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$y = \tan ({x^2} - x + c)$
- ✓$y = \tan \left( {\frac{{{x^2}}}{2} + x + c} \right)$
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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$\frac{\pi^4}{2}$
$\frac{\pi^4}{4}$
$0$