eigenvector but still there are two linearly independent solutions to the DE system.
     
    Last edited: Apr 28, 2014
  6. Apr 29, 2014 #5
    I see.

    Would my [itex]\vec{w}[/itex] then still be the same? I assume it would just be written in the second form that I stated.
     
  7. Apr 29, 2014 #6

    LCKurtz

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    Science Advisor
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    I don't have access to your notes, so I can't really comment on the method you are suggesting. In general for this problem I would expect a second solution of the form$$
    \vec a e^t + \vec b te^t$$which looks like it may be equivalent to what you have written. If you follow through with the methods in your notes you should be OK.

    As an aside, note that your original system can be very easily solved with non matrix methods. You could do that to check your answer when you are finished.
     
  8. Apr 29, 2014 #7
    Great! Thanks again.
     
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