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The family of curve whose differential equation is $(x y + 1) (3 x^{2} y d x - x^{3} d y) = (x^{6} + y^{2}) (x d y + y d x)$ is

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${sin}^{- 1} x^{3} = y ln (1 + x y) + c$

none of these

${tan}^{- 1} (\frac{x^{3}}{y}) = ln (1 + x y) + C$

${sin}^{- 1} (\frac{x^{3}}{y}) = ln (1 - x y) + c$

answer is A.

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Detailed Solution

$(x y + 1) (3 x^{2} y d x - x^{3} d y) = (x^{6} + y^{2}) (x d y + y d x) \Rightarrow (x y + 1) d (\frac{x^{3}}{y}) = ({(\frac{x^{3}}{y})}^{2} + 1) d (x y)$

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