Question:
$y=\cos x+\mathrm{C} \quad: y^{\prime}+\sin x=0$
Solution:
$y=\cos x+C$
Differentiating both sides of this equation with respect to x, we get:
$y^{\prime}=\frac{d}{d x}(\cos x+\mathrm{C})$
$\Rightarrow y^{\prime}=-\sin x$
Substituting the value of $y^{\prime}$ in the given differential equation, we get:
L.H.S. $=y^{\prime}+\sin x=-\sin x+\sin x=0=$ R.H.S.
Hence, the given function is the solution of the corresponding differential equation.
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