Question on separation of variables involving X-section of a rectangular bar and its temperature The temperature $T(x,y)$ of a long rectangular bar of X-sectional width $a$ and depth $b$ satisfies the equation $\nabla ^2 T=0$, and subject to the boundary conditions 
$$T(0,y)=T(a,y)=0,$$
$$T(x,0)=0, \;\;T(x,b)=T_b(x)$$
Use the method of separation of variables to express the solution to this problem in the form 
$$T(x,y)=\int^a_0 G(x,/xi;y)T_b(\xi)d\xi,$$
where the function $G$ is to be determined.

Assume $T(x,y)=X(x).Y(y) $
$$\nabla^2 T=0 \rightarrow \frac{\partial^2T}{\partial x^2}+\frac{\partial^2T}{\partial y^2}=0$$
$$\frac{X''}{X}=\frac{Y''}{Y}=-m^2$$
\begin{align}
 X&=A\cos{mx}+B\sin{mx},\\
   Y&=C \cosh{mt}+D\sinh{mt}
\end{align}
\begin{align}
T(0,y)&=0 \rightarrow X=B\sin{mx}\\
T(a,y)&=0 \rightarrow m=\frac{n\pi}{a} \\
T(x,0)&=0 \rightarrow Y=D \sinh{mt}
\end{align}
$$
\therefore T_b(x)=B\sin{\frac{m\pi x}{a}}\cdot D\sinh{\frac{m\pi t}{a}}
$$
$$T(x,y)=\sum ^{\infty}_{n=1} E_n T_b(x)$$
Then how do I determine G?

There are a few things to notice. Taking many shortcuts in the process doesn't usually pay; I know it doesn't for me. Starting with $T(x,y)=X(x)Y(y)$,
$$
              \nabla^2T=X''(x)Y(y)+X(x)Y''(y)=0 \\
                \frac{X''Y}{XY}+\frac{XY''}{XY}=0 \\
                 \frac{X''}{X}+\frac{Y''}{Y}=0 \\
                \frac{X''}{X}=-\frac{Y''}{Y} \\
              \frac{X''}{X} = -\lambda,\;\;\; \lambda = \frac{Y''}{Y}.
$$
Skipping steps led you to conclude that the equations in $X$ and $Y$ are identical. They are not. The conditions
$$
           T(0,y)=T(a,y)=0 \implies X(0)=X(a)=0.
$$
That can happen only for trigonometric functions, not the hyperbolic functions, which means $\lambda > 0$, and $X(0)=0$ gives
$$
        X_{\lambda}(x)=A\sin(\sqrt{\lambda}x)
$$
And $X(a)=0$ means $\sin(\sqrt{\lambda}a)=0$ so that
$$
          \sqrt{\lambda}a = n\pi,\;\;\; n=1,2,3,\cdots \\
              \lambda = \frac{n^2\pi^2}{a^2},\;\;  X_n(x) = A_n\sin(n\pi x/a).
$$
Then
$$
       Y_{n}(y) = B_n\sinh(n\pi y/a)+C_n\cosh(n\pi y/a)
$$
However, $Y_n(0)=0$ gives $C_n=0$, which leads to a solution
$$
              T(x,y)=\sum_{n=1}^{\infty}D_n\sin(n\pi x/a)\sinh(n\pi y/a)
$$
The remaining condition at $y=b$ determines the constants $D_n$:
$$
              T_b(x) = T(x,b) = \sum_{n=1}^{\infty}D_n\sin(n\pi x/a)\sinh(n\pi b/a).
$$
Multiplying by $\sin(n\pi x/a)$, integrating both sides over $[0,a]$, and
using the orthogonality of the sin functions on $[0,a]$ results in equations for the $D_n$:
\begin{align}
     \int_{0}^{a}T_b(x)\sin(n\pi x/a)dx&=D_n\sinh(n\pi b/a)\int_{0}^{a}\sin^2(n\pi x/a)dx \\
  &=D_n\sinh(n\pi b/a)\frac{1}{2}
\end{align}
Therefore,
$$
    T(x,y)=\sum_{n=1}^{\infty}\left[\frac{2}{\sinh(n\pi b/a)}\int_{0}^{a}T_b(x')\sin(n\pi x'/a)dx'\right]\sin(n\pi x/a)\sinh(n\pi y/a) \\
  = \int_{0}^{a}\left[\sum_{n=1}^{\infty}\sin(n\pi x'/a)\sin(n\pi x/a)
  \frac{2\sinh(n\pi y/a)}{\sinh(n\pi b/a)}\right]T_b(x')dx'
$$
The kernel is enclosed by square brackets in the final expression.
