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  • Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation. (iv) x^2 = 2 y^2 log y : (x^2 + y^2) dy/dx - xy = 0

Q2.    Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

            (iv)    x^2 = 2y^2\log y\qquad : \ (x^2 + y^2)\frac{dy}{dx} - xy = 0

Answers (1)

best_answer

Given,

x^2 = 2y^2\log y

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

\\ 2x = (2y^2.\frac{1}{y} + 2(2y)\log y)\frac{dy}{dx} = 2(y + 2y\log y)\frac{dy}{dx} \\ \implies \frac{dy}{dx} = \frac{x}{y(1+ 2\log y)}

Putting \frac{dy}{dx}\ and \ x^2 values in LHS

\\ (2y^2\log y + y^2)\frac{dy}{dx} - xy = y^2(2\log y + 1)\frac{x}{y(1+ 2\log y)} -xy \\ = xy - xy = 0 = RHS

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

Posted by

HARSH KANKARIA

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