How to solve an ODE to find its solution

[Hector Triana]

Salutations, I have a problem when I approach this ODE:

$$\left(\frac{y}{y'}\right)^2+y^2=b^2\left(x-\frac{y}{y'}\right)^2$$

I have done a series of steps as I show in this link:
https://drive.google.com/file/d/1Ht4xxUlm7vXqg4S5-wirKwm7vTESU3mU/view?usp=sharing
But I'm not convinced that those were the correct steps neither solutions were adequated, and my question is:
How would be the mathematical steps to apply to find the correct solution of the ODE?

So, I would like any guidance or starting steps or explanations to find the solution of this interesting problem.
Thanks for your attention.

 

[Delta2]

Well this seems like a hard non linear ODE. I tried it at wolfram and the solution seems quite complex and it is given in implicit (perplexed form) $G(y,x)=0$

I can comment only on one of your result, the $y=kx$ simply does not verify the ODE, cause in the left hand side we get $x^2(k^2+1)$ while in the right hand side we get 0.

You can check the solution of wolfram here https://www.wolframalpha.com/input/?i=solve+ODE+(y/y')^2+y^2=b^2(x-(y/y'))^2 but you can't see the step by step solution unless you are a subscribed user at wolfram.

 

[Hector Triana]

Thanks for your commentary @Delta2 about wolfram it's true, I'm not premium user, but thanks for it, indeed, y=kx is not solution, thanks for the link again.

 

[phyzguy]

y=kx is a solution if k=+/-i. In other words, y = i*x or y = -i*x are solutions.

 

[Delta2]

y=kx is a solution if k=+/-i. In other words, y = i*x or y = -i*x are solutions.

That is true, however in his/her workings he writes $y=kx, k\in\mathbb{R}$, and i also thought we are interested for real valued functions y.

 

[phyzguy]

That is true, however in his/her workings he writes $y=kx, k\in\mathbb{R}$, and i also thought we are interested for real valued functions y.

Sorry, I missed that.

 

[Hector Triana]

Hi, thanks for your commentaries, according the suggestions of @Delta2 about wolfram, I was exploring and I think I could solve it, I didn't realize I had a mistake with a sign in my procedure. I share my proposal solution in this link: https://drive.google.com/file/d/13kIlVTCUfbkCVpfjzzZtQe8nwt8SrsOT/view?usp=sharing
Thanks for your attention.

 

[bigfooted]

Hi, your ode splits into two odes:
$y' = \frac{(y*\sqrt((b^2-1)*y^2+b^2*x^2)-b^2*x*y)}{(y^2-b^2*x^2)}$
$y' = -\frac{(y*\sqrt((b^2-1)*y^2+b^2*x^2)+b^2*x*y)}{(y^2-b^2*x^2)}$
an integrating factor for the first one is
$\frac{1}{(b^2*(b^2-2)*y*(x*\sqrt((b^2-1)*y^2+b^2*x^2)-y^2))}$
and the second one is
$\frac{1}{(b^4*y*(x*\sqrt((b^2-1)*y^2+b^2*x^2)+y^2))}$

When trying to find the solution from this integrating factor, I couldn't get it to simplify to something readable...
I only managed to find the integrating factor by assuming that the ODE has a generalized rational point symmetry using Lie's symmetry method.