Let consider the following ODE: $$ y(x)'' = \gamma^2 y(x) $$ The general solution should be: $$ y(x) = C_1 e^{- \gamma x} + C_2 e^{\gamma x} $$
Is it correct to assume solutions of the type $C e^{-\gamma |x|}$ instead? I suppose the latter does not meet the equation in $x=0$, because here the absolute value is not derivable, right? Or is there a stricter motivation?
Thank you.
It depends on which domain you work. If you want a solution $y : \mathbb{R} \rightarrow \mathbb{R}$, then $f(x) = Ce^{-\gamma|x|}$ is not a solution, mainly for the two following reasons :
$f''(0)$ is not defined, so that this function doesn't solve the initial equation at $x=0$.
Even if we admit distributional solutions (and thus distributional derivatives), it still doesn't solve the problem at $x=0$; indeed, one has $f'(x) = -\gamma Ce^{-\gamma|x|}\,\mathrm{sgn}(x)$ and $f''(x) = \gamma^2Ce^{-\gamma|x|}\mathrm{sgn}^2(x) - 2\gamma Ce^{-\gamma|x|}\delta(x) \equiv \gamma^2\mathrm{sgn}^2(x)f(x)-2\gamma C\delta(x) \neq \gamma^2f(x)$ when $x=0$, where $\delta$ is the Dirac delta function.
However, if you are looking for a solution $y : D \rightarrow \mathbb{R}$, where $0 \not\in D$, then it is easy to check that $Ce^{\pm\gamma|x|}$ are solutions.