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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Verify that the given function(explicit or implicit)is a solution of the corresponding differential equation $y=\cos x+C\qquad:y'+\sin x=0$

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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.
answered Aug 19, 2013 by sreemathi.v
 

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