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

8.  In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

       y - cos y = x \qquad :(\ y\sin y + \cos y + x) y' = y

 

Answers (1)

best_answer

Given,

y - cos y = x

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

\frac{\mathrm{d}y }{\mathrm{d} x} +siny\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(x) = 1

\implies y' + siny.y' = 1

\implies y'(1 + siny) = 1

\implies y' = \frac{1}{1+siny}

Substituting the values of y and y' in LHS,

(\ (x+cosy)\sin y + \cos y + x) (\frac{1}{1+siny})

= [x(1+siny) + cosy(1+siny)]\frac{1}{1+siny}

= (x + cosy) = y = RHS

Therefore, the given function is a solution of the corresponding differential equation.

Posted by

HARSH KANKARIA

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