Solve the differential equation
$\frac{{dy}}{{dx}} = 1 + x + {y^2} + x{y^2},$ when $y = 0$ and $x = 0$.

Given that, $\frac{{dy}}{{dx}} = 1 + x + {y^2} + x{y^2}$

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

$\Rightarrow$ $\frac{{dy}}{{dx}} = \left( {1 + {y^2}} \right)(1 + x)$

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

On integrating both sides,

we get
${\tan ^{ - 1}}y = x + \frac{{{x^2}}}{2} + K$

……….(i)
When $y = 0$ and $x = 0,$
then substituting these values in Eq. (i), we get

${\tan ^{ - 1}}(0) = 0 + 0 + K$
$\Rightarrow$ $K = 0$

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

$\Rightarrow$ $y = \tan \left( {x + \frac{{{x^2}}}{2}} \right)$