Question

Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: $y=cos x + C : y'+ sin x=0$

Answer 1

y=cos x + C
Differentiating both sides of this equation with respect to x, we get:

$y'=\frac{d}{dx}(\cos x+ C)$

$\Rightarrow y' =-\sin x$
Substituting the value of y' in the given differential equation, we get:

L.H.S=$y'+\sin x=\sin x +\sin x=0$=R.H.S.

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