If x2 - y2 - t - 1-t and x4 - y4 - t2 - 1-t2 -then x3 y dy-dx equals

X^2 + y^2 = t 1/t and x^4 + y^4 = t_2 + 1/t^2 ∴ x^2 y^2 = 1 ⇒ x^2. 2y dy/dx + y^2 . 2x = 0 ⇒ x^3 y dy/dx = x^2 y^2 = 1 ...

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Formation of a Differential Equation from a General Solution

If x2 + y2 = ...

Question

If $x^{2} + y^{2} = t - \frac{1}{t}$ and $x^{4} + y^{4} = t_{2} + \frac{1}{t^{2}}$ ,then $x^{3} y \frac{d y}{d x}$ equals

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

x^{4} + y^{4} = t_{2} + \frac{1}{t^{2}}

x^{3} y \frac{d y}{d x}

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

x^{4} + y^{4} = t^{2} + \frac{1}{t^{2}}

x^{3} y \frac{d y}{d x}

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

x^{4} + y^{4} = t^{2} + \frac{1}{t^{2}}

x^{3} y \frac{d y}{d x}

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

x^{4} + y^{4} = t^{2} + \frac{1}{t^{2}}

x^{3} y \frac{d y}{d x} =

x^{2} + y^{2} = t + \frac{1}{t}, x^{4} + y^{4} = t^{2} + \frac{1}{t^{2}} \Rightarrow x^{3} y \frac{d y}{d x} =

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