What rationale am I missing with this simple boundary value problem? Consider $$u_{xy} = xy,\ u(0,\ y) = 0,\ u_x(x,\ 0) = 0$$
Antidifferentiating the PDE with respect to $y$ yields $u_x = \frac{xy^2}{2} + a(x)$, and antidifferentiating this equation with respect to $x$ yields $u = \frac{x^2 y^2}{4} + A(x) + b(y)$, where $A(x)$ is the antiderivative of $a(x)$ with respect to $x$.
Now the boundary conditions produce the system $\begin{cases}0 = A(0) + b(y) \\ 0 = a(x)\end{cases}$.  Antidifferentiating the second equation, $A(x) = C$, where $C$ is constant with respect to both $x$ and $y$ (because $A(x) = C(y)$ makes no sense, except in a trivial way when $C$ is constant).  Plugging this into the the first equation, $0 = C + b(y)$, or lumping the minus sign, $b(y) = C$.  This suggests the solution $u = \frac{x^2 y^2}{4} + C$, which might be thought to require another boundary condition to pin down $C$.
However, only $C = 0$ seems to work when checking the solution.  Where is the mistake in my reasoning which leads to the conclusion that any $C$ should solve the problem?

$$u(x,y) = \frac{x^2 y^2}{4} + A(x) + b(y) \quad\text{is OK.}$$
Condition $u(0,y) = 0$ :
$$u(0,y) =  A(0) + b(y) \quad\implies\quad b(y)=-A(0)=\text{constant}$$
$$u(x,y) = \frac{x^2 y^2}{4} + A(x) -A(0)$$
Condition $u_x(x, 0) = 0$ :
$u_x(x,y)=\frac{x y^2}{2} + A'(x)$
$$u_x(x,0)= A'(x)=0$$
$$A(x)=\text{constant}=C\quad\implies\quad A(0)=C$$
$$u(x,y) = \frac{x^2 y^2}{4} + C -C$$
$$\boxed{u(x,y) = \frac{x^2 y^2}{4}}$$
