Below is a question I am trying to solve, and my attempt.
$\int_0^T \frac{\dot{x}^2}{t^3} \mathrm{d} t$, where $x(0)=1 $ and $x(T)$ lies on the curve Transversal condition:
$$f-(\dot{c} -\dot{x})\frac{\partial f}{\partial \dot{x}}=0$$ $$\frac{\dot{x}^2}{t^3}-(2t-2-\dot{x})(\frac{2\dot{x}}{t^3})=0$$ $$\frac{\dot{x}^2}{t^3}-\left(\frac{4t\dot{x}-4\dot{x}-2\dot{x}^2}{t^3}\right)=0$$ $$3\dot{x}^2-4t\dot{x}+4\dot{x}=0$$ $$3\dot{x}-4t + 4=0$$ $$\dot{x}=\frac{4t-4}{3}$$ $$x=\frac{2t^2-4t}{3}+C$$ $$1=C$$ $$x=\frac{2t^2-4t}{3}+1$$
The apparent answer is $x(t) = \frac{t^4}{8}+1$, where have I gone wrong?
You need to first use $E-L$ and then sub in for $\dot{x}$ when you start using transversality conditions. You will see immediately that you will have some $T^4$ and that is where that comes from in the actual answer.
If you need further clarification ask and I will edit.