Question:
y = cosx + C : y^' + sinx = 0

Answer:

y = cosx + C

Differentiating both sides of this equation with respect to x, we get:

\begin{align}y^{'}=\frac{d}{dx}(cosx + C)\end{align}

=> y^' = - sinx

Substituting the value of y^' in the given differential equation, we get:

L.H.S. = y^' + sinx = - sinx + sinx = 0 = R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

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Question:
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Question:
y = cosx + C : y^' + sinx = 0