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  • Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: x + y = tan inverse y : y^2 y' plus y^2 + 1 = 0

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

        x + y = \tan^{-1}y\qquad :\ y^2y' + y^2 + 1 = 0

Answers (1)

best_answer

Given,

x + y = \tan^{-1}y

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

\\ 1 + \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1}{1 + y^2}\frac{\mathrm{d} y}{\mathrm{d} x}\\ \\ \implies1+y^2 = y'(1-(1+y^2)) = -y^2y' \\ \implies y' = -\frac{1+y^2}{y^2}

Substituting the values of y' in LHS,

y^2(-\frac{1+y^2}{y^2}) + y^2 + 1 = -1- y^2+ y^2 +1 = 0 = RHS

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

Posted by

HARSH KANKARIA

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