I'm currently working through a boundary value problem in my PDEs course. When searching for eigenvalues and eigenfunctions, I referred to my notes from class.
The following is exactly what my professor had for our particular problem.
$$\omega (x) = Ae^{x\sqrt{-\lambda}} + Be^{-x\sqrt{-\lambda}} $$ Recall the hyperbolic functions: $$ \sinh(z) = \frac{e^z-e^{-z}}{2} \\ \\ \cosh(z) = \frac{e^z+e^{-z}}{2} $$ Rewriting $\omega (x)$, $$ \omega (x) = C\cosh(x\sqrt{-\lambda}) + D\sinh(x\sqrt{-\lambda}) $$
In my attempt to figure out how my professor got the trigonometric version of $\omega$, I found that:
$$ \omega (x) = Ae^{x\sqrt{-\lambda}} + Be^{-x\sqrt{-\lambda}} $$ $$ = \frac{1}{2} (ae^{x\sqrt{-\lambda}}+be^{-x\sqrt{-\lambda}}) + \frac{1}{2} (ae^{x\sqrt{-\lambda}}+be^{-x\sqrt{-\lambda}}) $$ $$ = 2E\cosh(x\sqrt{-\lambda}) $$
Where $a$, $b$, and $E$ are just arbitrary constants.
Am I missing something, or did my professor make a mistake? My bet is that there is something I'm missing.
Thank you!
Hint:
Write $e^t=\cosh t+\sinh t$ and $e^{-t}=\cosh t -\sinh t$.
In you derivation, how can you make $\cosh$ suddenly appear? There is nothing in the preceding line that allows you to do that.