Solve $$\min\int_{0}^{1} \left( x^2 + 2tx + tx\dot{x} + \dot{x}^2 \right) {\rm d} t, \qquad x(0)=0, \quad x(1)=1$$
We need to solve the problem with the Euler Equation
$$\frac{\partial F}{\partial x}-\frac{d}{dt}\left(\frac{\partial F}{\partial\dot{x}}\right)=0$$
We define $F$ as
$$F(t,x,\dot{x})=x^2+2tx+tx\dot{x}+\dot{x}^2$$
Then subsequently we have:
$$ \begin{aligned} \frac{\partial F}{\partial x} &= 2x+2t+t\dot{x} \\ \frac{\partial F}{\partial\dot{x}} &= tx+2\dot{x} \end{aligned} $$
When substituting this into the above mentioned equation: $\dfrac{\partial F}{\partial x}-\dfrac{d}{dt}\left(\dfrac{\partial F}{\partial\dot{x}}\right)=0$, my final equation becomes $\ddot{x}-\frac{1}{2}x=t$ which is incorrect recording to the Student's Manual. The correct 'answer' should be $\ddot{x}-\frac{1}{2}x=\frac{1}{2}t$. Could someone point out where I made a small/big mistake?
$\frac12x^2+tx\dot x$ is a complete derivative, so is constant for all admissible functions $x$ satisfying the boundary conditions. So it can be removed from $F$. The task reduces to $$ \min\int_0^1(\tfrac12x^2+2tx+\dot x^2)dt $$ This reduced task has now no mixed terms between $\dot x$ and the state $(t,x)$, the Euler-Lagrange equation reads directly $$ 2\ddot x=x+2t. $$