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Find the equations of trajectories for the system $\;\;\dfrac{dx}{dt} = xy$, $\;\;\dfrac{dy}{dt} = -5x$.

I know that the equation $\dfrac{dy}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}$ is sometimes used to find equations for trajectories in the phase plane, so can I just say that the equations will all take the form:

$\dfrac{dy}{dx} = \dfrac{-5x}{xy} = \dfrac{-5}{y}\;\;$ for varying values of $y$?

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up vote 2 down vote accepted

Yes, that's right, though I don't know why you say "equations... for varying values of $y$": it's just one differential equation $\dfrac{dy}{dx} = \dfrac{-5}{y}$. Then solve this separable differential equation.

EDIT: Note, however, that we have a line of fixed points along $x=0$. The parts of the curves for $x > 0$ and the parts for $x < 0$ are actually separate trajectories.

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OK, so then we have equations $y = \pm \sqrt{2(-5x+c)}$, correct? –  user1038665 Nov 28 '12 at 23:52
    
Yes, that's right. –  Robert Israel Nov 28 '12 at 23:55

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