Find the Particular solution of the differential equations
$x^{2} d y + (x y + y^{2}) d x = 0; y = 1$ when , $x = 1$

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Solution

Given,

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

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

$\frac{d y}{d x} = \frac{- (x y + y^{2})}{x^{2}}$

substitute $y = v x \to \frac{d y}{d x} = x \frac{d v}{d x} + v$

$x \frac{d v}{d x} + v = \frac{- (x^{2} v + x^{2} v^{2})}{x^{2}}$

$x \frac{d v}{d x} + v = - (v + v^{2})$

$x \frac{d v}{d x} = - v - v^{2} - v$

$\frac{d v}{v^{2} + 2 v} = - \frac{d x}{x}$

integrating on both sides, we get,

$\int \frac{d v}{v^{2} + 2 v} = - \int \frac{d x}{x}$

$log \sqrt{\frac{v}{v + 2}} = - log x + log c$

$log \frac{x \sqrt{v}}{\sqrt{v + 2}} = log c$

$\frac{x \sqrt{v}}{\sqrt{v + 2}} = c$

re-substitute $y = v x$

$x^{2} y = c^{2} (y + 2 x)$

$y = 1, w h e n, x = 1$

$1^{2} (1) = c^{2} (1 + 2 (1))$

$c^{2} = \frac{1}{3}$

$∴ x^{2} y = \frac{1}{3} (y + 2 x)$

$y + 2 x = 3 x^{2} y$

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