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Flow line curve

  1. Jun 5, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the flow line curve [itex] c(t) [/itex] to the vector field [itex] F = (x,-y) [/itex] which passes through the point [itex] (1, 2) [/itex].

    3. The attempt at a solution

    So I let [itex] c(t) = (x(t), y(t)) [/itex].
    So [itex] c'(t) = ( \frac{dx}{dt} , \frac{dy}{dt} ) [/itex].

    Now, [itex] \frac{dx}{dt} = x [/itex] and [itex] \frac{dy}{dt} = -y [/itex].

    So [itex] \frac{dy}{dx} = -\frac{y}{x} [/itex]

    Solving the differential equation, I get

    [itex] ln(y) = -ln(x) + C [/itex]
    [itex] y = e^{-ln(x) + C} [/itex]
    [itex] y = \frac{A}{x}[/itex]
    [itex] y = \frac{2}{x} [/itex] by using the point given.

    This is not the answer given, I am not sure what they want. The answer given is
    [itex] c(t) = ( e^{t}, 2e^{-t} ) [/itex].

    Thanks.
     
    Last edited: Jun 5, 2010
  2. jcsd
  3. Jun 5, 2010 #2

    lanedance

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    you have given an explicit equation y(x).

    The answer given is parametric c(t) = (y(t), x(t))
     
  4. Jun 5, 2010 #3

    lanedance

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    there's an arithmetic mistep as well...
    so you could solve for the parametric form of y(t) and x(t) here rather than the explicit substitution

    use the given point as you initial conditions for each
    the next step isn't quite right either, it should go
    [itex] y = e^{-ln(x) + C} = e^C e^{ln(x^{-1})} = e^C (x^{-1}) [/itex]

     
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