MCQ
The solution of the differential equation $\frac{{dy}}{{dx}} = \frac{{1 + {y^2}}}{{1 + {x^2}}}$ is
  • A
    $1 + xy + c(y + x) = 0$
  • B
    $x + y = c(1 - xy)$
  • $y - x = c(1 + xy)$
  • D
    $1 + xy = c(x + y)$

Answer

Correct option: C.
$y - x = c(1 + xy)$
c
(c) $\frac{{dy}}{{dx}} = \frac{{1 + {y^2}}}{{1 + {x^2}}} \Rightarrow \frac{1}{{1 + {y^2}}}dy = \frac{1}{{1 + {x^2}}}dx$

Now on integrating both sides, we get

${\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}c$==> ${\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\frac{{x + c}}{{1 - cx}}} \right)$
 

==> $y = \frac{{x + c}}{{1 - cx}}$ ==> $y - x = c(1 + xy)$.

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