The solution of the differential equation $ydx = xdy + 3 x^{2} y^{2} e^{x^{3}} dx = 0$ satisfying y = 1 when x = 1 is

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$y (e^{x^{3}} - (1 + 2 e)) + x = 0$

$y (e^{x^{3}} + (1 + 2 e)) - x = 0$

$y (e^{x^{3}} + (1 - e)) + x = 0$

$y (e^{x^{3}} - (1 + 2 e)) - x = 0$

answer is B.

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Detailed Solution

$\begin{array}{l} (y - 3 x^{2} y^{2} e^{x^{3}}) d x = x d y \\ \frac{d y}{d x} = \frac{y}{x} - 3 x y^{2} e^{x^{3}} \\ \Rightarrow \frac{1}{y^{2}} \frac{d y}{d x} = \frac{1}{y} (\frac{1}{x}) - 3 x e^{x^{3}} \\ P u t \frac{1}{y} = v \\ \Rightarrow \frac{- 1}{y^{2}} \frac{d y}{d x} = \frac{d v}{d x} \\ ∴ \frac{- d v}{d x} = \frac{1}{x} v - 3 x e^{x^{3}} \\ \Rightarrow \frac{d v}{d x} + \frac{1}{x} v = 3 x e^{x^{3}} \end{array}$

$\begin{array}{l} I . F = e^{\int \frac{1}{x} d x} = e^{\log x} = x \\ S o l u t i o n i s \\ v (I . F) = \int 3 x e^{x^{3}} (I . F) d x \\ \Rightarrow \frac{x}{y} = \int 3 x e^{x^{3}} x d x \\ = \int 3 x^{2} e^{x^{3}} d x \\ = e^{x^{3}} + c \\ \Rightarrow x = y (e^{x^{3}} + c) \\ p u t x = 1, y = 1 \\ \Rightarrow 1 = e + c \\ \Rightarrow c = e - 1 \\ ∴ x = y (e^{x^{3}} + 1 - e) \end{array}$