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  • Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: xy equals log y plus C : y' = y^2/1 - xy (xy not equal to 1)

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

        xy = \log y + C\qquad :\ y' = \frac{y^2}{1 - xy}\ (xy\neq 1)

Answers (1)

best_answer

Given,

xy = \log y + C

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

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

Substituting the values of y' in LHS,

y' = \frac{y^2}{1-xy} = RHS

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

Posted by

HARSH KANKARIA

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