The solution of the differential equation $x^{2} dy + y (x + y) dx = 0$ is

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

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

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

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

answer is A.

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Detailed Solution

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

$\begin{array}{l} \Rightarrow \int \frac{(v + 2) - v}{v (v + 2)} d v = - \int \frac{2}{x} d x \\ \Rightarrow \int (\frac{1}{v} - \frac{1}{v + 2}) d v = - \int \frac{2}{x} d x \\ \Rightarrow \log v - \log (v + 2) = - 2 \log x + \log c \\ \Rightarrow \log (\frac{v}{v + 2}) = \log (\frac{c}{x^{2}}) \\ \Rightarrow x^{2} (\frac{\frac{y}{x}}{\frac{y}{x} + 2}) = c \\ \Rightarrow x^{2} (\frac{y}{y + 2 x}) = c \\ \Rightarrow x^{2} y = c (y + 2 x) \end{array}$