Solving $\min\int_{a}^{1}t\dot{x}^2 dt$, with $x(a)=0$ and $x(1)=1$

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We need to show that the problem above has a solution if $a\in(0,1)$, but not if $a=0$. We need to solve the problem with the Euler Equation

$$\frac{\partial F}{\partial x}-\frac{d}{dt}(\frac{\partial F}{\partial\dot{x}})=0$$

We define $F$ as

$$F(t,x,\dot{x})=t\dot{x}^2$$

Then subsequently we have: $$ \begin{aligned} \frac{\partial F}{\partial x} &= 0 \\ \frac{\partial F}{\partial\dot{x}} &= 2t\dot{x} \end{aligned} $$

When substituting this into the above mentioned equation: $\frac{\partial F}{\partial x}-\frac{d}{dt}(\frac{\partial F}{\partial\dot{x}})=0$, my final equation becomes $0-\frac{d}{dt}(2t\dot{x}) = -2\dot{x}-2t\ddot{x} = 0$. This is probably where I make a mistake since the Student's Manual informs me that the Euler Equation is; $(d/dt)(t\dot{x})=0$. Can someone point out my mistake?