(PDEs) Why doesn't my solution look like anything I can find online? I'm using the Fourier transform to solve PDEs.
I was able to solve the wave equation $u_{tt}=c^2u_{xx}$ and got 
$$u(x,t)=f(x-ct)+g(x+ct)$$
I tried to apply the same method for the Laplace equation $u_{xx}+u_{yy}=0$.
For a function of $u(x,y)$, we get the Fourier transform $\widehat u(k,y)$:
$$\widehat{u}(k,y)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} u(x,y)~\mathrm e^{-\mathrm i kx}~\mathrm dx$$ 
The inverse transform is given by
$$u(x,y) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} \widehat u(k,y)~\mathrm e^{\mathrm i xk}~\mathrm dk$$
The two important rules here are that $\widehat{(u_x)}=\mathrm ik\widehat u$ and $\widehat{(u_y)}=(\widehat u)_y$, so applying the Fourier transform to both sides of $u_{xx}+u_{yy}=0$, gives $-k^2\widehat u+(\widehat u)_{yy}=0$. which is a second order ODE in $\widehat u$ and $y$. Solving this, and inverting gives
$$\widehat u(k,y)=\widehat A(k)\mathrm e^{ky}+\widehat B(k)\mathrm e^{-ky}$$
$$\color{blue}{\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}}\widehat u(k,y)~\color{blue}{\mathrm e^{\mathrm ixk}~\mathrm dk} =
\color{blue}{\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}}\left(\widehat A(k)\mathrm e^{ky}+\widehat B(k)\mathrm e^{-\mathrm ky}\right)~\color{blue}{\mathrm e^{\mathrm ixk}~\mathrm dk}
$$
$$u(x,y)= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} \widehat A(k)\mathrm e^{\mathrm i(x-\mathrm iy)k}~\mathrm dk
+ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} \widehat B(k)\mathrm e^{\mathrm i(x+\mathrm iy)k}~\mathrm dk$$
$$u(x,y) = A(x-\mathrm iy)+B(x+\mathrm iy)$$
It seems to me that any function of the form $A(z^*)+B(z)$, where $z=x+\mathrm iy$ will solve the Laplace equation. Indeed, we can check: $u_{xx} = A''+B''$ and $u_{yy} = -A''-B''$.
This solution looks absolutely unlike anything I find online. I realise that I haven't applied any boundary conditions, and different BCs seem to use different methods for solving the PDE and have different solutions.
Is $A(z^*)+B(z)$ a recognised solution? Is it possible to apply boundary conditions now, or is it too late?
 A: Your solution is absolutely correct, but you cannot use it in the same way
as the solution to the wave equation. For the wave equation the analysis tells you that lines $x\pm ct=cte.$ have special properties. For the Laplace equation there are no real lines that are special in any way (that is, there are no waves) 
The $u(x,y)$ that you derive is a function of a complex variable and therefore complex itself ($u+iv$). Its real and imaginary parts both satisfy Laplace. They have different interpretations in different contexts. For example if you have inviscid fluid flow in two dimensions, the real part is the velocity potential and the imaginary part is the stream function.
Interesting flows can be found by superposing elementary solutions of the Laplace equation, such as $u+iv=1/(x+iy)$,and $ u+iv=\ln(x+iy)$
A powerful method is to note that $f(g(z))$ is a solution, where $g(z)$ is simple (like $u+iv=x+iy$) and $f(z)$ converts that into something more interesting (Look up conformal transformations)
