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

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Solution

Given, $(x^{2} + y^{2}) d x - 2 x y d y = 0$
$\Rightarrow (x^{2} + y^{2}) d x = 2 x y d y$
$\Rightarrow \frac{d y}{d x} = \frac{x^{2} + y^{2}}{2 x y}$ .... (i)
Let $y = v x$
Thus, $\frac{d y}{d x} = v + x \frac{d v}{d x}$
Thus, $v + x \frac{d v}{d x} = \frac{x^{2} + (v x)^{2}}{2 x (v x)}$
$\Rightarrow v + x \frac{d v}{d x} = \frac{1 + v^{2}}{2 v}$
$\Rightarrow x \frac{d v}{d x} = \frac{1 + v^{2}}{2 v} - v$
$\Rightarrow x \frac{d v}{d x} = \frac{1 + v^{2} - 2 v^{2}}{2 v}$
$\Rightarrow x \frac{d v}{d x} = \frac{1 - v^{2}}{2 v}$
$\Rightarrow \frac{d x}{x} = \frac{2 v}{1 - v^{2}} d v$
$\Rightarrow \frac{d x}{x} - \frac{2 v}{1 - v^{2}} d v = 0$ .... (ii)
Integrating both sides, we have
$log x + log (1 - v^{2}) = l o g C$
$\Rightarrow log x (1 - v^{2}) = log C$
$\Rightarrow x (1 - v^{2}) = C$
$\Rightarrow x (1 - \frac{y^{2}}{x^{2}}) = C$
$\Rightarrow x (\frac{x^{2} - y^{2}}{x^{2}}) = C$
$\Rightarrow x^{2} - y^{2} = C x$

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