Solve the following differential equation:
$(y - x y^{2}) d x - (x + x^{2} y) d y = 0$

x = c y e^{x y}

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y = c x e^{x y}

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y = - c x e^{x y}

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x = c y e^{- x y}

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Solution

The correct option is A $x = c y e^{x y}$
$(y - x y^{2}) d x - (x + x^{2} y) d y = 0$
Substituting $y = \frac{v - 1}{x} \Rightarrow \frac{d y}{d x} = \frac{\frac{d v}{d x}}{x} - \frac{v - 1}{x^{2}}$
$(y - x {(\frac{v - 1}{x})}^{2}) - (x + x^{2} (\frac{v - 1}{x})) ⎛ ⎜ ⎜ ⎜ ⎝ \frac{\frac{d v}{d x}}{x} - \frac{v - 1}{x^{2}} ⎞ ⎟ ⎟ ⎟ ⎠ = 0 \Rightarrow \frac{(- (x \frac{d v}{d x}) + 2) v - 2}{x} = 0 \Rightarrow \frac{d v}{d x} = \frac{2 (v - 1)}{x v} = \frac{2 (- \frac{1}{v} + 1)}{x} \Rightarrow \frac{\frac{d v}{d x}}{- \frac{1}{v} + 1} = \frac{2}{x}$
Integrating both sides w.r.t $x$ , we get
$\int \frac{\frac{d v}{d x}}{- \frac{1}{v} + 1} d x = \int \frac{2}{x} d x \Rightarrow log (v - 1) + v = 2 log x + c \Rightarrow log (x y) + x y + 1 = 2 log x + c \Rightarrow x = c y e^{x y}$

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