I guess the meaning is: even if you take the constant as $0$ (which indeed should be $ \pm {c \over 2}$ for some constant $c$), $u$ function you get will still satisfy the defining equations in the problem, hence, be a solution.

And, in fact, I think you will always get the same solution to $u$ for any integration constant you put (since the constant in $\psi $ and $\phi $ will cancel each other).