Answer
$y(x)=C_1e^{-\sqrt 2x}+C_2 e^{\sqrt 2x}$
Work Step by Step
Solve the characteristic equation for the differential equation. $$(D^2-2)y=0$$
Factor and solve for the roots. $$(r-\sqrt 2)(r+\sqrt 2)=0$$ $$r_1=-\sqrt 2, r_2=\sqrt 2$$ as roots.
This implies that there are two independent solutions to the differential equation $y_1(x)=e^{-\sqrt 2x}$ and $y_2= e^{\sqrt 2 x}$.
Therefore, the general equation is equal to $y(x)=C_1e^{-\sqrt 2x}+C_2 e^{\sqrt 2x}$.