Laplace's Equation with One Inhomogeneous Boundary Condition While solving Laplace's equation,
$$
\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}=0,
$$
with Dirichlet boundary conditions
$$\begin{align}
u(x,0)&=f_1(x),\\
u(x,b)&=0,\\
u(0,y)&=0,\\
u(a,y)&=0,
\end{align}$$
and assuming that the solution has the separable form $u(x,y)=X(x)Y(y)$, I ran into the case where I have to solve the ODE
$$
Y''(y)=\lambda Y(y),
$$
which has only one homogeneous boundary condition.
My book immediately concludes that its solution is
$$
Y(y)=c_1\sinh\frac{n\pi}a(y-b)\tag{1}
$$
where $\lambda=\left(\frac{n\pi}a\right)^2$ was previously computed.
I cannot understand how $(1)$ was obtained. Thanks in advance for your help!
Edit 1
If it helps to know, I computed that
$$
X_n(x)=A_n\sin\frac{n\pi}ax,
$$
where $n=1,2,\dots$.
 A: Your boundary condition is $Y(b) = 0$.   Looking for solutions of the differential equation of the form $u(y) = e^{ry}$, we find $r^2 = \lambda$, so $r = \pm \sqrt{\lambda}$.  Now
the general solution of the differential equation can  be written as $u(y) = c_1 e^{\sqrt{\lambda} y} + c_2 e^{-\sqrt{\lambda} y}$.  If the boundary condition was $Y(0) = 0$, we would immediately see that $c_1 + c_2 = 0$, so $u(y) = c_1 \left(e^{\sqrt{\lambda} y} - e^{-\sqrt{\lambda} y}\right) = 2 c_1 \sinh(\sqrt{\lambda} y)$.  But note that $v(y) = u(y-b)$ is a solution of the differential equation if $u$ is, and $v(b) = u(0)$.  So we get
$v(y) = 2 c_1 \sinh(\sqrt{\lambda} (y - b))$.  And now absorb the $2$ into the arbitrary 
constant $c_1$ to get the book's $Y(y)$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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*

*I like to take advantage of homogeneous boundary conditions:
\begin{align}
&\on{u}\pars{x,y} = \sum_{n = 1}^{\infty}\on{a}_{n}\pars{y}
\sin\pars{k_{n}x}\,,\qquad k_{n} \equiv n\,{\pi \over a}
\end{align}

*$\ds{\on{u}\pars{x,y}}$ must satisfies the Laplace differential equation:
$$
\sum_{n = 1}^{\infty}\bracks{\on{a}_{n}''\pars{y} - k_{n}^{2}\on{a}_{n}\pars{y}}\sin\pars{k_{n}x} = 0
$$

*Multiply both members by $\ds{2\sin\pars{k_{n}x}/a}$ and integrate over $\ds{x \in \pars{0,a}}$:
$$
\on{a}_{n}''\pars{y} - k_{n}^{2}\on{a}_{n}\pars{y} = 0
$$
The solutions are linear combinations of $\ds{\expo{\pm k_{n}y}}$. In order to satisfy $\ds{\on{u}\pars{x,b} = 0}$, I choose $\ds{b_{n}\sinh\pars{k_{n}\bracks{b - y}}}$ such that the solution becomes
$$
\on{u}\pars{x,y} =
\sum_{n = 1}^{\infty}b_{n}\sinh\pars{k_{n}\bracks{b - y}}
\sin\pars{k_{n}x}
$$

*It must satisfy the boundary condition
$\ds{\on{u}\pars{x,0} = \on{f}_{1}\pars{x}}$. Namely,
$$
\sum_{n = 1}^{\infty}b_{n}\sinh\pars{k_{n}b}\sin\pars{k_{n}x}
= \on{f}_{1}\pars{x}
$$
Again, multiply both members by $\ds{2\sin\pars{k_{n}x}/a}$ and integrate over $\ds{x \in \pars{0,a}}$:
\begin{align}
& b_{n}\sinh\pars{k_{n}b}
= {2 \over a}\int_{0}^{a}\on{f}_{1}\pars{x}\sin\pars{k_{n}x}
\,\dd x \equiv \varphi_{n}
\\ \implies &\
b_{n} = {\varphi_{n} \over \sinh\pars{k_{n}b}}
\\[5mm] \implies &\
\bbx{\on{u}\pars{x,y} =
\sum_{n = 1}^{\infty}{\varphi_{n}  \over \sinh\pars{k_{n}b}}
\sinh\pars{k_{n}\bracks{b - y}} 
\sin\pars{k_{n}x}} \\ &
\end{align}
