If the curve $y = ⨍ (x)$ passes through (1, 2) and satisfies the differential equation $y (1 + xy) dx - xdy = 0$ , then

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$⨍ (x)= \frac{4 x}{1 - 2 x^{2}}$

$⨍ (x)= \frac{x + 1}{x^{2} + 1}$

$⨍ (x)= \frac{2 x}{2 - x^{2}}$

$⨍ (x)= \frac{x - 1}{4 - x^{2}}$

answer is A.

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Detailed Solution

$\begin{array}{l} G i v e n y (1 + x y) d x = x d y \\ \Rightarrow \frac{d y}{d x} = \frac{y (1 + x y)}{x} \\ \Rightarrow \frac{d y}{d x} = \frac{y}{x} + y^{2} \\ \Rightarrow \frac{1}{y^{2}} \frac{d y}{d x} - \frac{1}{y} (\frac{1}{x}) = 1 \\ L e t \frac{1}{y} = v \\ \Rightarrow \frac{- 1}{y^{2}} \frac{d y}{d x} = \frac{d v}{d x} \\ \Rightarrow - \frac{d v}{d x} - \frac{1}{x} v = 1 \\ \Rightarrow \frac{d v}{d x} + (\frac{1}{x}) v = - 1 \end{array}$

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