Second-order ODE involving two functions I am wondering how to find a general analytical solution to the following ODE:
$\frac{dy}{dt}\frac{d^2x}{dt^2} = \frac{dx}{dt}\frac{d^2y}{dt^2}$
The solution method might be relatively simple; but right now I don't know how to approach this problem.
 A: Put $p=dx/dt$ and $q=dy/dt$ and you will have an equation with separate variables
A: I find it more convenient to rewrite the equation using Newton's notation. Instead of writing $\frac{\mathrm{d}x}{\mathrm{d}t},$ it is more helpful to write $x'.$ Thus, the equation is $$x'y''=x''y'.$$ Now, suppose $x'=0.$ Then $x''=0$ is trivial, so every differentiable function $y\in\mathbb{R}^{\mathbb{R}}$ satisfies the equation. This holds analogously if $y'=0.$ Otherwise, we can divide by $x'y',$ thus $$\frac{y''}{y'}=\frac{x''}{x'}.$$ There are four cases to consider from here: $x'\lt0$ and $y'\lt0$; $x'\lt0$ and $y'\gt0$; $x'\gt0$ and $y'\lt0$; and $x'\gt0$ and $y'\gt0$. These cases simplify the equation respectively $$\ln(-x')+C=\ln(-y')$$ $$\ln(-x')+C=\ln(y')$$ $$\ln(x')+C=\ln(-y')$$ $$\ln(x')+C=\ln(y')$$ which are equivalent to $$e^Cx'=y'$$ $$-e^Cx'=y'$$ $$-e^Cx'=y'$$ $$e^Cx'=y'.$$ These cases simply reduce to $$y'=Ax'$$ where $A\neq0.$ Therefore, we have that $y(t)=Ax(t)+B,$ where $A\neq0.$ Remember that this is in the case when neither $x$ nor $y$ is a constant function.
