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

3. Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

       y = \cos x + C\qquad :\ y' - \sin x = 0

Answers (1)

best_answer

Given,

y = \cos x + C

Now, differentiating both sides w.r.t. x,

\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(cosx + C) = -sinx

Substituting the values of y’ in the given differential equations,

y' - \sin x = -sinx -sinx = -2sinx \neq RHS.

Therefore, the given function is not the solution of the corresponding differential equation.

Posted by

HARSH KANKARIA

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