# Euler Lagrange equation issue with answers final form

1. Dec 28, 2016

### fahraynk

1. The problem statement, all variables and given/known data
For the following integral, find F and its partial derivatives and plug them into the Euler Lagrange equation
$$F(y,x,x')=y\sqrt{1+x'^2}\\$$

2. Relevant equations
Euler Lagrange equation : $$\frac{dF}{dx}-\frac{d}{dy}\frac{dF}{dx'}=0$$

3. The attempt at a solution
$$A=2\pi\int_{Y1}^{Y2} y\sqrt{1+x'^2}dy$$

$$F(y,x,x')=y\sqrt{1+x'^2}\\$$
$$\frac{dF}{dx}=0\\$$
$$\frac{dF}{dx'}=\frac{yx'}{\sqrt{1+x'^2}}\\$$
$$0-\frac{d}{dy}\frac{dF}{dx'} = -\frac{d}{dy}\frac{yx'}{\sqrt{1+x'^2}}=0$$
The books answer i s this :
$$\frac{yx'}{\sqrt{1+x'^2}}=C\\ x=acosh^-1\frac{y}{a}+b$$
I dont understand how they get a cosh function. The integral according to wolfram alpha is $$\frac{yx*x'}{\sqrt{x'y^2+1}} + c$$
Is there some identity to turn wolframs into the books answer?

2. Dec 28, 2016

### Ray Vickson

You cannot just blindly apply Wolfram Alpha to the problem, because it is not a simple integration problem. It is a differential-equation solving problem. If you write your equation as $y x' = C\sqrt{1+x'^2}$ that really says the following:
$$y \frac{dx}{dy} = C \sqrt{ 1 + \left(\frac{dx}{dy}\right)^2}$$

3. Dec 28, 2016

### vela

Staff Emeritus
That should be
$$\frac{\partial F}{\partial x}-\frac{d}{dy}\frac{\partial F}{\partial x'}=0.$$ Partial derivatives and total derivatives aren't interchangeable.

Try solving for $x'$ first and then integrating.

4. Dec 29, 2016

### fahraynk

$$c(1+\frac{dx}{dt})^2=yx'\\ \int{c(1+\frac{dx}{dt})^2dx} = \int{yx'dx}\\ cx + c\int{\frac{dx}{dt}^2dx}=yx$$
I'm not sure here. I suspect it has something to do with a function of e, since I know cosh is like e^s+e^-s or something similar... but first off, I am confused about this integration.

5. Dec 29, 2016

### vela

Staff Emeritus
I repeat, solve for $x'$ first, remembering that $x' = dx/dy$. Also, there's no $t$ in this problem, right?

6. Dec 30, 2016

### fahraynk

$$C=\frac{yx'}{\sqrt{1+x'^2}}\\ yx'=c\sqrt{1+x'^2}\\ (yx')^2=c^2+(cx')^2\\ x'^2[y^2-c^2]=c^2\\ x'=\frac{c}{\sqrt{y^2-c^2}}\\ \frac{x}{c}=\int \frac{1}{y^2-c^2}dy\\ \frac{x}{c}=ln|y+\sqrt{y^2-c^2}|$$
Now what? was that right? I could set it as e^x=y+sqrt(y^2-c^2) but still thats not the right form...

7. Dec 30, 2016

### vela

Staff Emeritus
I'd do the integration a bit differently to get to the result in the form you want. You got to
$$x = \int \frac{c}{\sqrt{y^2-c^2}}\,dy = \int \frac{1}{\sqrt{(y/c)^2-1}}\,dy.$$ At this point, try the subtitution $y/c = \cosh u$.

8. Dec 30, 2016

### Ray Vickson

You forgot the constant of integration:
$$\frac{x}{c}= ln\left|y+\sqrt{y^2-c^2}\right| + \text{const.}$$

Anyway, $\text{arccosh}(u) = \ln(u + \sqrt{u^2-1})$.