For the PDE:$$u_{xx}+u_{yy}=0$$ $$u(0,y)=\sin\pi y, \ u(1,y)=0$$ $$u(x,0)=u(x,1)=0.$$
I have that $u(x,y)=X(x)Y(y)$, then $$-Y(y)=\mu Y(y), \ Y(0)=Y(1)=0$$ $$X''(x)=\mu X(x), \ X(1)=0.$$
Thus, $$\mu_n=(n\pi)^2, \ Y_n(y)=\sin n\pi y, \ n = 1,2,...$$ and then $$X_n(x)=A_ne^{n\pi(x-1)}+B_ne^{-n\pi(x-1)}$$ satisfies $X_n(1)=0$ if $A_n = -B_n.$
But why is $X_n(x)=A_n\sin\text hn\pi(x-1)$?
For the equation $u_{xx} + u_{yy} = 0$ with $u(0,y)=\sin\pi y$, $u(1,y)=0$, and $u(x,0)=u(x,1)=0$ it can be seen that for $u(x,y) = F(x) G(y)$ then \begin{align} G'' + \mu^{2} G &= 0 \\ F'' - \mu^{2} F &= 0. \end{align} The solutions are $F(x) = A_{1} \cosh(\mu x) + A_{2} \sinh(\mu x)$ and $G(y) = B_{1} \cos(\mu y) + B_{2} \sin(\mu y)$. For the case of $G(0) = G(1) = 0$ then $B_{1} = 0$ and $\sin(\mu) = 0$ which yields $\mu = n \pi$, $n \geq 0$. The y-solution is $G(y) = \sin(n \pi y)$. The x-equation becomes $F''- n^{2} \pi^{2} F = 0$ with $F(1) = 0$. The solution of which yields $0 = A_{1} \cosh(n \pi) + A_{2} \sinh(n \pi)$. Solving for $A_{2}$ this yields \begin{align} A_{2} = - A_{1} \coth(n \pi) \end{align} and \begin{align} F(x) &= A_{1} \cosh(n \pi x) - A_{1} \coth(n \pi) \, \sinh(n \pi x) \\ &= \frac{A_{1}}{\sinh(n \pi)} \left( \sinh(n \pi) \, \cosh(n \pi x) - \sinh(n \pi x ) \, \cosh(n \pi) \right) \\ &= A \sinh(n\pi(x-1)). \end{align} The general solution is then \begin{align} u(x,y) = \sum_{n=1}^{\infty} A_{n} \sinh(n\pi(x-1)) \, \sin(n \pi y). \end{align} The last condition is $u(0,y) = \sin(\pi y)$ for which \begin{align} \sin(\pi y) = - \sum_{n=1}^{\infty} A_{n} \sinh(n \pi) \, \sin(n \pi y). \end{align} By comparing both sides it is determined that the coefficients are given by $A_{1} = - (\sinh(\pi))^{-1}$ and $A_{n \geq 2} = 0$. The solution is now given by \begin{align} u(x,y) = \frac{\sinh(\pi(1- x))}{\sinh(\pi)} \, \sin(\pi y). \end{align}