The solution of the differential equation (1 + x^2 ) dy dx = x(1 … - Vidyadip

MCQ

The solution of the differential equation $(1 + {x^2})\frac{{dy}}{{dx}} = x(1 + {y^2})$ is

✓
$2{\tan ^{ - 1}}y = \log (1 + {x^2}) + c$
B
${\tan ^{ - 1}}y = \log (1 + {x^2}) + c$
C
$2{\tan ^{ - 1}}y + \log (1 + {x^2}) + c = 0$
D
None of these

✓

Answer

Correct option: A.

$2{\tan ^{ - 1}}y = \log (1 + {x^2}) + c$

a
(a) $(1 + {x^2})\frac{{dy}}{{dx}} = x(1 + {y^2})$ ==>$\frac{1}{{1 + {y^2}}}dy = \frac{x}{{1 + {x^2}}}dx$

On integrating, we get ${\tan ^{ - 1}}y = \frac{1}{2}\log (1 + {x^2}) + c$

==> $2{\tan ^{ - 1}}y = \log (1 + {x^2}) + c$.

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