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.


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:

$$u(x) = e^{\gamma x + x_0 + q} + e^{-\gamma x - x0 + q}.$$


$$u(x) = 2e^q \left(\frac{e^{\gamma x + x_0} + e^{-\gamma x - x_0}}{2}\right) = 2e^q \cosh(\gamma x + x_0) = D\cosh(\gamma x + x_0).$$

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:

$$ D = 2e^{\frac{1}{2}\log\left(AB\right)} = 2\sqrt{AB}.$$


$$u(x) = 2\sqrt{AB} \cosh\left(\gamma x + \frac{1}{2}\log\left(\frac{A}{B}\right) \right).$$

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.


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