Find the equation of the curve which passes through the point (2, 2) and satisfies the differential equation $y - x \frac{d y}{d x} = y^{2} + \frac{d y}{d x}$ .

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Solution

We have,
$y - x \frac{d y}{d x} = y^{2} + \frac{d y}{d x}$

$\Rightarrow \frac{d y}{d x} (x + 1) = y (1 - y) \Rightarrow \frac{d y}{y (1 - y)} = \frac{d x}{(x + 1)}$

$Integrating both sides, we get \int \frac{d y}{y (1 - y)} = \int \frac{d x}{x + 1} \Rightarrow \int (\frac{1}{y} + \frac{1}{1 - y}) d y = \int \frac{d x}{x + 1} \Rightarrow \log |y| - \log |1 - y| = \log |x + 1| + C . . . . . (1) Since the curve passes throught the point (2, 2), it satisfies the equation of the curve . \Rightarrow \log |2| - \log |1 - 2| = \log |2 + 1| + C \Rightarrow C = \log |\frac{2}{3}| Putting the value of C in (1), we get \log |y| - \log |1 - y| = \log |x + 1| + \log |\frac{2}{3}| \Rightarrow \log |\frac{y}{(1 - y)}| = \log |\frac{2 (x + 1)}{3}| \Rightarrow |\frac{y}{(1 - y)}| = |\frac{2 (x + 1)}{3}| \Rightarrow \frac{y}{(1 - y)} = \pm \frac{2 (x + 1)}{3} \Rightarrow \frac{y}{(1 - y)} = \frac{2 (x + 1)}{3} or \frac{y}{(1 - y)} = - \frac{2 (x + 1)}{3} Here, given point (2, 2) does not satisfy \frac{y}{(1 - y)} = \frac{2 (x + 1)}{3} But it satisfy \frac{y}{(1 - y)} = - \frac{2 (x + 1)}{3} ∴ \frac{y}{(1 - y)} = - \frac{2 (x + 1)}{3} \Rightarrow \frac{y}{(y - 1)} = \frac{2 (x + 1)}{3} \Rightarrow 3 y = 2 (x + 1) (y - 1) \Rightarrow 3 y = 2 x y - 2 x + 2 y - 2 \Rightarrow 2 x y - 2 x - y - 2 = 0$

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