Question

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

Answer 1

Let
$y=\cos x+C$
Differentiating both sides w.r.t. $x$ then we get,
$y^{\prime }=(\cos x+c{)}^{\prime }$
$y^{\prime }=-\sin x$
Taking LHS
$y^{\prime }+\sin x=-\sin x+\sin x$
$y^{\prime }+\sin x=0$
Thus $LHS=RHS$
Hence verified.

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