I am looking at the $2$-D Laplace's equation $$\nabla^2u=u_{xx}+u_{yy}=0$$ $$u(x,0)=f(x), x \in (0,a)$$ $$u(x,b)=0, x \in (0, a)$$ $$u(0,y)=u(a,y)=0, y \in (0,b)$$ The solution is in the form $u(x,y)=X(x)Y(y)$.
By using the separation of variables, we get:
$$(1):\left\{\begin{matrix} X''+\lambda X=0, 0<x<a\\ X(0)=X(a)=0 \end{matrix}\right.$$
$$(2):\left\{\begin{matrix} Y''-\lambda Y=0, 0<y<b\\ Y(b)=0 \end{matrix}\right.$$ $$u(x,0)=X(x)Y(0)=f(x):(3)$$ $$$$ From $(1)$ we get $$\lambda_n=(\frac{n \pi}{a})^2$$ $$$$ From $(2)$ we get: $$Y(y)=C_ne^{-\frac{n \pi}{a}y}+D_ne^{\frac{n \pi}{a}y}$$
$\sinh{(x)}=\frac{e^x-e^{-x}}{2}, \cosh{(x)}=\frac{e^x+e^{-x}}{2}$
Since $Y(b)=0$:
$$Y(y)=C_n \cosh{(\frac{n \pi}{a}(b-y))}+D_n \sinh{(\frac{n \pi}{a}(y-b))}$$
Could you explain to me how we got to the last relation? I got stuck right now. :/
Let's redo this from the beginning.
We have $$\nabla^2 u = u_{xx} + u_{yy} = 0$$ With $u=X(x)Y(y)$ this becomes: $$X''Y + XY''=0$$ $$\frac{X''}X = - \frac {Y''}Y$$ Since either side is independent from the other, this has to be constant. So let's define: $$\lambda = \frac{Y''}Y$$ Then we get: $$X''+\lambda X = 0, \quad Y'' -\lambda Y=0$$
Solving this and with the boundary conditions for x, this becomes: $$\lambda = \left(\frac{n\pi}a\right)^2, \quad X = B \sin\left(\frac{n\pi x}a\right), \quad Y = C e^{-n\pi y / a} + D e^{n\pi y / a}$$
With the boundary condition $Y(b)=0$, we get: $$C e^{-n\pi b / a} + D e^{n\pi b / a} = 0$$ $$D = -C e^{-2n\pi b / a}$$ \begin{array}{rcl} Y &=& C e^{-n\pi y / a} -C e^{-2n\pi b / a} e^{n\pi y / a} \\ Y &=& -C e^{-n\pi b / a}\left(e^{n\pi y / a -n\pi b / a } - e^{-n\pi y / a + n\pi b / a}\right) \\ Y &=& C'\left(e^{n\pi (y-b) / a} - e^{-n\pi (y - b) / a }\right) \\ Y &=& C''\sinh\left(\frac{n\pi (y-b)}a\right) \end{array}