I don't understand this PDE solution involving Fourier coefficients and orthonormal eigenfunctions. $u_{xx} + u_{yy} = 0$ with $x \in (0,\pi)$ and $y \in (0, \pi)$
Initial Conditions:
$$
u(x,0) = x^2
$$
$$
u(x,\pi) = 0
$$
Boundary conditions:
$$
u_{x}(0,y) = 0 = u_{x}(\pi, y)
$$
I performed separation of variables, which yielded $\frac{Y''}{Y} = \frac{-X''}{X} = \lambda$
From this we derive two ODE's:
$$
X'' + \lambda X = 0
$$
$$
Y'' - \lambda Y = 0
$$
For the first ODE, we know the general solution is $X(x) = A\sin(\sqrt \lambda x) + B\cos(\sqrt \lambda x)$
Using the boundary conditions I differentiated my general solution which yielded $X'(x) = \sqrt \lambda A \cos ( \sqrt \lambda x ) - \lambda B \sin (\sqrt \lambda x)$ or with the first boundary condition when $x = 0$, yields simply $X'(0) = \sqrt \lambda A = 0$ which tells us A = 0, right?
Well I am just having real trouble understanding what is the logic behind the solution to this problem that my professor posted. Where do the orthonormal eigenfunctions come from? Why is there no talk of solving for the constants that come out of the general solution? Where do the solutions for Y come from? I would really appreciate if someone could work through this problem step by step because I am really lost as to where the thinking comes to take the steps taken.
Please help! Thank you in advance.
 A: 
Well I am just having real trouble understanding what is the logic
  behind the solution to this problem that my professor posted. Where do
  the orthonormal eigenfunctions come from? 

You are finding them by solving the $X$ problem, i.e. by finding all values of the parameter $\lambda$ for which there exists nontrivial solutions to the $X$ ODE and the $X$ boundary conditions. Note that you have only applied the BC at $x=0$. You need to apply the BC at $x=\pi$ to complete your analysis. Also, note that you have assumed $\lambda>0$ since you got sines and cosines for your general solution. Based on where it sounds you are in your understanding, you need to consider the cases $\lambda=0$ and $\lambda<0$ as well.

Why is there no talk of
  solving for the constants that come out of the general solution? 

Well, you have to have the general solution (and $\lambda$ values) before you can do this. Once you find the eigenvalues $\lambda$ and eigenfunctions $X(x)$ (there will be an infinite sequence of each), then you can use the orthogonality of those eigenfunctions to determine the values of the constants in the general solution.

Where do the solutions for Y come from? 

Once $\lambda$ is known, you simply solve the $Y$ ODE for those $\lambda$ values.

Edit: A little more to help you along (for the $\lambda>0$ case)... $X(x)=A\sin(\sqrt{\lambda}x)+B\cos(\sqrt{\lambda}x)\implies X'(x)=A\sqrt{\lambda}\cos(\sqrt{\lambda}x)-B\sqrt{\lambda}\sin(\sqrt{\lambda}x)$. Then $X'(0)=0\implies A\sqrt{\lambda}=0\implies A=0\text{ or }\sqrt{\lambda}=0\implies A=0$ since we are assuming $\lambda>0$.
On the other hand, $X'(\pi)=0\implies -B\sqrt{\lambda}\sin(\sqrt{\lambda}\pi)=0\implies B=0, \sqrt{\lambda}=0,\text{ or }\sin(\sqrt{\lambda}\pi)=0$. If $B=0$, we already had $A=0$ so $X(x)=0$, but we were seeking nontrivial solutions. Also, $\sqrt{\lambda}=0$ is impossible since we are assuming $\lambda>0$. That leaves us with $\sin(\sqrt{\lambda}\pi)=0\implies \sqrt{\lambda}=n\in\mathbb{Z}\implies \lambda=\lambda_n=n^2$, $n=1,2,\dots$ are the eigenvalues with associated eigenfunctions $X_n(x)=B_n\cos(\sqrt{\lambda_n}x)=B_n\cos(nx)$, $n=1,2,\dots$
Now that the $\lambda_n$ are known, solve the $Y$ problem...
