Question
In each of the verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
$\text{y} = \text{cos} \ \text{x} + \text{C} \ :\ \text{y}' + \text{sin} \ \text{x} = 0$

Answer

Given: y = cos x + C .....(i)
To prove: y is a solution of the differential equation y' + sin x = 0 .....(ii)
Proof: From eq. (i), y' = - sin x
$\therefore$ L.H.S. of eq. (ii), y' + sin x = - sin x + sin x = 0 = R.H.S.

Hence, y given by eq. (i) is a solution of y' + sin x = 0.

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