I'm having a difficult time understanding how one would take $ y = Ae^{\sqrt{\lambda}ix} + Be^{-\sqrt{\lambda}ix} $ and end up with $ y = A \cos{\sqrt{\lambda}x} + B\sin{\sqrt{\lambda}x} $ knowing that $ \lambda > 0 $.
Would it be easiest to start out by just plugging in Euler's Formulas $e^{i\theta} = \cos(\theta) + i\sin(\theta)$ and $e^{-i\theta} = \cos(\theta) - i\sin(\theta)$? The only problem I have is I end up with this very long equation of $ y = A \cos{\sqrt{\lambda}x} + Ai\sin{\sqrt{\lambda}x} + B\cos{\sqrt{\lambda}x} - Bi\sin{\sqrt{\lambda}x}$. I'm really confused as to how this long expression can be reduced to the $ y = A \cos{\sqrt{\lambda}x} + B\sin{\sqrt{\lambda}x} $ it is asking for.
Any help is appreciated!
You can just add the complex coeffiecients together so that $$y = A \cos(t) + Ai\sin(t) + B\cos(t) - Bi \sin(t)= (A + B)\cos(t) + (Ai - Bi)\sin(t).$$