Viable method to solving this first order system of linear DE? Given the following system of differential equations
\begin{align}
\frac{dy}{dt} &= x
\\ \\
\frac{dx}{dt} &= y
\end{align}
is the following operation allowed?
\begin{align}
\frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}  &= \frac{x}{y}
\\\\
\frac{dy}{dx} &= \frac{x}{y}
\end{align}
 A: More useful to substitute $y= \frac{dx}{dt}$ to get $\frac{d \frac{dx}{dt}}{dt} = x$
$\frac{d^2x}{dt^2} = x$
and solve the resulting second order de
A: The action is "allowed", as long as you restrict $y > 0$, or $y < 0$ (but not both at once, since we want differentiable solutions). Doing this you can then get different sets of solutions for each case. You'll probably notice that they are "the same functions", and that they are defined at $y=0$, but you can't rigorously justify that the concatenation of the three is a solution to the ODE for all possible values of $y$. You can, however, use this as a heuristic for finding orbits $y(x)$, and justifying uniqueness (given an initial value) by noticing that the system is linear.
But still, beginning from the observation that the system is linear, we can use more solid methods to find even the trajectories $x(t), y(t)$, which is much better than just having $y$ as a function of $x$.
A: what you have done is correct. that is the trajectories of $$\frac{dy}{dt} = x, \frac{dx}{dt} = y\tag 1$$ coincides with that of $$\frac{dy}{dx} = \frac xy \tag 2$$ that is the solutions of $(2)$ are the $xy$ projection of the solution $(x(t), y(t), t).$   
we can say even more; that  we can replace $(1)$ by the more general $$\frac{dy}{dt} = xf(x,y, t), \frac{dx}{dt} = yf(x,y,t)\tag 3$$ with mild constraints on $f$ so that the solution exists for all time.
