General form of solution to 2-D Laplace's equation I'm reading through Fourier Series and Orthogonal Polynomials to get a better sense of the mathematical background to a lot of the physics and electrical engineering that I am doing as an undergraduate. 
Of the general solution to Laplace's equation, the author states:
"Since the conditions thus far satisfied are  homogeneous, containing only terms which are of the first degree 
as to their dependence on the unknown function and its 
derivatives, any constant multiple of a solution is a solution, and the sum of any two solutions is a solution." 
Why is this? This is a fact I've just accepted up until now during the course of my degree, but since this is something I'm doing on my own I really want to 
understand why. 
 A: The Laplacian is a linear differential operator, thus if $\nabla^2 f =0$ for real a we would have $ \nabla^2(af) = a \nabla^2 f =0 $ and if  $\nabla^2 g =0$, we would have $\nabla^2 (f+g) = \nabla^2 f + \nabla^2 g =0$. That is "any linear combinations of solutions to Laplace's equation is itself a solution to Laplace's equation." A real advantage of this is property, as the author states, is that we can find a general solution to Laplace's equation and get the specific solution to a certain boundary value problem by simply finding the coefficients that correspond to each term. I can give an example if that would also be helpful.
Alright, here's an example: 
suppose we wish to find harmonic $u(r,\theta)$ in the domain of the unit disk (|x|<1) subject to the Dirichlet BC $u(1,\theta) = h(\theta)$. 
We have (and you can verify for yourself) that  $u_n = r^n [A_n cos(\theta) + B_n sin(\theta)]$ is indeed a solution of Laplace's equation in polar coordinates for $n \geq 0$. Since the Laplacian is linear it follows that $\Sigma u_n$ is also a solution to Laplace's equation. By the uniqueness of solutions to the Dirichlet problem, we only need to find one harmonic function that satisfies the BC, which will thus be our solution. 
We then wish to find coefficients $A_n$ and $B_n$ such that $\Sigma u_n (1, \theta) = A_n cos(\theta) + B_n sin(\theta) = h(\theta)$. This is simply finding the Fourier series for h. By then substituting the A's and B's back into our general solution, we have thus found a harmonic function that satisfies our BC, hence the solution for $u(r, \theta )$
This method can be applied for solving Laplace's equations in different geometries and for different BC's (e.g. Neumann, Robin, etc.), however the only difference is there would be a different general solution.
