$y=e^{x}+1 \quad: \quad y^{\prime \prime}-y^{\prime}=0$
$y=e^{x}+1$
Differentiating both sides of this equation with respect to x, we get:
$\frac{d y}{d x}=\frac{d}{d x}\left(e^{x}+1\right)$
$\Rightarrow y^{\prime}=e^{x}$ ...(1)
Now, differentiating equation (1) with respect to x, we get:
$\frac{d}{d x}\left(y^{\prime}\right)=\frac{d}{d x}\left(e^{x}\right)$
$\Rightarrow y^{\prime \prime}=e^{x}$
Substituting the values of $y^{\prime}$ and $y^{\prime \prime}$ in the given differential equation, we get the L.H.S. as:
$y^{\prime \prime}-y^{\prime}=e^{x}-e^{x}=0=$ R.H.S.
Thus, the given function is the solution of the corresponding differential equation.
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