Differential equation must satisfy its edge conditions. I have this variation problem
$$\text{Minimize} \; \int_0^1 \left( 12xt- \dot{x}^2-2 \dot{x} \right) \; dt$$
With the edge conditions $x(0)=0$ and $x(1)$ is "free".
And from here solve it:
$$x(t)\to -t^3 +c_1t+c_2$$
From here it should've been correctly. Now I must solve the equation and compute $c_1$ and $c_2$
However I'm not aware of the edge condition $x(1)$ as free. How do I solve this, and what does it exactly mean by "free"?
Result: $c_1$ should be $2$ and $c_2$ should be $0$.
(Ps. If you can show it with Mathematica It would be great!)

Denote the functional as $J(x)$:
$$
J(x) = \int_0^1 \left( 12xt- \dot{x}^2-2 \dot{x} \right)
$$
Then the minimizer $x$ satisfies the following (perturbing the minimum with $\epsilon y$):
$$ \frac{d}{d\epsilon} J(x + \epsilon y)\Big\vert_{\epsilon = 0} =0$$
Simplifying above gives us:
$$
\int^1_0 (12 y t - 2\dot{x}\dot{y} - 2\dot{y})dt = 0
$$
Integration by parts yields:
$$
\int^1_0 (12 y t + 2\ddot{x}y)dt - 2(\dot{x}+1)y\big|^1_0= 0
$$
Let's look at the boundary term: $(\dot{x}+1)y\big|^1_0 = (\dot{x}(1)+1)y(1) - (\dot{x}(0)+1)y(0)$. The term "free" from what I know would be a natural boundary condition. The essential boundary condition is $x(0)=0$ hence test function $y(0)=0$, the second term vanishes. On the natural boundary $t=1$, we do not impose anything on $y$'s value, hence we have to let $\dot{x}(1)+1 = 0$ to make variational problem well-posed, thus to get the differential equation $6 t + \ddot{x} = 0$.
And the final answer is: "$x(1)$ is free" leads us to the natural boundary condition
$$
\dot{x}(1)+1 = 0
$$
thus the coefficients are $c_1 = 2, c_2 = 0$.
