Answer
$\{e^{\sqrt 2x},e^{-\sqrt 2x}\}$
Work Step by Step
We are given:
$y''-2y=0$
Let $y=e^{ux}$
$y''=u^2 e^{ux}$
The equation becomes:
$u^2e^{ux}-2e^{ux}=0$
$(u^2-2)e^{ux}=0$
$u^2-2=0$
$u=\pm \sqrt 2$
Since $u$ can not be negative, $\rightarrow u=\sqrt 2$
Therefore $y=e^{\sqrt 2x}c_1+e^{-\sqrt 2x}c_2$
Hence, the basis for the given equation is $\{e^{\sqrt 2x},e^{-\sqrt 2x}\}$
