Does $\dot{x}t/\sqrt{c^2 - x^2} + x - \omega t = 0$ have a closed form?

Question

The differential equation $$\frac{\dot{x}t}{\sqrt{c^2 - x^2}} + x - \omega t = 0$$ Is related to another problem known to have a closed form. I have attempted finding integrating factors, changing dependent variables, and other techniques.

Answer 1

Hint:

Let $x=c\sin\theta$ ,

Then $\dot{x}=c\cos\theta~\dot\theta$

$\therefore\dfrac{tc\cos\theta~\dot\theta}{\sqrt{c^2-c^2\sin^2\theta}}+c\sin\theta-\omega t=0$

$\dot\theta=\omega-\dfrac{c\sin\theta}{t}$

Similar to solving ODE : $xy'=\sin(x+y)$:

Follow the method in http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=223:

Let $u=\tan\dfrac{\theta}{2}$ ,

Then $\dot{u}=\dfrac{\omega u^2}{2}-\dfrac{cu}{t}+\dfrac{\omega}{2}$

Which reduces to a Riccati ODE.

Answer 2

This differential equation appears not to have closed-form solutions. Maple doesn't find any, nor any symmetries, so it's extremely unlikely that it will yield to elementary methods.

Answer 3

Try the substitution $x = c\sin(\theta)$. Then the equation becomes

$$ t\dot{\theta} + c\sin(\theta) - \omega t = 0 $$

There's no analytic solution, but you at least have an easier form to work with (for numerical solving)

One neat thing: $\theta$ becomes asymptotically linear as $t \to \infty$. Hence, the steady-state solution is sinusoidal, which is what you might expect.