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

Answer

Correct option: D.
$y = \tan \left( {\frac{{{x^2}}}{2} + x + c} \right)$
d
(d) $\frac{{dy}}{{dx}} = (1 + x)(1 + {y^2})$ ==> $\frac{{dy}}{{1 + {y^2}}} = (1 + x)dx$

On integrating both sides, we get

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

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