The general solution of differential equation $x (1 + y^{2}) d x + y (1 + x^{2}) d y = 0$ is/are:

\frac{1}{2} ln (1 + x^{2}) (1 + y^{2}) = k

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(1 + x^{2}) (1 + y^{2}) = c

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(1 + y^{4}) = c (1 + x^{2})

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ln [\frac{(1 + x^{2})}{(1 + y^{2})}] = k

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Solution

The correct options are
A $\frac{1}{2} ln (1 + x^{2}) (1 + y^{2}) = k$
B $(1 + x^{2}) (1 + y^{2}) = c$
Given equation is $x (1 + y^{2}) d x + y (1 + x^{2}) d y = 0$

$\Rightarrow x (1 + y^{2}) d x = - y (1 + x^{2}) d y$

integrating on both sides

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

Put $1 + x^{2} = t \Rightarrow 2 x d x = d t$

and $1 + y^{2} = u \Rightarrow 2 y d y = d u$

$\frac{1}{2} \int \frac{d t}{t} = - \frac{1}{2} \int \frac{d u}{u}$

$\Rightarrow \frac{1}{2} l n (1 + x^{2}) = - \frac{1}{2} l n (1 + y^{2}) + k$

$\Rightarrow l n (1 + x^{2}) (1 + y^{2}) = 2 k$

$\Rightarrow (1 + x^{2}) (1 + y^{2}) = e^{2 k}$

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