If xy 2 -1- then prove that 2 dy dx - y 3 -0-

We have, xy^2=1 ⇒ x=1y^2⇒dxdy= 2y^3⇒dydx= y^32⇒ 2dydx+y^3=0 ...

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Power of a Power

If xy 2 =1,...

Question

If ${x y}^{2} = 1$ , then prove that $2 \frac{d y}{d x} + y^{3} = 0$ .

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x y^{2} = 1,

2 \frac{d y}{d x} + y^{3} = 0

y = x^{3} log (\frac{1}{x})

x \frac{d^{2} y}{d x^{2}} - \frac{2 d y}{d x} + 3 x^{2} = 0

y = {tan}^{- 1} \frac{u}{\sqrt{1 - u^{2}}}

x = {sec}^{- 1} \frac{1}{2 u^{2} - 1}

u \in (0, \frac{1}{\sqrt{2}})

\cup (\frac{1}{\sqrt{2}}, 1)

2 \frac{d y}{d x} + 1 = 0

\frac{d y}{d x}

x^{3} + x^{2} y + x y^{2} + y^{3} = 81

\sqrt{1 - x^{6}} + \sqrt{1 - y^{6}} = a^{3} (x^{3} - y^{3})

\frac{d y}{d x} = \frac{x^{2}}{y^{2}} \sqrt{\frac{1 - y^{6}}{1 - x^{6}}}

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