$y = {e^x} + 1:y'' - y' = 0$

.: We have $y = {e^x} + 1$

…(1)
Differentiating (1) w.r.t. $x$,

we get
$y' = \cfrac{d}{{dx}}({e^x} + 1) = {e^x}$

and $y'' = \cfrac{d}{{dx}}({e^x}) = {e^x} \Rightarrow y'' - y' = 0$

Hence, $y = {e^x} + 1$ is a solution of the given
differential equation.