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 $n=1,2,\dots$.