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Finding a solution at a point

  1. Nov 4, 2006 #1
    Give the general solution of the set of equations below:


    Which I found to be:

    Here's where I'm stuck. They want me to find the solution at t=0, (x,y,z)=(2,3,-1)

    Which the professor hasn't told us how to do this.
  2. jcsd
  3. Nov 4, 2006 #2


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    I'm sure you know how to do this. If t=0, what is x(0)? There are two ways to express this... one is from the equation for x you solved for, the other is from the point you were given. Since t=0, all those nasty exponentials go away, and you have a system of three variables with constant coefficients (which is remarkably simpler than the initial set of differential equations you solved)
  4. Nov 4, 2006 #3
    ok, so what I understand from this is that I plug t=0 into all the t's of the general solutions.


    And then do I set these equal to (x,y,z)=(2,3,-1). What is the solutions form suppose to look like?
  5. Nov 4, 2006 #4


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    Let's call [tex]C_1 C_2[/tex] and [tex]C_3[/tex] A, B, and C to make it easier.

    You know
    x(0)=3A - 3B - C=2
    y(0)=6B = 3
    z(0)=2A + 8B + 3C=-1

    So you should be able to solve for A, B and C as numbers. For example, if 6B=3, B=2. Now you know 3A-6-C = 2, and 2A + 16 + 3C = -1
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