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.

I am wondering how to find a general analytical solution to the following ODE:

The solution method might be relatively simple; but right now I don't know how to approach this problem.

Charlie Haley

Beginner2022-04-01Added 14 answers

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

${x}^{\prime}y{}^{\u2033}=x{}^{\u2033}{y}^{\prime}.$

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

$\frac{y{}^{\u2033}}{{y}^{\prime}}=\frac{x{}^{\u2033}}{{x}^{\prime}}$ .

Step 2

There are four cases to consider from here:${x}^{\prime}<0\text{}\text{and}\text{}{y}^{\prime}0;{x}^{\prime}0\text{}\text{and}\text{}{y}^{\prime}0;{x}^{\prime}0\text{}\text{and}\text{}{y}^{\prime}0$ ; and ${x}^{\prime}>0\text{}\text{and}\text{}{y}^{\prime}0$ . These cases simplify the equation respectively

$\mathrm{ln}(-{x}^{\prime})+C=\mathrm{ln}(-{y}^{\prime})$

$\mathrm{ln}(-{x}^{\prime})+C=\mathrm{ln}\left({y}^{\prime}\right)$

$\mathrm{ln}\left({x}^{\prime}\right)+C=\mathrm{ln}(-{y}^{\prime})$

$\mathrm{ln}\left({x}^{\prime}\right)+C=\mathrm{ln}\left({y}^{\prime}\right)$

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.

I find it more convenient to rewrite the equation using Newton's notation. Instead of writing

Now, suppose

Step 2

There are four cases to consider from here:

which are equivalent to

These cases simply reduce to

where

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