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System of equations

  1. Apr 14, 2009 #1
    1. The problem statement, all variables and given/known data
    solve for x and y:

    tx' = x + y
    ty' = -3x + 5y

    2. Relevant equations



    3. The attempt at a solution


    normally if the t was not in front of the primes then i'd set it up as:
    (D-1)x -y = 0
    (D+3)x -5y=0

    but im not sure what to do when the variable t is there. can somebody help me set this up?
     
  2. jcsd
  3. Apr 14, 2009 #2
    Sorry, just to be clear, are there 5 variables in this problem? t, x', y', x, y, or are some of those constants?
     
  4. Apr 15, 2009 #3
    yes but x' is dx/dt and y' is dy/dt
     
  5. Apr 18, 2009 #4
    nobody knows how to do this? i'll put it in a different form if it makes more sense this way:

    t*(dx/dt)= x + y
    t*(dy/dt) = -3x + 5y

    then solve for y(t) and x(t)
     
  6. Apr 18, 2009 #5
    Would this work?

    dx/(x+y)=dt/t

    and

    dy/(-3x+5y) = dt/t.

    Therefore set

    dx/(x+y)=dy/(-3x+5y)

    and solve for y=y(x) or x=x(y). I can do that by using y=vx. Not sure what would come next, though.
     
  7. Apr 18, 2009 #6
    thanks but i dont think thats what they're looking for.

    in the book it says set t = e^w and it should become a linear system with constant coefficients.
    then i would get:

    x'e^w = x + y
    y'e^w = -3x + 5y

    can i treat e^w as a constant and solve the problem then replace w with ln(t) ?

    i guess i'll try this and see what happens

    edit: ok i figured out how to get the right answer. i dont understand the theory behind it but i checked it with both equations and it works, so im happy
     
    Last edited: Apr 18, 2009
  8. Apr 18, 2009 #7

    Mark44

    Staff: Mentor

    Since t = e^w, you can't treat e^w as a constant if t isn't a constant.
    The idea of this substitution is to get rid of the t factors on the left sides.

    t = e^w ==> dt = e^w * dw
    Your system is
    dx/dt * e^w = x + y
    dy/dt * e^w = -3x + 5y

    or
    dx/(e^w * dw) * e^w = x + y
    dy/(e^w * dw) * e^w = -3x + 5y

    On the left sides, the e^w factors cancel, so you get
    dx/dw = x + y
    dy/dw = -3x + 5y

    Presumably you can solve this system of linear first-order DEs.
     
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