The curve satisfying the differential equation, $ydx - (x + 3 y^{2}) dy = 0$ and passing through the point (1,1), also passes through the point :

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$(\frac{1}{4}, - \frac{1}{2})$

$(\frac{1}{3}, - \frac{1}{3})$

$(- \frac{1}{3}, \frac{1}{3})$

answer is A.

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Detailed Solution

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

$\begin{array}{l} \Rightarrow \frac{x}{y} = 3 y + c \\ I t p a s s e s t h r o u g h (1, 1) \\ \Rightarrow 1 = 3 + c \\ \Rightarrow c = - 2 \\ \Rightarrow \frac{x}{y} = 3 y - 2 \\ I t a l s o p a s s e s t h r o u g h (\frac{- 1}{3}, \frac{1}{3}) \end{array}$