I am not sure of how to go about obtain the $g\left(y\right)$ in terms of a hyperbolic function. The partial differential equation I am dealing with is
Find an infinite series representation for the solution to the equilibrium problem:
$$u_{xx}+u_{yy}=0:0<x<a, 0<y<b$$ $$u_{x}\left(0,y\right)=u_{x}=0:0<y<b$$ $$u\left(x,0\right)=f\left(x\right), u\left(x,b\right)=0:0<x<a$$
I have to solve this using separation of variables, and I have to find the solution of $y$ in terms of a hyperbolic function. What I've gotten for my separation of variables is (the book I go by makes the right side as $-\lambda$ instead of just $\lambda$, so I have followed it for consistency):
$$\frac{p^{\prime\prime}\left(x\right)}{p\left(x\right)}=\frac{-g^{\prime\prime}\left(y\right)}{g\left(y\right)}=-\lambda$$
For $\lambda>0$, I've gotten that $\lambda_{n}=\frac{n^2\pi^2}{a^2}$, $p_{n}\left(x\right)=\cos\left(\frac{n\pi x}{a}\right)$. For $g\left(y\right)$,
$$g\left(y\right)=c_{1}e^{\frac{-n\pi y}{a}}+c_{2}e^{\frac{n\pi y}{a}}$$
How do I go about converting $g\left(y\right)$ into terms of a hyperbolic function? My textbook says that $$g_{n}\left(y\right)=c_{n}\sinh\left(\frac{n\pi y}{a}\right)+d_{n}\cosh\left(\frac{n\pi y}{a}\right)$$ but I am not sure how they got to this.
Hint: $e^x = \cosh x + \sinh x$, $e^{-x} = \cosh x - \sinh x$.