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Describe the phase portrait of the nonlinear system when $x'$ and $y'$ are dependent on only one of the variables. For example, for the system $x'=y^2$, $y'=y^2$, both $x'$ and $y'$ are dependent only on $y$. Then I get that the equilibrium solution is $\begin{bmatrix} x\\0 \end{bmatrix}$ for $x\in \mathbb{R}$.

For the phase portrait, we only have the line $y=0$. When $y<0$, $x'>0, y'<0$. When $y>0$, $x'>0, y'>0$. I am not sure how to draw the $x'$ lines or the $y'$ lines on this graph. Which one corresponds to the line $y=0$? And how do you draw the lines corresponding to the other one?

Housefire
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    Do you mean the nullclines? They are just the line $y=0$, there are no other nullclines. – Ian Nov 14 '22 at 19:47
  • Yes the nullclines, I just know what the phase portrait will look like – Housefire Nov 14 '22 at 19:52
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    Well, you have the nullcline now, and so the remaining question is what happens on each side of it, i.e. if you start at $y<0$ do you approach the nullcline and same for starting at $y>0$. – Ian Nov 14 '22 at 20:03
  • But what about on the nullcline itself? When y=0, we get no information about x' or y' except that they equal zero. Does that mean y=0 has neither source nor sink on it? – Housefire Nov 14 '22 at 20:08
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    Every point on $y=0$ is an equilibrium; to see whether you have a source or a sink, you need to look at $y$ close to zero and see what the behavior is. Ordinarily you would use the linearization for that but in this problem the linearization is useless. – Ian Nov 14 '22 at 20:26
  • Yes I think I drew it correctly – Housefire Nov 16 '22 at 03:52

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