If $x^2-y^2=t-\frac{1}{t}$, and $x^2 + y^2 = t^2 +\frac{1}{t^2}$ then which of the following is correct?

$x^3y\frac{dy}{dx}+1=0$

The correct answer is Option (4) → $x^3y\frac{dy}{dx}+1=0$

Given:

$x^2 - y^2 = t - \frac{1}{t}$

$x^2 + y^2 = t^2 + \frac{1}{t^2}$

Add the two equations:

$2x^2 = t^2 + t + \frac{1}{t^2} - \frac{1}{t}$

Subtract the two equations:

$2y^2 = t^2 - t + \frac{1}{t^2} + \frac{1}{t}$

Differentiate both original equations w.r.t. t:

$2x \frac{dx}{dt} - 2y \frac{dy}{dt} = 1 + \frac{1}{t^2}$

$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 2t - \frac{2}{t^3}$

Add them:

$4x \frac{dx}{dt} = 2t + 1 + \frac{1}{t^2} - \frac{2}{t^3}$

Subtract them:

$4y \frac{dy}{dt} = 2t - 1 + \frac{1}{t^2} + \frac{2}{t^3}$

Now use:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

After simplification, the relation becomes:

$x^3 y \frac{dy}{dx} + 1 = 0$

This matches the correct option.

Final Answer: $x^3 y \frac{dy}{dx} + 1 = 0$