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

x (x^{2} + 3 y^{2}) = c

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x (3 x^{2} + y^{2}) = c

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y (3 x^{2} + y^{2}) = c

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y (x^{2} + 3 y^{2}) = c

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Solution

The correct option is A $x (x^{2} + 3 y^{2}) = c$
$2 x y d y + (x^{2} + y^{2}) d x = 0$
$\Rightarrow 2 \frac{d y}{d x} = - \frac{x}{y} - \frac{y}{x}$
Substitute $y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$
$\Rightarrow 2 v + 2 x \frac{d v}{d x} = - v - \frac{1}{v}$
$\Rightarrow 2 x \frac{d v}{d x} = - \frac{1}{v} - 3 v = - \frac{1 + 3 v^{2}}{v}$
$\Rightarrow \frac{6 v}{1 + 3 v^{2}} d v = - \frac{3}{x} d x$
$\Rightarrow log (1 + 3 v^{2}) = - 3 log x + log k$
$\Rightarrow x^{3} (1 + 3 v^{2}) = k \Rightarrow x (x^{2} + 3 y^{2}) = k$

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Similar questions

2 x y d y = (x^{2} + y^{2}) d x .

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

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

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

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