Finding a solution to a PDE EDIT: This is for a production scheduling problem with quadratic production and linear inventory costs. 
The goal is to \begin{equation*}
  \max_{u} \int_{0}^{T} -(c_{1}u^{2} + c_{2}y) dt
\end{equation*}
subject to
\begin{align*}
  y' &= u,\\
  y(0) &= 0,\\
  y(T) &=B.\\
  \end{align*}
So the Bellman equation is: \begin{equation*}
  \max_{u} {-c_{1}u^{2} + c_{2}y + \frac{\partial V}{\partial y}u + \frac{\partial V}{\partial y}} =0.
\end{equation*}
By first-order condition, we obtain $u^{*} = \frac{1}{2c_{1}}\left ( \frac{\partial V}{\partial y} \right )$.
We insert this $u^{*}$ into the Bellman equation to obtain:
$$0 = -c_{1}\left (\frac{1}{2c_{1}} \frac{\partial V}{\partial y}  \right )^{2} + \frac{\partial V}{\partial y} \left (\frac{1}{2c_{1}} \frac{\partial V}{\partial y}  \right ) + \frac{\partial V}{\partial t}$$ 
And then I obtain the following PDE: $$0 = \frac{1}{4c_{1}}\left (\frac{\partial V}{\partial y}  \right )^{2} + \frac{\partial V}{\partial t}$$
I have to solve for $V(y,t)$.
My work: $V = e^{ay + bt}$. 
Then I find $V_{y} = ae^{ay+bt}$, and $V_{t} = be^{ay+bt}$.
Then I put it values of $V_{y}$ and $V_{t}$ in the PDE to solve for $b$: 
$$\frac{1}{4c_{1}}(ae^{ay+bt})^{2} + be^{ay+bt} = 0$$ 
$$\Rightarrow a^{2}e^{2(ay+bt)} + be^{ay+bt} = 0$$
$$\Rightarrow a^{2}e^{2} e^{(ay+bt)} = -be^{ay+bt}$$
Assuming that $e^{ay+bt} \neq 0$, we obtain $b = -a^{2}e^{2}$.
Now, substitute the value of $b$ back in the equation for $V$, to obtain $$V(y,t) = e^{ay - a^{2} e^{2}t} = e^{a(y-aAt)},$$ where $A = e^{2}$.
Does this look correct?
Also, the boundary conditions are given with $y(0)=0$, and $y(T)=B$. 
 A: The solution which is just a little better than trivial is
$$
V = \alpha y + \beta t
$$
This leads to
$$
\frac{\alpha^2}{4c_1} + \beta  = 0
$$
The boundary conditions do not make sense to me. As I was expecting $V(\text{some t},\text{some y})$?
A: $\dfrac{\partial V}{\partial t}+\dfrac{1}{4c_1}\left(\dfrac{\partial V}{\partial y}\right)^2=0$
$4c_1\dfrac{\partial V}{\partial t}+\left(\dfrac{\partial V}{\partial y}\right)^2=0$
$4c_1\dfrac{\partial^2V}{\partial y\partial t}+2\dfrac{\partial V}{\partial y}\dfrac{\partial^2V}{\partial y^2}=0$
$2c_1\dfrac{\partial^2V}{\partial y\partial t}+\dfrac{\partial V}{\partial y}\dfrac{\partial^2V}{\partial y^2}=0$
Let $U=\dfrac{\partial V}{\partial y}$ ,
Then $2c_1\dfrac{\partial U}{\partial t}+U\dfrac{\partial U}{\partial y}=0$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dt}{ds}=2c_1$ , letting $t(0)=0$ , we have $t=2c_1s$
$\dfrac{dU}{ds}=0$ , letting $U(0)=U_0$ , we have $U=U_0$
$\dfrac{dy}{ds}=U=U_0$ , letting $y(0)=f(U_0)$ , we have $y=f(U_0)+U_0s=f(U)+\dfrac{Ut}{2c_1}$ , i.e. $U=F(2c_1y-Ut)$
