# When solving 2 ODEs by eliminating time is valid?

When solving a 2nd order ODE, say $$$$\tag{*}\begin{cases}\frac{dx}{dt}=f(x,y)\\\frac{dy}{dt}=g(x,y),\end{cases}$$$$ it is common to eliminate time and solve the resulting 1st order ODE $$$$\tag{**}\frac{dy}{dx}=\frac{g(x,y)}{f(x,y)}$$$$ that gives a dependence $$y=y(x)$$.

I wonder what are the conditions for this approach to be valid at an equilibrium point $$(x^*,y^*)$$? At this point, the fraction $$\frac{g(x^*,y^*)}{f(x^*,y^*)}=\frac{0}{0}$$. Potentially, this indeterminacy can be resolved using a sort of multivalued L'Hopital rule, but that is quite tricky.

Intuitively, I understand that the answer depends on the structure of the eigenvalues of the linearization of (*) at $$(x^*,y^*)$$, but I cannot formulate this quite well.

Let, say, the linearized system have a saddle at $$(x^*,y^*)$$. The equation ($$**$$) has two solutions corresponding to the stable and the unstable manifolds (they seem to be both unstable as they go away from $$(0,0)$$). Does it imply that the DE ($$**$$) isn't well posed?

Indeed, it doesn't work at an equilibrium point, but you don't really need it there: $$(x,y) = (x^*, y^*)$$ is the solution with initial conditions $$x(0)=x^*, y(0)=y^*$$.
It's also not defined on the curve $$f(x,y) = 0$$ (although there, when $$g(x,y) \ne 0$$, you could look at $$x$$ as a function of $$y$$). However, it is OK everywhere else, and it can be useful to study limits of these solutions as $$x \to x^*$$.
• I don't think so: the stable and unstable manifolds would have the same initial condition $y(x^*) = y^*$. The equilibrium should correspond to a limit. For example, the linear saddle-point system $\dot{x} = y$, $\dot{y} = x$ leads to $\dfrac{dy}{dx} = \dfrac{x}{y}$, not defined at $(0,0)$, but the solutions $y=x$ and $y = -x$ have limits $(0,0)$ as $x \to 0$. Dec 26, 2018 at 0:52