The question is about the system of equations

$$\frac{dx}{dt} = y - 1, \frac{dy}{dt} = -xy$$

and I'm trying to make a phase picture by finding the orbits. I'm not sure if orbit is the correct term, but I mean the following: if $\gamma: t \mapsto (x(t), y(t))$ is a solution of the system with initial values $x(t_0) = x_0, y(t_0) = y_0$ for $t \in I$ for some interval $I$, then the orbit is $\gamma[I]$. The solution is given locally by solving $\frac{dy}{dx} = \frac{-xy}{y - 1}$, which gives $y \ln|y| = y_0 \exp(y_0 + \frac{1}{2}(x_0^2 - x^2))$. I just don't see how we can extract the orbit from this.

The stationary points are $(0, 1)$ because that is the solution of $y - 1 = 0$ and $-xy = 0$.

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    $\begingroup$ Can't you just do an implicit plot of the solution with the desired initial condition? Your means of solution of the original DE has essentially removed the parameter $t$, but an orbit the way you've defined it is, I think, just an $(x,y)$ plot. $\endgroup$ – Adrian Keister Feb 8 '18 at 21:28
  • $\begingroup$ @AdrianKeister It is an $(x, y)$ plot, but I want an explicit expression for the orbit. If that is not possible, how would I make a sketch of the orbit? I'm not allowed to use a computer simulation for this class. $\endgroup$ – Pel de Pinda Feb 8 '18 at 21:31
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    $\begingroup$ You could do a slope field from your DE $\displaystyle\frac{dy}{dx}=-\frac{xy}{y-1}.$ Tedious, but it would work. $\endgroup$ – Adrian Keister Feb 8 '18 at 21:36
  • $\begingroup$ @AdrianKeister that's a pity, but thanks. $\endgroup$ – Pel de Pinda Feb 8 '18 at 21:43

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