So im trying to solve an equation $$ u_{xx}+4u_{tt}=0, \ \ \ \ \ 0<x < \pi, \ \ \ \ 0 < t < 2 $$ $$ u(0,y)=u(\pi,t)=0, \ \ \ \ \ 0 \le t \le 2 $$ After seperation of variables i get that: $$ G(t)=c_1 e^{\sqrt{\frac{k}{4}}t}+c_1 e^{-\sqrt{\frac{k}{4}}t} $$ When i look at what the answer should be it says $$ G(t)=A_n \cosh(\frac{n}{2}t)+ B_n \sinh(\frac{n}{2}t) $$ I have no errors in my speration of variables i think, so no need to write all of that. And i also have correct $k = -n^2$ for $n = 1,2,3...$. So im Wondering about how I can get it on that other form from what i origonaly got.
Answer should also be here, Problem number 3a.
In general,
$$e^{x}=\cosh x+\sinh x\quad\text{ and }\quad e^{-x}=\cosh x-\sinh x$$
We have $G''+\frac k4G=0$. Suppose the solution is of the form $e^{\gamma t}$. Then $e^{\gamma t}(\gamma^2+\frac k4)=0$. Therefore, $\gamma=\pm\sqrt{-\frac k4}$.
Therefore, $G(t)=c_1e^{\sqrt{-\frac k4}t}+c_2e^{-\sqrt{-\frac k4}t}$
From here, we get
\begin{align*} c_1e^{\sqrt{-\frac k4}t}+c_2e^{-\sqrt{-\frac k4}t}&=c_1e^{\sqrt{\frac {n^2}4}t}+c_2e^{-\sqrt{\frac {n^2}4}t}\\ &=c_1e^{\frac {n}2t}+c_2e^{-\frac {n}2t}\\ &=c_1\left(\cosh\frac n2 t+\sinh\frac n2 t\right)+c_2\left(\cosh\frac n2 t-\sinh\frac n2 t\right)\\ &=C_1\cosh\frac n2 t+C_2\sinh\frac n2 t. \end{align*}