Given F:
$$ F(x,y,y\prime) = 2\cdot \pi \cdot y \cdot \sqrt{1+(y\prime)^2} $$
We can derive the following Euler-Lagrange equation (I know how to do this part):
$$ \frac{d}{dx}\left(\frac{y\cdot y\prime}{\sqrt{1+(y\prime)^2}}\right) - \left(\sqrt{1+(y\prime)^2}\right) = 0 $$
Can someone please show me how to use:
$$ \frac{1}{y\prime}\left[\frac{d}{dx}\left(F-\frac{\partial{F}}{\partial{y\prime}}\cdot\frac{dy}{dx}\right)-\frac{\partial{F}}{\partial{x}}\right] = 0$$
To derive:
$$ y\cdot y\prime\prime - \left(y\prime\right)^2 - 1 = 0 $$
Here is my working so far but it is wrong. Can someone please show me the error of my ways and provide a step by step method please?
I have $$ \frac{\partial{F}}{\partial{x}} = 0, $$
$$ \frac{\partial{F}}{\partial{y\prime}} \cdot \frac{dy}{dx} = \left(\frac{y\cdot y\prime}{\sqrt{1+(y\prime)^2}}\right) \cdot y\prime = \left(\frac{y\cdot y\prime^2}{\sqrt{1+(y\prime)^2}}\right), $$
$$ \left(F-\frac{\partial{F}}{\partial{y\prime}}\cdot\frac{dy}{dx}\right) = 2\cdot \pi \cdot y \cdot \sqrt{1+(y\prime)^2} - \left(\frac{y\cdot y\prime^2}{\sqrt{1+(y\prime)^2}}\right) $$
Differntiating with respect to x:
$$ \frac{d}{dx}\left(2\cdot \pi \cdot y\prime \cdot \sqrt{1+(y\prime)^2} - \left(\frac{y\cdot y\prime^2}{\sqrt{1+(y\prime)^2}}\right)\right) = 2 \cdot \pi \cdot y \cdot \sqrt{1+(y\prime)^2} + \frac{2 \cdot \pi \cdot y \cdot y\prime \cdot y\prime\prime}{\sqrt{1+y\prime^2}} - \frac{y\prime^3}{\sqrt{1+y\prime^2}} - \frac{2 \cdot y \cdot y\prime \cdot y\prime\prime}{\sqrt{1+y\prime^2}} + \frac{y \cdot y\prime^2 \cdot y\prime\prime}{(1+y\prime^2)^{3/2}} $$
So...
$$ \frac{1}{y\prime}\left[\frac{d}{dx}\left(F-\frac{\partial{F}}{\partial{y\prime}}\cdot\frac{dy}{dx}\right)-\frac{\partial{F}}{\partial{x}}\right] = 2 \cdot \pi \cdot y\prime \cdot \sqrt{1+(y\prime)^2} + \frac{2 \cdot \pi \cdot y \cdot y\prime \cdot y\prime\prime}{\sqrt{1+y\prime^2}} - \frac{y\prime^3}{\sqrt{1+y\prime^2}} - \frac{2 \cdot y \cdot y\prime \cdot y\prime\prime}{\sqrt{1+y\prime^2}} + \frac{y \cdot y\prime^2 \cdot y\prime\prime}{(1+y\prime^2)^{3/2}} $$
$$ = 2 \cdot \pi \cdot \sqrt{1+(y\prime)^2} + \frac{2 \cdot \pi \cdot y \cdot y\prime\prime}{\sqrt{1+y\prime^2}} - \frac{y\prime^2}{\sqrt{1+y\prime^2}} - \frac{2 \cdot y \cdot y\prime\prime}{\sqrt{1+y\prime^2}} + \frac{y \cdot y\prime \cdot y\prime\prime}{(1+y\prime^2)^{3/2}}$$
So...
How on earth do I get from here to $$ y\cdot y\prime\prime - \left(y\prime\right)^2 - 1 = 0 $$
???
It's at this point I'm failing to continue. I'm sure I'm being daft but I would really appreciate some assistance from this point forward if anyone can help.
Thankyou.
