How Can I Solve This Differential Equation for x(y)?

[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.