Class 12 Mathematics - Chapter Differential Equations NCERT Solutions | y = cosx + C : y' + sinx = 0&nb
y = cosx + C : y' + sinx = 0
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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