x ^ 2 (y+1)dx+ y ^ 2 (x 1)dy=0 ⇒ x ^ 2 (y+1)dx= y ^ 2 (x 1)dy ⇒ (x+ x x 1 )dx=( y+ y y+1 )dy ⇒∫ (x+ x x 1 )dx =∫ ( y+ y y+1 )dy ...

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Variable Separable Method

x2y+1dx+y2x-1...

Question

$x^{2} (y + 1) d x + y^{2} (x - 1) d y = 0$ .

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Solution

$x^{2} (y + 1) d x + y^{2} (x - 1) d y = 0 \Rightarrow x^{2} (y + 1) d x = - y^{2} (x - 1) d y \Rightarrow (x + \frac{x}{x - 1}) d x = (- y + \frac{y}{y + 1}) d y \Rightarrow \int (x + \frac{x}{x - 1}) d x = \int (- y + \frac{y}{y + 1}) d y \Rightarrow \int (x + 1 + \frac{1}{x - 1}) d x = \int (- y + 1 - \frac{1}{y + 1}) d y \Rightarrow \frac{x^{2}}{2} + x + log (x - 1) = \frac{- y^{2}}{2} + y - log (y + 1) + C$

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