The general solution of the differential equation $(x - y^{2}) d x + y (5 x + y^{2}) d y = 0$ is:

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$|{(y^{2} + 2 x)}^{3}| = C {(2 y^{2} + x)}^{4}$

${(y^{2} + x)}^{4} = C |{(y^{2} + 2 x)}^{3}|$

$|{(y^{2} + x)}^{3}| = C {(2 y^{2} + x)}^{4}$

${(y^{2} + 2 x)}^{4} = C |{(y^{2} + x)}^{3}|$

answer is A.

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Detailed Solution

$\begin{array}{l} (x - y^{2}) dx + y (5 x + y^{2}) dy = 0 \\ \frac{dy}{dx} = \frac{y^{2} - x}{y (5 x + y^{2})} . Let y^{2} = v \\ \frac{2 ydy}{dx} = 2 (\frac{y^{2} - x}{5 x + y^{2}}) \\ \frac{dv}{dx} = 2 (\frac{v - x}{5 x + v}) v = kx \\ k + x \frac{dk}{dx} = 2 (\frac{kx - x}{5 x + kx}) \\ x \frac{dk}{dx} = - \frac{(k^{2} + 3 k + 2)}{k + 5} \\ \int \frac{(5 + k)}{(k + 1) (k + 2)} dk = \int - \frac{dx}{x} \\ \int (\frac{4}{k + 1} - \frac{3}{k + 2}) dk = - \int \frac{dx}{x} \\ 4 \ln (k + 1) - 3 \ln (k + 2) = - \ln x + \ln c \\ \frac{(k + 1)^{4}}{(k + 2)^{3}} = - \ln x + \ln c \\ c {(y^{2} + 2 x)}^{3} = {(y^{2} + x)}^{4} \end{array}$