$Solve \frac{d y}{d x} = x^{3} y^{3} - x y$

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$(y^{2} + 1 + c e^{y^{2}}) = 1$

$(x^{2} + 1 + c e^{x^{2}}) y^{2} = 1$

$(x + 1 + c e^{y^{2}}) x^{2} = 1$

$(y^{2} + 1 + c e^{y^{2}}) = x$

answer is B.

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Detailed Solution

$\begin{array}{l} G.E \frac{d y}{d x} + x y = x^{3} y^{3} \Rightarrow \frac{1}{y^{3}} \frac{d y}{d x} + \frac{x}{y^{2}} = x^{3} \\ Put \frac{1}{y^{2}} = z \Rightarrow \frac{d z}{d x} - 2 x z = - 2 x^{3}; I . F = e^{- x^{2}} \\ General solution z e^{- x^{2}} \int - 2 x^{3} e^{- x^{2}} d x \\ Put x^{2} = t \Rightarrow z e^{- x^{2}} = - \int t e^{- t} d t \\ \Rightarrow \frac{1}{y^{2}} e^{- x^{2}} = x^{2} e^{- x^{2}} + e^{- x^{2}} + c \\ \Rightarrow (x^{2} + 1 + c e^{x^{2}}) y^{2} = 1 \end{array}$