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Background:

In section 5.1 Linear systems in A First Course in Differential Equations, Second Edition by J. David Logan , there is an example, Example 5.4. Here you consider

$ x'=2y\\ y'=x $

and want to find the shape of the orbits the solutions have in the phase plane. In the book, they have simply divided the two equations to get

$\frac{y'}{x'}=\frac{dy/dt}{dx/dt}=\frac{dy}{dx}=\frac{x}{2y}$.

The author also writes that he has used the chain rule: "Along an orbit $x=x(t), y=y(t)$ we also have $y$ as a function of $x$, or $y=y(x)$. Then the chain rule requires:

$\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$ or $\frac{dy/dt}{dx/dt}=\frac{dy}{dx}$.

We can obtain the orbits by solving the differential equation above by separation of variables, as is done in the book. One then obtains the curves

$\frac{1}{2}x^2 -y^2=C$.

This corresponds to a phase plot with a saddle point. The plot is similiar to:

Phase plot illustatiton1

Question:

The curves that crosses the positive $x_1$ axis does not appear to be functions of $x_1$. How can then the above use of the chain rule be justified, where we have used that $x_2=f(x_1)$ or equivalently $y=f(x)$?

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  • $\begingroup$ Is this connected with the Implicit function theorem? $\endgroup$ – FredikLAa Jul 13 '15 at 15:47
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The "trick" here is that for curves that cross the positive $x_1$ axis it's more convenient to think of them as of functions of $x_2$. If you still insist on considering them as functions of $x_1$ you should take only one "branch" of the curve and you'll still have a some kind of problem at the point where branch crosses $x_1$-axis. If you are considering a point where either $x'_1$ or $x'_2$ doesn't vanish (it's a necessary and sufficient condition for existence of inverse for $\mathbb{R} \rightarrow \mathbb{R}$ functions) you have an inverse function $t = \tilde{f}(x_1)$ (or $t = \tilde{f}(x_2)$) and hence $x_2 = \tilde{g}(x_1)$ (or $x_1=\tilde{g}(x_2)$). And here you can again apply the chain rule.

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