Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:
$y = e^{x} + 1 : y " - y^{'} = 0$

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Solution

Let
$y = e^{x} + 1$
Differentiating both sides w.r.t. $x,$
We get,
$y^{'} = (e^{x} + 1)^{'} = e^{x} + 0$
$y^{'} = e^{x}$
Again, Differentiating both sides w.r.t. x,
We get.
$y "= (e^{x})^{'}$
$y^{''} = e^{x}$
Taking LHS\\
$y^{''} - y^{'} = e^{x} - e^{x}$
$y^{''} - y^{'} = 0$
Thus $L H S = R H S$
Hence verified.

Final answer:
Hence, the function $y = e^{x} + 1$ is a solution of the differential equation $y^{''} - y^{'} = 0.$

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