The general solution of the differential equation $(x + y) d x + x d y = 0$ is

x^{2} + y^{2} = C

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2 x^{2} - y^{2} = C

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x^{2} + 2 x y = C

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y^{2} + 2 x y = C

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Solution

The correct option is C $x^{2} + 2 x y = C$
Given differential equation is

$(x + y) d x + x d y = 0$

$\Rightarrow x d x = - (x + y) d y$

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

It is a homogeneous differential equation.

So, putting $y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$ , we get

$v + x \frac{d v}{d x} = - \frac{x + v x}{x} = - \frac{1 + v}{1}$

$\Rightarrow x \frac{d v}{d x} = - 1 - 2 v$

$\Rightarrow \int \frac{d v}{1 + 2 v} = - \int \frac{d x}{x}$

$\Rightarrow log (1 + 2 v) = - log x + log C_{1}$

$\Rightarrow log (1 + \frac{2 y}{x}) = 2 log \frac{C_{1}}{x}$

$\Rightarrow \frac{x + 2 y}{x} = {(\frac{C_{1}}{x})}^{2}$

$\Rightarrow x^{2} + 2 x y = C$ (where, $C = C_{1}^{2}$ )

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