How to solve $\frac{\partial V}{\partial t} + x + \frac{\partial V}{\partial x}- \frac{1}{2} \frac{1}{\left(\frac{\partial V}{\partial x}\right)} = 0$ I'm solving this problem of optimal control using the dynamic programming:
$$
\begin{cases}
\min \displaystyle \int_0^2(x-u)dt + x(2) \\
\dot x = 1+u^2 \\
x(0) = 1
\end{cases}
$$
Then solving the Bellman-Hamilton-Jacobi equation I found the following PDE:
$$\frac{\partial V}{\partial t} + x + \frac{\partial V}{\partial x}- \frac{1}{4} \frac{1}{\left(\frac{\partial V}{\partial x}\right)} = 0$$
The problem gives an hint: In order to solve BHJ equation, we suggest to find the solution in the family of functions $\mathcal{F} = \{V(t,x) = A +Bt + Ct^2 + D\log(3-t) + E(3-t)x\}$ where $A,B,C,D,E$ are all real costants.
My question is:

How can someone derives that all the solution of the BHJ equation are in that family of functions? In other words, how could I manage to solve the BHJ without any hint?

 A: I find that this problem can be more easily solved using Pontryagin's maximum principle, which gives the following Hamiltonian
$$
H(t,x,\lambda) = x - u + \lambda\,(1 + u^2),
$$
such that
$$
\dot{\lambda} = -1, \quad u = \frac{1}{2\,\lambda}
$$
and from the terminal cost is follows that $\lambda(2) = 1$. In this case solving for the co-state as a function of time is easy, namely $\lambda(t) = 3 - t$ and thus
$$
u(t) = \frac{1}{2\,(3 - t)}. \tag{1}
$$
Note that this solution is still independent of $x(0)$.

When formulating the PDE one also obtains that
$$
u = \frac{1}{2\,V_x}, \tag{2}
$$
with $V_x$ shorthand notation for the partial derivative of $V(t,x)$ with respect to $x$. Equating $(2)$ to $(1)$ yields
$$
V_x = 3 - t. \tag{3}
$$
Therefore, the final expression for $V(t,x)$ should be of the form
$$
V(t,x) = x\,(3 - t) + U(t), \tag{4}
$$
with $U(t)$ a yet unknown function of only $t$ and no $x$. Substituting $(4)$ together with $(3)$ in the PDE yields
\begin{align}
0 &= -x + \dot{U} + x + 3 - t - \frac{1}{4} \frac{1}{3 - t}, \\
&= \dot{U} + 3 - t - \frac{1}{4} \frac{1}{3 - t}.
\end{align}
From this it becomes hopefully clear where the function family comes from. Namely, the last term is derived from $(3)$, the second order polynomial comes from integrating $t - 3$ and the logarithm from integrating $(3 - t)^{-1}$.
I am not sure how one could spot this family of functions without Pontryagin's maximum principle. Though, I suspect this is also why the exercise gave the function family, because solving nonlinear PDE's is hard.
