The family of curve satisfying differential equation $(x + x^{2} y) (2 y d x - x d y) = (x^{4} + y^{2}) (x d y + y d x)$ is

{tan}^{- 1} (\frac{x^{2}}{y}) = ln | 1 + x y | + c

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{sin}^{- 1} (\frac{x^{2}}{y}) = ln | 1 + x y | + c

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{sin}^{- 1} x^{2} = ln | 1 + x y | + c

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None of these

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Solution

The correct option is B ${tan}^{- 1} (\frac{x^{2}}{y}) = ln | 1 + x y | + c$
$\begin{matrix} (x + x^{2} y) (2 y d x - x d y) = (x^{4} + y^{2}) (x d y + y d x) \Rightarrow x (1 + x y) (2 y d x - x d y) = (x^{4} + y^{2}) d (x y) \Rightarrow \frac{x (2 y d x - x d y)}{x^{4} + y^{2}} = \frac{d (x y)}{1 + x y} \Rightarrow \frac{2 x y d x - x^{2} d y}{y^{2}} . \frac{y^{2}}{x^{4} + y^{2}} = \frac{d (x y)}{1 + x y} \Rightarrow d (\frac{x^{2}}{y}) . \frac{1}{\frac{x^{4}}{y^{2}} + 1} = \frac{d (x y)}{1 + x y} \Rightarrow \frac{d (\frac{x^{2}}{y})}{{(\frac{x^{2}}{y})}^{2} + 1} = \frac{d (x y)}{1 + x y} \Rightarrow {tan}^{- 1} (\frac{x^{2}}{y}) = ln | 1 + x y | + c \end{matrix}$

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