Your general solution is not correct.
The differential equation you are dealing with is the Laplace equation.
For the general theory of Laplace equation, please have a look at the introductory exposition in
http://en.wikipedia.org/wiki/Laplace's_equation
Once the setting becomes familiar, you should get used to the method called "separation of variables" and the Fourier series expansions of periodic functions.
Some good references (to start with):
http://en.wikipedia.org/wiki/Separation_of_variables (paragraph: PDEs)
http://en.wikipedia.org/wiki/Fourier_series
Edit I would add some explicit fomulae. Let us assume a separable solution
$u(x,y)=X(x)Y(y)$: the Laplace equation leads to $X''(x)=k X(x)$ and $Y''(y)=-k Y(y)$
for some real constant $k$. Depending whether $k<0, k=0$ or $k>0$ the solutions to the ordinary diff. eqs for $X$ and $Y$ are different. If we had some extra boundary conditions (periodic for example) we could select a choice for $k$ at this stage. In our case, however, we should consider all the combinations of $u(x,y)=X(x)Y(y)$ for $k<0, k=0$ or $k>0$. Any of these choices must satisfy the boundary condition $\frac{\partial u(x,0)}{\partial y}=h(x)$, for all $x\in\mathbb R$. Depending on the form of $h(x)$ we could, in general, determine the "right" $k$. As you can see I am pretty vague, as Iwe do not know anything about $h(x)$ and we cannot choose $k$ using other boundary conditions.
Let us move to a more explicit example. We select the case $k:=\omega^2>0$, obtaining through superposition the separable solution $u(x,y)=\sum_{i}(A_ie^{\omega_i x}+B_ie^{-\omega_i x})(C_ie^{i\omega_i y}+D_ie^{-i\omega_i y})$, for real coefficients $A_i,B_i,C_i,D_i$. We have to impose the boundary condition obtaining
$\sum_i(A_ie^{\omega_i x}+B_ie^{-\omega_i x})(C_i-D_i)i\omega_i=h(x)$,
for all $x\in \mathbb R$. At this stage a Fourier analysis of $h(x)$ is needed. Note that the Fourier analysis is done using
$\sum_i(\tilde{A}_ie^{\omega_i x}+\tilde{B}_ie^{-\omega_i x})=h(x)$,
where $\tilde{A}_i=i\omega_iA_i(C_i-D_i)$ and $\tilde{B}_i=i\omega_iB_i(C_i-D_i)$.