If xy^2 = 1, prove that 2 dy dx +y^3=0 - Vidyadip

Question

If $xy^2 = 1$, prove that $2\frac{\text{dy}}{\text{dx}}+\text{y}^3=0$

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Answer

We have $xy^2 = 1 .....(i)$
Differentiating with respect to x, we get,
$\frac{\text{d}}{\text{dx}}(\text{xy}^2)=\frac{\text{d}}{\text{dx}}(1)$
$\Rightarrow\text{x}\frac{\text{d}}{\text{dx}}(\text{y}^2)+\text{y}^2\frac{\text{d}}{\text{dx}}(\text{x})=0$
$\Rightarrow\text{x}(2\text{y})\frac{\text{d}}{\text{dx}}+\text{y}^2(1)=0$
$\Rightarrow2\text{xy}\frac{\text{dy}}{\text{dx}}=-\text{y}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\text{y}^2}{2\text{xy}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\text{y}}{2\text{x}}$
Put $\text{x}=\frac{1}{\text{y}^2}$ from equation (i)
$ \Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\text{y}}{2\Big(\frac{1}{\text{y}^2}\Big)}$
$\Rightarrow2\frac{\text{dy}}{\text{dx}}=-\text{y}^3$
$\Rightarrow2\frac{\text{dy}}{\text{dx}}+\text{y}^3=0$

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