Tristan,

I don't think you are making too many assumptions if your goal is to find *one* possible solution, but it is valid to be concerned if it there are leaps in logic for a general solution.

Your solutions (I) and (II) are interpreted to be the general solutions to u_{xx}=0 and u_{yy}=0, respectively. I believe your solution steps for u_{x} and u_{y} for respective constraints to be correct. However, I have modified the final steps for the general solutions:

(I) : u(x,y) = xϕ(y)+ψ(y)+bx+C

(II) : u(x,y) = yα(x)+β(x)+cy+C

Note that if you take the derivative of (I) with respect to x, b gets absorbed into C. Analogous conditions apply to (II).

When constrained to variables x and y, C is a constant (and so are b and c).

Finally, note that the general solution which obeys both u_{xx}=0 and u_{yy}=0 is equal to both solution (I) and solution (II).

Thus, solve: xϕ(y)+ψ(y)+bx+C = yα(x)+β(x)+cy+C

Let's assume none of the terms are equal to 0. This will allow us to impose constraints that will deny trivial solutions. If any of ϕ(y), ψ(y), α(x), and β(x) are equal to 0, the general solution should still be valid.

Thus, xϕ(y) = yα(x).

Thus, ϕ(y)=ay and α(x)=ax. (This is pretty much what you said, except I imposed the extra constraint that the functions share the same coefficient). Check that xϕ(y) = yα(x) = axy.

So now we have axy + ψ(y) + bx + C = axy + β(x) + cy + C

Thus, ψ(y) = cy and β(x) = bx.

Thus, the general solution is axy + bx + cy + C.