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

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Solution

Let
$y = cos x + C$
Differentiating both sides w.r.t. $x$ then we get,
$y^{'} = (cos x + c)^{'}$
$y^{'} = - sin x$
Taking LHS
$y^{'} + sin x = - sin x + sin x$
$y^{'} + sin x = 0$
Thus $L H S = R H S$
Hence verified.

Final answer:
Hence, the function $y = cos x + C$ is a solution of the differential equation.

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