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$y=e^{x}+1 \quad: \quad y^{\prime \prime}-y^{\prime}=0$



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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