Steady state heat equation problem The problem is :
Let $$HS = \{(x,y):0\leq x\leq 1, y\geq 0\}$$ 
and consider the steady state heat problem on $HS$ with boundary conditions $u(0,y)=u(1,y)=0$ and $u(x,0) = x-x^2$.
Use separation of variables to obtain a family of basic solutions.
so I let 
$$u(x,y) = X(x)Y(y)$$
Since $u$ is the solution of heat equation,
$$u_{xx} + u_{yy} = X''Y + XY'' = 0$$
Manipulate this..
$$\frac{X''}{X} = -\frac{Y''}{Y}$$
But I don't know how to continue
 A: We have $X''(x)/X(x) = -Y''(y)/Y(y)$ for all $x \in [0,1]$ and all $y \ge 0$.
Then there is a constant $c$ such that  $X''(x)/X(x) =c$ for all $x \in [0,1]$ and $Y''(y)/Y(y)=-c$ for all $y \ge 0$.
Hence we get the differential equations $X''=cX$ and $Y''=-cY$.
Can you take it from here ?
A: Assume a solution of the form 
$$
u(x,y)=X(x)Y(y)
$$
and as you did, find 
$$
\frac{X''(x)}{X(x)}=-\frac{Y''(y)}{Y}
$$
which then implies both quantities are some constant, say $\lambda$. Solve the two second order linear ODE's
$$
Y''(y)+\lambda Y=0\\
X''(x)-\lambda X=0\\
\implies Y(y)=ae^{\sqrt{-\lambda}y}+ce^{-\sqrt{-\lambda}y}\\
X(x)=b_1e^{\sqrt{\lambda}x}+b_2e^{-\sqrt{\lambda}x}
$$
making your solution the product of the two solutions, 
$$
u(x,y)=X(x)Y(y)=(b_1e^{\sqrt{\lambda}x}+b_2e^{-\sqrt{\lambda}x})(ae^{\sqrt{-\lambda}y}+ce^{-\sqrt{-\lambda}y})
$$
Now use your boundary condition to figure out all the constants, 
$$
u(0,y)=(b_1+b_2)(ae^{\sqrt{-\lambda}y}+ce^{-\sqrt{-\lambda}y})=0\\
u(1,y)=(b_1e^{\sqrt{\lambda}}+b_2e^{-\sqrt{\lambda}})(ae^{\sqrt{-\lambda}y}+ce^{-\sqrt{-\lambda}y})=0
$$
in order for these to vanish identically for all $y$, we need 
$$
b_1=-b_2\\
b_1e^{2\sqrt{\lambda}}=-b_2\implies e^{2\sqrt{\lambda}}=1
$$
unless $b_1=0=b_2$ and we have the zero solution. 
We could pick $\lambda=0$, but this would again give us the zero solution. So we take 
$$
2\sqrt{\lambda}=2\pi i k\implies \lambda=-\pi^2k^2
$$
for any integer valued $k$. 
Where has this left us, now we have 
$$
u(x,y)=X(x)Y(y)=(b_1e^{\pi i kx}-b_1e^{-\pi i kx})(a_1e^{\pi |k|y}+a_2e^{-\pi|k|y})=2b_1i\sin(\pi kx)(ae^{\pi |k|y}+ce^{-\pi|k|y})
$$
for any integer $k$. Then using superposition, we may sum over all integer $k$ to build the solution, with coefficients depending on $k$, 
$$
u(x,y)=\sum_{k\in \mathbb{Z}}B_k\sin(\pi kx)(a_ke^{\pi |k|y}+c_ke^{-\pi|k|y})
$$ 
Let us now impose your final condition, 
$$
u(x,0)=x-x^2=\sum_{k\in \mathbb{Z}}B_k\sin(\pi kx)(a_k+c_k)
$$
where we can integrate away our troubles. Using orthogonality of trig functions, we integrate against $\sin(nx)$ over $[-1,1]$, with $\frac{1}{2}$ a normalizing constant. This yields, 
$$
\frac{1}{2}\int_{-1}^1(x-x^2)\sin(\pi n x)\mathrm dx=B_n(a_n+c_n)\\
\stackrel{\text{symmetry}}{\implies}
\int_{0}^1x\sin(\pi n x)\mathrm dx=\frac{-1}{\pi n}\cos(\pi n)=B_n(a_n+c_n)
$$
Here I think we are stuck unless we impose a condition at $y=\infty$, namely to insure that 
$$
\lim_{y\to \infty}|u(x,y)|
$$
exists.
This requires $a_k=0$ for any $k\in \mathbb{Z}$. Yielding for our coefficient
$$
\frac{-1}{\pi n}\cos(\pi n)=B_nc_n
$$
and our solution 
$$
u(x,y)=\sum_{k\in \mathbb{Z}}\frac{-1}{\pi k}\cos(\pi k)\sin(\pi kx)e^{-\pi|k|y}
$$
