Find the equation of the curve satisfying the differential equation $y (x + y^{3}) d x = x (y^{3} - x) d y,$ and passing through the point (4,2)

- \frac{1}{x y} = \frac{1}{2} {(\frac{y}{x})}^{2} - \frac{1}{4}

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\frac{1}{x y} = \frac{1}{2} {(\frac{y}{x})}^{2} - \frac{1}{4}

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\frac{1}{x y} = \frac{1}{2} {(\frac{y}{x})}^{2} + \frac{1}{4}

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- \frac{1}{x y} = \frac{1}{2} {(\frac{y}{x})}^{2} + \frac{1}{4}

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Solution

The correct option is A $- \frac{1}{x y} = \frac{1}{2} {(\frac{y}{x})}^{2} - \frac{1}{4}$
$y (x + y^{3}) d x = x (y^{3} - x) d y,$
$x (y d x + x d y) = y^{3} (x d y - y d x)$
$x d (x y) = y^{3} x^{2} d (\frac{y}{x})$
$∴ \frac{d (x y)}{x^{2} y^{2}} = \frac{y}{x} d (\frac{y}{x})$
Integrating we get,
$- \frac{1}{x y} = \frac{1}{2} {(\frac{y}{x})}^{2} + c$
Given it passes through (4,2). $\Rightarrow - \frac{1}{8} = \frac{1}{8} + c ∴ c = - \frac{1}{4}$
Hence required solution is,
$- \frac{1}{x y} = \frac{1}{2} {(\frac{y}{x})}^{2} - \frac{1}{4}$

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