Standard hyperbolic solution to 2nd order ODE, equivalent forms. Consider the following 2nd order ODE:
$$\frac{d^2u}{dx^2}-\gamma^2u=0.$$
This equation has solutions of the form $$u(x)=Ae^{\gamma x}+Be^{-\gamma x},$$
or equivalently $$u(x)=(A+B)\cosh{\gamma x}+(A-B)\sinh{\gamma x}\equiv C_1\cosh{\gamma x}+C_2\sinh{\gamma x}.$$
I would like to demonstrate, that both are equivalent to $$u(x)=D\cosh{(\gamma x +x_0)}.$$
How do you go about doing this (if it is indeed possible)? I tried to do it the same way you do it for regular trig functions but cosh is undefined for values less than 1. 
 A: Notice that:
$$u(x) = Ae^{\gamma x}+Be^{-\gamma x} = e^{\log(A) + \gamma x} + e^{\log(B) - \gamma x}.$$
We want to write the previous as follows:
$$u(x) = e^{\gamma x + x_0 + q} + e^{-\gamma x - x0 + q}.$$
Then:
$$u(x) = 2e^q \left(\frac{e^{\gamma x + x_0} + e^{-\gamma x - x_0}}{2}\right) = 2e^q \cosh(\gamma x + x_0) = D\cosh(\gamma x + x_0).$$
To this aim, we must solve the following linear system:
$$\begin{cases}
\log(A) + \gamma x = \gamma x + x_0 + q\\
\log(B) - \gamma x = -\gamma x - x_0 + q
\end{cases}\Rightarrow \begin{cases}
x_0 + q = \log(A) \\
-x_0 + q = \log(B)
\end{cases} \Rightarrow 
\begin{cases}
x_0 = \frac{1}{2}\log\left(\frac{A}{B}\right)\\
q = \frac{1}{2}\log\left(AB\right)
\end{cases}.$$
Since $D= 2e^q$, then:
$$ D = 2e^{\frac{1}{2}\log\left(AB\right)} = 2\sqrt{AB}.$$
Finally:
$$u(x) = 2\sqrt{AB} \cosh\left(\gamma x + \frac{1}{2}\log\left(\frac{A}{B}\right) \right).$$
Notice that, in order to have well defined real solution, then you must pay care to the sign of $A$ and $B$. Indeed, they must be both positive or both negative. Also, $A=0$ and/or $B = 0$ are forbidden.
