Verify that the given function $y=e^{x}+1$ is a solution of the differential equation $y^{\prime \prime}-y^{\prime}=0$.

(A) Given function: $y=e^{x}+1$
Differentiating both sides with respect to $x$,we get:
$\frac{dy}{dx} = \frac{d}{dx}(e^{x}+1)$
$\Rightarrow y^{\prime} = e^{x}$ --- $(1)$
Now,differentiating equation $(1)$ with respect to $x$,we get:
$\frac{d}{dx}(y^{\prime}) = \frac{d}{dx}(e^{x})$
$\Rightarrow y^{\prime \prime} = e^{x}$
Substituting the values of $y^{\prime \prime}$ and $y^{\prime}$ in the given differential equation $y^{\prime \prime}-y^{\prime}=0$:
$L.H.S. = y^{\prime \prime}-y^{\prime} = e^{x} - e^{x} = 0$
$R.H.S. = 0$
Since $L.H.S. = R.H.S.$,the given function is a solution of the differential equation.