Browse Questions

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

Toolbox:
• Differentiation of $\cos x$ is $- \sin x$
Step 1:
Given $y = \cos x + C$
Differentiating on both sides we get,
$y' = -\sin x + 0$
Step 2:
Bringing all the terms to the LHS
$y' + \sin x = 0$
Hence the answer is the solution for the given function.