Zoie Phillips

2022-03-31

Second-order ODE involving two functions
I am wondering how to find a general analytical solution to the following ODE:
$\frac{dy}{dt}\frac{{d}^{2}x}{{dt}^{2}}=\frac{dx}{dt}\frac{{d}^{2}y}{{dt}^{2}}$
The solution method might be relatively simple; but right now I don't know how to approach this problem.

Charlie Haley

Step 1
I find it more convenient to rewrite the equation using Newton's notation. Instead of writing $\frac{dx}{dt}$, it is more helpful to write x'. Thus, the equation is
${x}^{\prime }y{}^{″}=x{}^{″}{y}^{\prime }.$
Now, suppose ${x}^{\prime }=0$. Then $x0$ is trivial, so every differentiable function $y\in {\mathbb{R}}^{\mathbb{R}}$ satisfies the equation. This holds analogously if ${y}^{\prime }=0$. Otherwise, we can divide by x'y', thus
$\frac{y{}^{″}}{{y}^{\prime }}=\frac{x{}^{″}}{{x}^{\prime }}$.
Step 2
There are four cases to consider from here: ; and . These cases simplify the equation respectively
$\mathrm{ln}\left(-{x}^{\prime }\right)+C=\mathrm{ln}\left(-{y}^{\prime }\right)$
$\mathrm{ln}\left(-{x}^{\prime }\right)+C=\mathrm{ln}\left({y}^{\prime }\right)$
$\mathrm{ln}\left({x}^{\prime }\right)+C=\mathrm{ln}\left(-{y}^{\prime }\right)$
$\mathrm{ln}\left({x}^{\prime }\right)+C=\mathrm{ln}\left({y}^{\prime }\right)$
which are equivalent to
${e}^{C}{x}^{\prime }={y}^{\prime }$
$-{e}^{C}{x}^{\prime }={y}^{\prime }$
$-{e}^{C}{x}^{\prime }={y}^{\prime }$
${e}^{C}{x}^{\prime }={y}^{\prime }$.
These cases simply reduce to ${y}^{\prime }=A{x}^{\prime }$
where $A\ne 0$. Therefore, we have that $y\left(t\right)=Ax\left(t\right)+B$, where $A\ne 0$. Remember that this is in the case when neither x nor y is a constant function.

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