Minimization of $\dot{x} = 2x + u, \; J(u) = \int_0^1 (x^2 + xu +u^2)\mathrm{d}t$ I am interested in the solution approach to the following Minimization problem (Optimal Control subject) :

Exercise : Solve the problem $\dot{x} = 2x + u, \; J(u) = \int_0^1 (x^2 + xu +u^2)\mathrm{d}t \to \min$ with $x(0) \in \mathbb R$ considered as known and $u(\cdot)$ with no restrictions.

Now, in similar I know that either the Maximum Principle approach can be utilized (via the Hamiltonian Operator) or a Riccati ODE approach. Both of them, though, require $u(\cdot)$ to belong in an interval, for example $-1 \leq u(\cdot) \leq 1$.
BUT in this particular problem, there are no restrictions as mentioned and I find the methods stated above unable to yield me an intuition. I will appreciate very much anyone who could provide me with an insight of an approach on such problems.
 A: For the Hamiltonian approach you get
$$
I(x,u,\lambda)=\int_0^1(x^2+xu+u^2+λ(2x+u-\dot x))dt=\int_0^1 (H(x,u,λ)-λ\dot x)\,dt
$$
The optimal control then minimizes $H(x,u,λ)$ for fixed $u$ and $λ$. As this is a quadratic convex function, the minimum is easy to compute as $x+2u+λ=0\implies u=-\frac{x+λ}2$.
The saddle point can be computed by setting the variation to zero
$$
\delta I=[λδx]_0^1-\int_0^1(H_xδx+H_uδu+H_λδλ-\dot xδλ+\dot λδx)\,dt
$$
where $δx(0)=0$. Then the ensuing equations are
\begin{align}
\dot x&=H_λ=2x+u,&x(0)&=x_0\\
\dot λ&=-H_x=-(2x+u+2λ),&λ(1)&=0
\end{align}
and inserting the optimal control
\begin{align}
\dot x&=\frac{3x-λ}2,&x(0)&=x_0\\
\dot λ&=-\frac{3(x+λ)}2,&λ(1)&=0
\end{align}
This now is a perfectly solvable linear system.
\begin{align}
λ&=3x-2\dot x\\
0&=2\dot λ+3x+3λ=2(3\dot x-2\ddot x)+3x+3(3x-2\dot x)
\\&=-4\ddot x+12x\\
x&=A\sinh(\sqrt3 x)+B\cosh(\sqrt3 x)\\
λ&=(3A-2B)\sinh(\sqrt3 x)+(3B-2A)\cosh(\sqrt3 x)\\
B&=x_0\\
0&=(3A-2x_0)\sinh(\sqrt3)+(3x_0-2A)\cosh(\sqrt3)\\
A&=\frac{3\cosh(\sqrt3)-2\sinh(\sqrt3)}{2\cosh(\sqrt3)-3\sinh(\sqrt3)}x_0
\end{align}
