# Showing equivalence of plane autonomous system

Given plane autonomous system $\frac{dy}{dt} = Y(x,y), \frac{dx}{dt}=X(x,y)$

From this I can get a first order ODE $\frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx}$, and we are given that $x(t)$ is $C^1$ and nonzero so we can say $\frac{dt}{dx} = \frac{1}{\frac{dx}{dt}}$. So we have $\frac{dy}{dx} = \frac{Y(x,y)}{X(x,y)}$.

Now the book I am reading says that these two ODEs are equivalent. I've shown one direction, but how do I get the other direction?

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Did you use equality $=$? If so, by the definition and properties of equality it is enough. When we use only definitions we are doing both directions at once because definitions are if and only if statements. – glebovg Nov 6 '12 at 3:41
I don't think so... I'm actually thinking they aren't really equivalent now. How do you show it? – StuartHa Nov 6 '12 at 3:49
It is just the relationship between ODE and its phase portrait, notice that the change of time scale doesn't matter in solving the ODE. – Ziqian Xie Nov 6 '12 at 3:53
After you solve the second ODE, substitute the solution into the first ODE, solve it and you can get the answer. – Ziqian Xie Nov 6 '12 at 4:31

Consider the following system: $$\begin{cases} \dot x=X(x,y)\\ \dot y=Y(x,y) \end{cases}$$ and the first order differential equation $$\frac{dy}{dx}=\frac{Y(x,y)}{X(x,y)}.\$$ These are equivalent in a neighborhood of a point $(x_0,y_0)$ if $X(x_0,y_0)\neq 0$ in the following sense: the integral curves (the graphs of the solutions) of the first order equation are exactly the phase curves (images of the solutions of the system) of our planar system in this this neighborhood. If at the point $(x_0,y_0)$ $X(x_0,y_0)=0$ but $Y(x_0,y_0)\neq 0$ then the phase curves will correspond to the integral curves of $$\frac{dx}{dy}=\frac{X(x,y)}{Y(x,y)}.\$$ And the problems start when both $X(x_0,y_0)=0$ and $Y(x_0,y_0)=0$ then the phase portrait has an equilibrium point there, but integral curves are simply not defined at this particular point.