Help me solve differential equation please involving a fraction and a square root

[gionole]

Homework Statement

Help me solve differential equation

Relevant Equations

##\frac{\dot x}{\sqrt{y(1+\dot x^2)}} = \text{const}##

I'm trying to solve the following differential:

##\frac{\dot x}{\sqrt{y(1+\dot x^2)}} = \text{const}##

##\dot x## is the derivative with respect to ##y##.

How do I solve it so that I end up with ##x(y)## solution ? You can find this here, but there're 2 problems: 1) I don't understand what ##a## is and how author solves it 2) Author solves it as moving into ##\theta##, which I don't want. I prefer to know how I solve it to get ##x(y)##.

 

[erobz]

Let ##k## be the constant. If ##\dot x = \frac{dx}{dy}##, then try using algebra to solve for ##\frac{dx}{dy}##.

 

[gionole]

@erobz

I ended up as well(thanks to you) with ##x = \int_{y_1}^{y_2} \frac{ydy}{\sqrt{2ay - y^2}}##. Now author moves to ##\theta##, but as I told you, I want to end up with ##x(y)## and not ##x(\theta)##. Thoughts ?

 

[gionole]

well, by using ##k##, we get ##\dot x = \sqrt{\frac{k^2y}{1-k^2y}}##.. Now, we say that ##x## is the integration of RHS, right ? but using integral calculator, I end up with huge answer. I guess, that's the downside of using ##x(y)## ? and how do I find ##k## ?

 

[erobz]

well, by using ##k##, we get ##\dot x = \sqrt{\frac{k^2y}{1-k^2y}}##.. Now, we say that ##x## is the integration of RHS, right ? but using integral calculator, I end up with huge answer. I guess, that's the downside of using ##x(y)## ? and how do I find ##k## ?

I got mixed up thinking ##y## was not under the root. But I'm getting something different than you.

$$ x' = k \sqrt{y\left( 1+ x'^2\right) } $$

Square both sides

$$ x'^2 = k^2 y\left( 1+ x'^2\right) $$

$$ x'^2 = \frac{k^2 y}{1-k^2y} $$

$$ x' = k \sqrt{ \frac{y}{1-k^2y}} $$

?

Never mind. I see we agree now.

 

[gionole]

Thanks very much. It makes sense and I realized solving this in terms of ##x(y)## is super complicated. All good.

 

[erobz]

##\theta## is just a dummy variable for the solution technique of the integral (trigonometric substitution). Whatever your end result is in terms of ##\theta## you would invert ##\theta(y)## via:

$$ \tan \theta = \frac{ky}{\sqrt{y - k^2 y^2 } }$$

to get ##x(y)##...it's going to be quite the mess I think!

EDIT: Looking at the form in the paper it wouldn't be terrible, but ##y## is going to be inside an ##\arctan## function as well as outside of it in the ##\sin\theta ## term of the solution.

 

[Dinda556]

I'll sugguest assuame k as constant. Then take derivative of this like dx/dy.