General solution to eigenvalue problem when $\lambda $ is negative.

Question

I have a simple one for you guys.

So I was reading this PDEs book which regularly discusses the eigenvalue problem

$F''(x)+\lambda F(x)=0$.

For $\lambda =-\mu^2 $, i.e negative eigenvalues, that is if

$F''(x)-\mu^2F(x)=0$

then the general solution is

$F(x)=Ae^{\mu x}+Be^{-\mu x}\quad\quad(1)$.

The author then expresses the solution as

$F(x)=C\cosh\mu x+D\sinh\mu x\quad\quad (2)$

for convenience in applying the BCs.

How did he convert the exponential solution (1) into the solution (2) involving hyperbolic functions ?

Thanks.

Answer 1

Let $Ae^{\mu x}+Be^{-\mu x}\equiv C\frac{e^{\mu x}+e^{-\mu x}}{2}+ D\frac{e^{\mu x}-e^{-\mu x}}{2}$ then $$\begin{cases} A=\frac{C+D}{2}\\B=\frac{C-D}{2} \end{cases}$$ or, in an alternative notation $$\begin{cases} A+B=C\\A-B=D \end{cases}$$ so $A,B$ in (1) and $A,B$ in (2) are just different.