Consider a $2D$ version of Laplace's Equation, for example, $\triangle u = 0, u(x, 0) = 3, u(x, 1) = 3, u(0, y) = 0, u_x (1, y) = 0$.
Separation of Variables leads to the general solution $u = (Ae^{-\sqrt\lambda x} + Be^{\sqrt\lambda x})(Ce^{-\sqrt{-\lambda} y} + De^{\sqrt{-\lambda} y})$, which has five constants to be determined, $A$, $B$, $C$, $D$, and $\lambda$. We can use the two homogeneous boundary conditions to solve for $A$ in terms of $B$, and also solve for $\lambda$, yielding $u = (Be^{\frac{(2n + 1)\pi x i}{2}} - Be^{-\frac{(2n + 1)\pi x i}{2}})(Ce^{-\frac{(2n + 1)\pi y}{2}} + De^{\frac{(2n + 1)\pi y}{2}})$.
The next step is to lump $B$ into $C$ and $D$. Then the two inhomogeneous boundary conditions can be used to solve for the new $C$ and $D$, in general via Fourier Series, but in this case via ordinary algebra. Optionally, Euler's formula can be applied around this point to get trig answers instead of complex exponentials.
However, if we fail to lump $B$ into $C$ and $D$ before applying the inhomogeneous boundary conditions, then it is possible to solve for $C$ and $D$, but leave $B$ in the final solution.
If this "mistake" has been made, is there a way to reason though what $B$ should be using the boundary conditions or other information about the problem, without retracing previous steps in the solution technique to lump away $B$, or has the information to pin down this degree of freedom genuinely been lost as a result of doing the algebra in a suboptimal manner, even though all of the information given has been applied? If the latter, is this phenomenon special in the algebra of differential equations, or is there a high-school algebra or linear algebra equivalent that would provide a more intuitive understanding of the nature this type of information loss?