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Matrices and general solutions

  1. Nov 13, 2006 #1
    I'm stuck on this problem that involves drawing a phase portrait but I'm at a stand still.

    Find the general solution to the following differential equation:
    [1 -12 -14]
    x'= [1 2 -3]*x
    [1 1 -2]

    the eigeinvalues that I found are 1 , +5i, -5i

    the general solution that I found is

    c1e^e[25] + c2[cos5t+5sin5t] + c2[-5cos5t+sin5t] + c3[2] + c3t[1-5i]
    [-7] [cos5t ] [sin5t ] [2] [1 ]
    [6 ] [cos5t ] [sin5t ] [1] [1 ]

    Am I suppose to have imaginary values in my solution?
    I'm supposed to find the solution if x(0)=2i-5j+3k
    I just plug 0 in for all the t's and solve for the constants which is no problem for me. The problem that I'm having is the fact that I have an imaginary value in the solution.

    I'm also suppose to find the phase portrait.
     
  2. jcsd
  3. Nov 14, 2006 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    No, since this problem has only real entries, you should not have any complex numbers in your solution- and I don't see why you should. If your eigenvalues are 1, 5i, and -5i, then your solutions will involve ex, cos(5x), and sin(5x).

    I have trouble understanding what you mean by
    c1e^e[25] + c2[cos5t+5sin5t] + c2[-5cos5t+sin5t] + c3[2] + c3t[1-5i]
    [-7] [cos5t ] [sin5t ] [2] [1 ]
    [6 ] [cos5t ] [sin5t ] [1] [1 ]
    Is that the "vector" solution? Is [-7][cos5t][sin5t][2][1] 5 separate functions or is that a product? Each row should be a single function like your first row.
     
  4. Nov 14, 2006 #3
    It's the vector solution. I just realized that this message board gets rid of the extra spaces. 25, -7, 6 is one column matrix. cos5t+5sin5t, cos5t, cos5t is one column matrix. -5cos55t+sin5t, sin5t, sin5t is one column matrix. 2, 2, 1 is one column matrix. 1-5i, 1, 1 is one column matrix. And that's what I thought...about the imaginary values. How should look in the general solution?
     
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