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System of Differential equations with a singular coefficient matrix, help?

  1. Jul 13, 2004 #1
    I tried using undetermined coefficients to solve this problem, but I know that I am missing something and i cannot find any reference material on this. If you help me, thank you.

    The homogeneous equation for the system is:

    y' = A*y

    where y = [tex] \left[ \begin{array}{c} y_1 \\ y_2 \end{array} \right] [/tex]

    and A = [tex] \left[ \begin{array}{cc} -2 & 1 \\ -1 & 0 \end{array} \right] [/tex]

    I end up with only one eigenvector of course, and I'm trying to use a solution that ends up as C1*V*[tex] e^t[/tex] + c2*V*[tex] t*e^t [/tex] where V is the only eigenvector of A, but that is not a complete solution.

    What am I missing?
  2. jcsd
  3. Jul 13, 2004 #2
    I suppose there is a general way to solve this type of linear constant coefficient differential equations.

    I suppose you mean A is not diagonalisable? But A can always be reduced to a matrix in Jordan form. Do you know what is "matrix exponential"? The "fundamental solution" of the system is given by exp(tA). (The fundamental solution is a matrix whose columns are solutions of the system and linearly independent.) When A is reduced to a matrix in Jordan form, the "matrix exponential" is readily computable. So I think that is the solution you want?

    And I think there is a way to solve this without using Jordan form? your equation reads
    That is
    which is readily solvable. From this you might obtain two independent solutions. Plug them into the other equation to obtain the solution to the system.
    Last edited: Jul 13, 2004
  4. Jul 13, 2004 #3
    I found out what the problem was, I had to create a new vector U instead of just using V.

    Thanks Wong, but i'm taking an advanced math class where they are teaching us how to use linear algebraic methods to solve differential equations.
  5. Jul 14, 2004 #4
    Initial conditions perhaps?
  6. Jul 15, 2004 #5
    initital condition only solve for coefficient of the Y solution. like his c1, c2

    first find ur eigenvalues then find ur eigenvectors and use that egeinvector in the Y solution.
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