Question 3

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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