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System of 3 Linear DEs in three variables-elimination

  1. Nov 8, 2008 #1
    System of 3 Linear DEs in three variables--elimination

    1. The problem statement, all variables and given/known data
    Solve the given system of linear DEs by systematic elimination.

    Dx = y
    Dy = z
    Dz = x

    What I wanted to do is solve this like you would any other system of three eqns, so I wrote:

    Dx - y + 0z = 0
    0x +Dy - z = 0
    -x +0y +Dz = 0

    and then I attempted to take two of the equations and eliminate one variable and take another two and eliminate the same variable and then combine those two. But this doesn't work because in each equation only two of the variables are present. Any pointers would be greatly appreciated. Thanks.



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 8, 2008 #2

    gabbagabbahey

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    Re: System of 3 Linear DEs in three variables--elimination

    The key phrase here is "systematic elimination"...let's look at your first two equations:

    (1)Dx=y and (2)Dy=z.....how could you go about eliminating 'y' from both of these equations? hint: what is D^2x? :wink:
     
  4. Nov 8, 2008 #3

    HallsofIvy

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    Re: System of 3 Linear DEs in three variables--elimination

    In other words, one variable has already been eliminated from one equation- part of your work has already been done! Just eliminate that variable from the other two equations. For example, if you decided to eliminate z, notice that z does not appear in the first equation. So you only need to eliminate z from equations 2 and 3: Dy= z and Dz= -y. As gabbagabbahey suggested, Differentiate the equation Dy- z= 0 to get "Dz" and replace "Dz" in the last equation.
    Any pointers would be greatly appreciated. Thanks.



    2. Relevant equations



    3. The attempt at a solution[/QUOTE]
     
  5. Nov 9, 2008 #4
    Re: System of 3 Linear DEs in three variables--elimination

    Here's what I tried

    eliminating z from the last two equations:

    D^2y - Dz = 0
    -x + Dz = 0

    = D^2y - x = 0

    Then I tried to add that to the first eqn.

    -y + Dx
    D^3y + -Dx

    (D^3 - 1)y = 0
    y = c1e^t + c2te^t + c3t^2e^t, but I know I must have done something wrong because y in my textbook has trig functions in it.
    I don't think I am understanding what you are saying I should do.
     
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