Solution to second order non constant differential equation without first derivative Cheers!
So I am studying the book ‘Introduction to quantum mechanics’ by David J. Griffiths for my introductory course of quantum mechanics. On page 51 at the bottom it introduces the differential expression:
\begin{equation}
\Phi’’ = \xi^2 \Phi
\end{equation}
Without saying much further, it presents the general solution as:
\begin{equation}
\Phi = A e^{-\xi^2/2} + B e^{\xi^2/2}
\end{equation}
Indeed, when I check this solution with the differential equation, I can see that it can match it. However I am still completely missing on how one would arrive from the differential equation to the general solution. 
I have already been trying several different methods, all unsuccessful, so I would indeed appreciate any help I could get.

For solving $y''-(\xi^2-K)y=0$ we have solution
$$y=Ae^{\sqrt{\xi^2-K}}+Be^{-\sqrt{\xi^2-K}}~~~~~,~~~~~\xi^2-K>0$$
with $\xi>>K$
$$\sqrt{\xi^2-K}=\sqrt{-K}\sqrt{1-\dfrac{\xi^2}{K}}=\sqrt{-K}\left(1-\dfrac{\xi^2}{K}\right)^\frac12\approx\sqrt{-K}\left(1-\frac12\dfrac{\xi^2}{K}\right)\approx\sqrt{-K}\left(-\frac12\dfrac{\xi^2}{K}\right)\approx\dfrac{\xi^2}{2i\sqrt{K}}\approx\dfrac{\xi^2}{2}$$
