I have this question:
By solving the Euler-Lagrange equation, show that the geodesics are straight line
My attempt at solveing the question is attached below:
However i do not know to how to solve the equation $\frac{d}{dx}(4\dot{y}^3)$
My attempt at solving it gives:
As a forewarning, I am switching notation from $\frac{dy}{dx}=\dot{y}$ to $\frac{dy}{dx}=y'$ because a dot is usually used to denote a "time" (or parametric) derivative.
Make sure we are starting off with a valid Euler-Lagrange equation. In case you are unaware, it is easy to convince yourself that when minimizing length as the integral of a function $$f(x,y,y') = \sqrt{1+(y')^2}$$ it will be equivalent to minimize the integral of the square $$f^2(x,y,y') = 1+(y')^2 .$$
The Euler-Lagrange equation we want to use is $$\frac{\partial f}{\partial y} - \frac{d}{dx}\bigg[\frac{\partial f}{\partial y'}\bigg]=0 \text{ .}$$
Applying this to our function $f$, we have
$$0 - \frac{d}{dx}\bigg[2y'\bigg]=0 $$ so clearly
$$ \frac{d}{dx}\bigg[2y'\bigg]=\frac{d}{dx}\bigg[2\frac{dy}{dx}\bigg]=2\frac{d^2y}{dx^2}=0 .$$
We are left with the second order ODE
$$\frac{d^2y}{dx^2} = 0$$
which models a line in the plane $y(x) = c_1 x + c_2 \text{ .}$