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Differential equation, linear sys. of eq. problem

  1. Oct 22, 2013 #1
    1. The problem statement, all variables and given/known data
    Right so I had a standard systems of diff eqs involving repeated eigenvalues. When I find the last vector I have the equation

    [tex]\begin{pmatrix}
    1 & 1\\
    -1 & -1
    \end{pmatrix}

    \begin{pmatrix}
    u_1\\
    u_1
    \end{pmatrix}

    = \begin{pmatrix}
    1\\
    -1
    \end{pmatrix}[/tex]


    3. The attempt at a solution

    [tex]u_1 + u_2 = 1[/tex]

    Setting [tex]u_2 = 1 - u_1[/tex]

    [tex]\bar{u} = \begin{pmatrix}
    u_1\\
    1-u_1
    \end{pmatrix}[/tex]

    [tex]\bar{u} = \begin{pmatrix}
    u_1\\
    1-u_1
    \end{pmatrix}

    =

    \begin{pmatrix}
    0\\1

    \end{pmatrix}

    +

    \begin{pmatrix}
    u_1\\-u_1

    \end{pmatrix}[/tex]

    Where the latter vector corresponds to the first answer I got from solving x' = Ax with only one eigenvalue, so ignoring that.

    Now my question, the vector (0,1) might as well have been (1,0) which would have solved the system just as well. But this rather radically affects how my general solution looks.
    Would I be correct in saying it doesn't matter if I chose (0,1) or (1,0) because

    [tex]\bar{x} = c_1e^{rt}\begin{pmatrix}
    1\\-1

    \end{pmatrix}

    + c_2 ( te^{rt}

    \begin{pmatrix}
    1\\-1

    \end{pmatrix}
    +
    e^{rt}\begin{pmatrix}
    0\\1

    \end{pmatrix}
    )[/tex]

    Will still span the same "e^(rt) space" depending on c_1 and c_2? Just seems odd that the general solution could be so different. ie the above being equivelant to

    [tex]\bar{x} = c_1e^{rt}\begin{pmatrix}
    1\\-1

    \end{pmatrix}

    + c_2 ( te^{rt}

    \begin{pmatrix}
    1\\-1

    \end{pmatrix}
    +
    e^{rt}\begin{pmatrix}
    1\\0

    \end{pmatrix}
    )[/tex]
    ?
     
    Last edited: Oct 22, 2013
  2. jcsd
  3. Oct 22, 2013 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You mean
    [tex]\begin{pmatrix}u_1\\ u_2\end{pmatrix}[/tex]
    right?

    I have no idea what you are doing here! Yes, it is true that [tex]u_2= 1- u_1[/tex].
    But that because [tex]u_1= 1[/tex] and [tex]u_2= 0[/tex]!

    No, it doesn't. If your vector is (0, 1), that is if [tex]u_1= 0[/tex] and [tex]u_2= 1[/tex], the first equation, [tex]u_1+ u_2= 1[/tex] (which is the only one you have looked at so far) but your second equation,n [tex]-u_1+ u_2= -1[/tex] would NOT be satisfied: -0+ 1= 1, not -1.

    What "general solution"? The equation
    [tex]\begin{pmatrix}1 & 1 \\ -1 & 1 \end{pmatrix}\begin{pmatrix}u_1 \\ u_2\end{pmatrix}= \begin{pmatrix}1 \\ -1\end{pmatrix}[/tex]
    has the unique solution [itex](u_1, u_2)= (1, 0)[/tex]. There is NO "general solution".

     
  4. Oct 22, 2013 #3
    Christ, no matter how many times I look it over longer latex statements get mistakes. Sorry, corrected the matrix now. I'm referring to the general solution of the differential equation I started with which I didn't detail as everything but the u vector was correct in my solution.
     
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