The solution of the differential equation $(x^{2} + y^{2}) d x - 2 x y d y = 0$ is

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

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

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

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Solution

The correct option is A
$x^{2} - y^{2} = C x$

Explanation of the correct option.
Compute the required value.
Given : $(x^{2} + y^{2}) d x - 2 x y d y = 0$
$\Rightarrow$ $(x^{2} + y^{2}) d x = 2 x y d y$
$\Rightarrow$ $\frac{d y}{d x} = \frac{x^{2} + y^{2}}{2 x y}$ ………….. $(1)$
Let $y = v x$
Differentiate with respect to $x$ ,
$\Rightarrow$ $\frac{d y}{d x} = v + x \frac{d v}{d x}$
Substitute the value in the equation $(1)$ .
$\Rightarrow$ $v + x \frac{d v}{d x} = \frac{x^{2} + {(v x)}^{2}}{2 x (v x)}$
$\Rightarrow$ $v + x \frac{d v}{d x} = \frac{1 + v^{2}}{2 v}$
$\Rightarrow$ $x \frac{d v}{d x} = \frac{1 + v^{2}}{2 v} - v$
$\Rightarrow$ $x \frac{d v}{d x} = \frac{1 + v^{2} - 2 v^{2}}{2 v}$
$\Rightarrow$ $\frac{d x}{x} = \frac{2 v}{1 - v^{2}} d v$
Integrate both sides,
$\Rightarrow$ $\log (x) = - \log (1 - v^{2}) + \log (C)$
$\Rightarrow$ $\log (x (1 - v^{2})) = \log (C)$
$\Rightarrow$ $x (1 - \frac{y^{2}}{x^{2}}) = C$
$\Rightarrow$ $x (\frac{x^{2} - y^{2}}{x^{2}}) = C$
$\Rightarrow$ $x^{2} - y^{2} = C x$
Hence option $(A)$ is the correct option.

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